# (Chiral) Supersymmetry in Different Dimensions

This week I’m at the CERN winter school on supergravity, strings and gauge theory. Jonathan Heckman’s talks about a top-down approach to 6D SCFTs have particularly caught my eye. After all the $\mathcal{N}=(2,0)$ theory is in some sense the mother of my favourite playground $\mathcal{N}=4$ in four dimensions.

Unless you’re a supersymmetry expert, the notation should already look odd to you! Why do I write down two numbers to classify supersymmetries in 6D, but one suffices for 4D. The answer comes from a subtlety in the definition of the superalgebra, which isn’t often discussed outside of lengthy (and dull) textbooks. Time to set the record straight!

At kindergarten we learn that supersymmetry adds fermonic generators to the Poincare algebra yielding a “unique” extension to the possible spacetime symmetries. Of course, this hides a possible choice – there are many fermionic representations of the Lorentz algebra one could choose for the supersymmetry generators.

Fortunately, mathematical consistency restricts you to simple options. For the algebra to close, the generators must live in the lowest dimensional representations of the Lorentz algebra – check Weinberg III for a proof. You’re still free to take many independent copies of the supersymmetry generators (up to the restrictions placed by forbidding higher spin particles, which are usually imposed).

Therefore the classification of supersymmetries allowed in different dimensions reduces to the problem of understanding the possible spinor representations. Thankfully, there are tables of these.

Reading carefully, you notice that dimensions $2$, $6$ and $10$ are particularly special, in that they admit Majorana-Weyl spinors. Put informally, this means you can have your cake and eat it! Normally, the minimal dimension spinor representation is obtained by imposing a Majorana (reality) or Weyl (chirality) condition. But in this case, you can have both!

This means that in $D=2,\ 6$ or $10$, the supersymmetry generators can be chosen to be chiral. The stipulation $\mathcal{N}=(1,0)$ says that $Q$ should be a left-handed Majorana spinor, for instance. In $D = 4$ a Majorana spinor must by necessity contain both left-handed and right-handed pieces, so this choice would be impossible! Or, if you like, should I choose $Q$ to be a left-handed Weyl spinor, then it’s conjugate $Q\dagger$ is forced to be right-handed.

# The Theorem of The Existence of Zeroes

It’s time to prove the central result of elementary algebraic geometry. Mostly it’s referred to as Hilbert’s Nullstellensatz. This German term translates precisely to the title of this post. Indeed ‘Null’ means ‘zero’, ‘stellen’ means to exist and ‘Satz’ means theorem. But referring to it merely as an existence theorem for zeroes is inadequate. Its real power is in setting up a correspondence between algebra and geometry.

Are you sitting comfortably? Grab a glass of water (or wine if you prefer). Settle back and have a peruse of these theorems. This is your first glance into the heart of a magical subject.

(In many texts these theorems are all referred to as the Nullstellensatz. I think this is both pointless and confusing, so have renamed them! If you have any comments or suggestions about these names please let me know).

Theorem 4.1 (Hilbert’s Nullstellensatz) Let $J\subsetneq k[\mathbb{A}^n]$ be a proper ideal of the polynomial ring. Then $V(J)\neq \emptyset$. In other words, for every nontrivial ideal there exists a point which simulataneously zeroes all of its elements.

Theorem 4.2 (Maximal Ideal Theorem) Every maximal ideal $\mathfrak{m}\subset k[\mathbb{A}^n]$ is of the form $(x-a_1,\dots,x-a_n)$ for some $(a_1,\dots,a_n)\in \mathbb{A}^n$. In other words every maximal ideal is the ideal of some single point in affine space.

Theorem 4.3 (Correspondence Theorem) For every ideal $J\subset k[\mathbb{A}^n]$ we have $I(V(J))=\sqrt{J}$.

We’ll prove all of these shortly. Before that let’s have a look at some particular consequences. First note that 4.1 is manifestly false if $k$ is not algebraically closed. Consider for example $k=\mathbb{R}$ and $n=1$. Then certainly $V(x^2+1)=\emptyset$. Right then. From here on in we really must stick just to algebraically closed fields.

Despite having the famous name, 4.1 not really immediately useful. In fact we’ll see its main role is as a convenient stopping point in the proof of 4.3 from 4.2. The maximal ideal theorem is much more important. It precisely provides the converse to Theorem 3.10. But it is the correspondence theorem that is of greatest merit. As an immediate corollary of 4.3, 3.8 and 3.10 (recalling that prime and maximal ideals are radical) we have

Corollary 4.4 The maps $V,I$ as defined in 1.2 and 2.4 give rise to the following bijections

$\{\textrm{affine varieties in }\mathbb{A}^n\} \leftrightarrow \{\textrm{radical ideals in } k[\mathbb{A}^n]\}$
$\{\textrm{irreducible varieties in }\mathbb{A}^n\} \leftrightarrow \{\textrm{prime ideals in } k[\mathbb{A}^n]\}$
$\{\textrm{points in }\mathbb{A}^n\} \leftrightarrow \{\textrm{maximal ideals in } k[\mathbb{A}^n]\}$

Proof We’ll prove the first bijection explicitly, for it is so rarely done in the literature. The second and third bijections follow from the argument for the first and 3.8, 3.10. Let $J$ be a radical ideal in $k[\mathbb{A}^n]$. Then $V(J)$ certainly an affine variety so $V$ well defined. Moreover $V$ is injective. For suppose $\exists J'$ radical with $V(J')=V(J)$. Then $I(V(J'))=I(V(J))$ and thus by 4.3 $J = J'$. It remains to prove that $V$ surjective. Take $X$ an affine variety. Then $J'=I(X)$ an ideal with $V(J')=X$ by Lemma 2.5. But $J'$ not necessarily radical. Let $J=\sqrt{J'}$ a radical ideal. Then by 4.3 $I(V(J'))=J$. So $V(J) = V(I(V(J')) = V(J') = X$ by 2.5. This completes the proof. $\blacksquare$

We’ll see in the next post that we need not restrict our attention to $\mathbb{A}^n$. In fact using the coordinate ring we can gain a similar correspondence for the subvarieties of any given variety. This will lead to an advanced introduction to the language of schemes. With these promising results on the horizon, let’s get down to business. We’ll begin by recalling a definition and a theorem.

Definition 4.5 A finitely generated $k$-algebra is a ring $R$ s.t. $R \cong k[a_1,\dots,a_n]$ for some $a_i \in R$. A finite $k$-algebra is a ring $R$ s.t. $R\cong ka_1 + \dots ka_n$.

Observe how this definition might be confusing when compared to a finitely generated $k$-module. But applying a broader notion of ‘finitely generated’ to both algebras and modules clears up the issue. You can check that the following definition is equivalent to those we’ve seen for algebras and modules. A finitely generated algebra is richer than a finitely generated module because an algebra has an extra operation – multiplication.

Definition 4.6 We say an algebra (module) $A$ is finitely generated if there exists a finite set of generators $F$ s.t. $A$ is the smallest algebra (module) containing $F$. We then say that $A$ is generated by $F$.

Theorem 4.7 Let $k$ be a general field and $A$ a finitely generated $k$-algebra. If $A$ is a field then $A$ is algebraic over $k$.

Okay I cheated a bit saying ‘recall’ Theorem 4.7. You probably haven’t seen it anywhere before. And you might think that it’s a teensy bit abstract! Nevertheless we shall see that it has immediate practical consequences. If you are itching for a proof, don’t worry. We’ll in fact present two. The first will be due to Zariski, and the second an idea of Noether. But before we come to those we must deduce 4.1 – 4.3 from 4.7.

Proof of 4.2 Let $m \subset k[\mathbb{A}^n]$ be a maximal ideal. Then $F = k[\mathbb{A}^n]/m$ a field. Define the natural homomorphism $\pi: k[\mathbb{A}^n] \ni x \mapsto x+m \in F$. Note $F$ is a finitely generated $k$-algebra, generated by the $x_i+m$ certainly. Thus by 4.7 $F/k$ is an algebraic extension. But $k$ was algebraically closed. Hence $k$ is isomorphic to $F$ via $\phi : k \rightarrowtail k[\mathbb{A}^n] \xrightarrow{\pi} F$.

Let $a_i = \phi^{-1}(x_i+m)$. Then $\pi(x_i - a_i) = 0$ so $x_i - a_i \in \textrm{ker}\pi = m$. Hence $(x_1-a_1, \dots, x_n-a_n) \subset m$. But $(x_1-a_1, \dots, x_n-a_n)$ is itself maximal by 3.10. Hence $m = (x_1-a_1, \dots, x_n-a_n)$ as required. $\blacksquare$

That was really quite easy! We just worked through the definitions, making good use of our stipulation that $k$ is algebraically closed. We’ll soon see that all the algebraic content is squeezed into the proof of 4.7

Proof of 4.1 Let $J$ be a proper ideal in the polynomial ring. Since $k[\mathbb{A}^n]$ Noetherian $J\subset m$ some maximal ideal. From 4.2 we know that $m=I(P)$ some point $P\in \mathbb{A}^n$. Recall from 2.5 that $V(I(P)) = \{P\} \subset V(J)$ so $V(J) \neq \emptyset$. $\blacksquare$

The following proof is lengthier but still not difficult. Our argument uses a method known as the Rabinowitsch trick.

Proof of 4.3 Let $J\triangleleft k[\mathbb{A}^n]$ and $f\in I(V(J))$. We want to prove that $\exists N$ s.t. $f^N \in J$. We start by introducing a new variable $t$. Define an ideal $J_f \supset J$ by $J_f = (J, ft - 1) \subset k[x_1,\dots,x_n,t]$. By definition $V(J_f) = \{(P,b) \in \mathbb{A}^{n+1} : P\in V(J), \ f(P)b = 1\}$. Note that $f \in I(V(J))$ so $V(J_f) = \emptyset$.

Now by 4.1 we must have that $J_f$ improper. In other words $J_f = k[x_1,\dots, x_n, t]$. In particular $1 \in J_f$. Since $k[x_1,\dots, x_n, t]$ is Noetherian we know that $J$ finitely generated by some $\{f_1,\dots,f_r\}$ say. Thus we can write $1 = \sum_{i=1}^r g_i f_i + g_o (ft - 1)$ where $g_i\in k[x_1,\dots , x_n, t]$ (*).

Let $N$ be such that $t^N$ is the highest power of $t$ appearing among the $g_i$ for $=\leq i \leq r$. Now multiplying (*) above by $f^N$ yields $f^N = \sum_{i=1}^r G_i(x_1,\dots, x_n, ft) f_i + G_0(x_1,\dots,x_n,ft)(ft-1)$ where we define $G_i = f^N g_i$. This equation is valid in $k[x_1,\dots,x_n, t]$. Consider its reduction in the ring $k[x_1,\dots,x_n,t]/(ft - 1)$. We have the congruence $f_N\equiv \sum_{i=1}^r h_i (x_1,\dots,x_n) f_i \ \textrm{mod}\ (ft-1)$ where $h_i = G_i(x_1,\dots,x_n,1)$.

Now consider the map $\phi:k[x_1,\dots, x_n]\rightarrowtail k[x_n,\dots, x_n,t]\xrightarrow{\pi} k[x_n,\dots, x_n,t]/(ft-1)$. Certainly nothing in the image of the injection can possibly be in the ideal $(ft - 1)$, not having any $t$ dependence. Hence $\phi$ must be injective. But then we see that $f^N = \sum_{i=1}^r h_i(x_1,\dots, x_n) f_i$ holds in the ring $k[\mathbb{A}^n]$. Recalling that the $f_i$ generate $J$ gives the result. $\blacksquare$

We shall devote the rest of this post to establishing 4.7. To do so we’ll need a number of lemmas. You might be unable to see the wood for the trees! If so, you can safely skim over much of this. The important exception is Noether normalisation, which we’ll come to later. I’ll link the ideas of our lemmas to geometrical concepts at our next meeting.

Definition 4.8 Let $A,B$ be rings with $B \subset A$. Let $a\in A$. We say that $a$ is integral over $B$ if $a$ is the root of some monic polynomial with roots in $B$. That is to say $\exists b_i \in B$ s.t. $a^n + b_{n-1}a^{n-1} + \dots + b_0 = 0$. If every $a \in A$ is integral over $B$ we say that $A$  is integral over $B$ or $A$ is an integral extension of $B$.

Let’s note some obvious facts. Firstly we can immediately talk about $A$ being integral over $B$ when $A,B$ are algebras with $B$ a subalgebra of $A$. Remember an algebra is still a ring! It’s rather pedantic to stress this now, but hopefully it’ll prevent confusion if I mix my termin0logy later. Secondly observe that when $A$ and $B$ are fields “integral over” means exactly the same as “algebraic over”.

We’ll begin by proving some results that will be of use in both our approaches. We’ll see that there’s a subtle interplay between finite $k$-algebras, integral extensions and fields.

Lemma 4.9 Let $F$ be a field and $R\subset F$ a subring. Suppose $F$ is an integral extension of $R$. Then $R$ is itself a field.

Proof Let $r \in R$. Then certainly $r \in F$ so $r^{-1} \in F$ since $F$ a field. Now $r^{-1}$ integral over $R$ so satisfies an equation $r^-n = b_{n-1} r^{-n+1} +\dots + b_0$ with$b_i \in R$. But now multiplying through by $r^{n-1}$ yields $r^{-1} = b_{n-1} + \dots + b_0 r^{n-1} \in R$. $\blacksquare$

Note that this isn’t obvious a priori. The property that an extension is integral contains sufficient information to percolate the property of inverses down to the base ring.

Lemma 4.10 If $A$ is a finite $B$ algebra then $A$ is integral over $B$.

Proof Write $A = Ba_1 + \dots +Ba_n$. Let $x \in A$. We want to prove that $x$ satisfies some equation $x^n + b_{n-1}x^n{n-1} + \dots + b_0 = 0$. We’ll do so by appealing to our knowledge about determinants. For each $a_i$ we may clearly write $xa_i = \sum_{i=1}^{n} b_{ij}a_j$ for some $b_ij \in B$.

Writing $\vec{a} = (a_1, \dots, a_n)$ and defining the matrix $(\beta)_{ij} = b_{ij}$ we can express our equation as $\beta a = xa$. We recognise this as an eigenvalue problem. In particular $x$ satisfies the characteristic polynomial of $\beta$, a polynomial of degree $n$ with coefficients in $B$. But this is precisely what we wanted to show. $\blacksquare$

Corollary 4.11 Let $A$ be a field and $B\subset A$ a subring. If $A$ is a finite $B$-algebra then $B$ is itself a field.

Proof Immediate from 4.9 and 4.10. $\blacksquare$

We now focus our attention on Zariski’s proof of the Nullstellensatz. I take as a source Daniel Grayson’s excellent exposition.

Lemma 4.12 Let $R$ be a ring an $F$ a $R$-algebra generated by $x \in F$. Suppose further that $F$ a field. Then $\exists s \in R$ s.t. $S = R[s^{-1}]$ a field.  Moreover $x$ is algebraic over $S$.

Proof Let $R'$ be the fraction field of $R$. Now recall that $x$ is algebraic over $R'$ iff $R'[x] \supset R'(x)$. Thus $x$ is algebraic over $R'$ iff $R'[x]$ is a field. So certainly our $x$ is algebraic over $R'$ for we are given that $F$ a field. Let $x^n + f_{n-1}x^{n-1} + \dots + f_0$ be the minimal polynomial of $x$.

Now define $s\in R$ to be the common denominator of the $f_i$, so that $f_0,\dots, f_{n-1} \in R[s^{-1}] = S$. Now $x$ is integral over $S$ so $F/S$ an integral extension. But then by 4.9 $S$ a field, and $x$ algebraic over it. $\blacksquare$

Observe that this result is extremely close to 4.7. Indeed if we take $R$ to be a field we have $S = R$ in 4.12. Then lemma then says that $R[x]$ is algebraic as a field extension of $R$. Morally this proof mostly just used definitions. The only nontrivial fact was the relationship between $R'(x)$ and $R'[x]$. Even this is not hard to show rigorously from first principles, and I leave it as an exercise for the reader.

We’ll now attempt to generalise 4.12 to $R[x_1,\dots,x_n]$. The argument is essentially inductive, though quite laborious. 4.7 will be immediate once we have succeeded.

Lemma 4.13 Let $R = F[x]$ be a polynomial ring over a field $F$. Let $u\in R$. Then $R[u^{-1}]$ is not a field.

Proof By Euclid, $R$ has infinitely many prime elements. Let $p$ be a prime not dividing $u$. Suppose $\exists q \in R[u^{-1}]$ s.t. $qp = 1$. Then $q = f(u^{-1})$ where $f$ a polynomial of degree $n$ with coefficients in $R$. Hence in particular $u^n = u^n f(u^{-1}) p$ holds in $R$ for $u^n f(u^{-1}) \in R$. Thus $p | u^n$ but $p$ prime so $p | u$. This is a contradiction. $\blacksquare$

Corollary 4.14 Let $K$ be a field, $F\subset K$ a subfield, and $x \in K$. Let $R = F[x]$. Suppose $\exists u\in R$ s.t. $R[u^{-1}] = K$. Then $x$ is algebraic over $F$. Moreover $R = K$.

Proof Suppose $x$ were transcendental over $F$. Then $R=F[x]$ would be a polynomial ring, so by 4.12 $R[u^{-1}]$ couldn’t be a field. Hence $x$ is algebraic over $F$ so $R$ is a field. Hence $R=R[u{-1}]=K$. $\blacksquare$

The following fairly abstract theorem is the key to unlocking the Nullstellensatz. It’s essentially a slight extension of 4.14, applying 4.12 in the process. I’d recommend skipping the proof first time, focussing instead on how it’s useful for the induction of 4.16.

Theorem 4.15 Take $K$ a field, $F \subset K$ a subring, $x \in K$. Let $R = F[x]$. Suppose $\exists u\in R$ s.t. $R[u^{-1}] = K$. Then $\exists 0\neq s \in F s.t. F[s^{-1}]$ is a field. Moreover $F[s^{-1}][x] = K$ and $x$ is algebraic over $F[s^{-1}]$.

Proof Let $L=\textrm{Frac}(F)$. Now by 4.14 we can immediately say that $L[x]=K$, with $x$ algebraic over $L$. Now we seek our element $s$ with the desired properties. Looking back at 4.12, we might expect it to be useful. But to use 4.12 for our purposes we’ll need to apply it to some $F' = F[t^{-1}]$ with $F'[x] = K$, where $t \in F$.

Suppose we’ve found such a $t$. Then 4.12 gives us $s' \in F'$ s.t. $F'[s'^{-1}]$ a field with $x$ algebraic over it. But now $s' = qt^{-m}$ some $q \in F, \ m \in \mathbb{N}$. Now $F'[s'^{-1}]=F[t^{-1}][s'^{-1}]=F[(qt)^{-1}]$, so setting $=qt$ completes the proof. (You might want to think about that last equality for a second. It’s perhaps not immediately obvious).

So all we need to do is find $t$. We do this using our first observation in the proof. Observe that $u^{-1}\in K=L[x]$ so we can write $u^{-1}=l_0+\dots +l_{n-1}x^{n-1}$, $l_i \in L$. Now let $t \in F$ be a common denominator for all the $l_i$. Then $u^{-1} \in F'=F[t^{-1}]$ so $F'[x]=K$ as required. $\blacksquare$

Corollary 4.16 Let $k$ a ring, $A$ a field, finitely generated as a $k$-algebra by $x_1,\dots,x_n$. Then $\exists 0\neq s\in k$ s.t. $k[s^{-1}]$ a field, with $A$ a finite algebraic extension of $k[s^{-1}]$. Trivially if $k$ a field, then $A$ is algebraic over $k$, establishing 4.7.

Proof Apply Lemma 4.15 with $F=k[x_1,\dots,x_{n-1}]$, $x=x_n$, $u=1$ to get $s'\in F$ s.t. $A' = k[x_1,\dots,x_{n-1}][s'^{-1}]$ is a field with $x_n$ algebraic over it. But now apply 4.15 again with $F=k[x_1,\dots,x_{n-2}]$, $u = s'$ to deduce that $A''=k[x_1,\dots, x_{n-2}][s''^{-1}]$ is a field, with $A'$ algebraic over $A''$, for some $s'' \in F$. Applying the lemma a further $(n-2)$ times gives the result. $\blacksquare$

This proof of the Nullstellensatz is pleasingly direct and algebraic. However it has taken us a long way away from the geometric content of the subject. Moreover 4.13-4.15 are pretty arcane in the current setting. (I’m not sure whether they become more meaningful with a better knowledge of the subject. Do comment if you happen to know)!

Our second proof sticks closer to the geometric roots. We’ll introduce an important idea called Noether Normalisation along the way. For that you’ll have to come back next time!

I’ll start this post by tying up some loose ends from last time. Before we get going there’s no better recommendation for uplifting listening than this marvellous recording. Hopefully it’ll help motivate and inspire you (and I) as we journey deeper into the weird and wonderful world of algebra and geometry.

I promised a proof that for algebraically closed fields $k$ every Zariski open set is dense in the Zariski topology. Quite a mouthful at this stage of a post, I admit. Basically what I’m showing is that Zariski open sets are really damn big, only in a mathematically precise way. But what of this ‘algebraically closed’ nonsense? Time for a definition.

Definition 3.1 A field $k$ is algebraically closed if every nonconstant polynomial in $k[x]$ has a root in $k$.

Let’s look at a few examples. Certainly $\mathbb{R}$ isn’t algebraically closed. Indeed the polynomial $x^2 + 1$ has no root in $\mathbb{R}$. By contrast $\mathbb{C}$ is algebraically closed, by virtue of the Fundamental Theorem of Algebra. Clearly no finite field is algebraically closed. Indeed suppose $k=\{p_1,\dots ,p_n\}$ then $(x-p_1)\dots (x-p_n) +1$ has no root in $k$. We’ll take a short detour to exhibit another large class of algebraically closed fields.

Definition 3.2 Let $k,\ l$ be fields with $k\subset l$. We say that $l$ is a field extension of $k$ and write $l/k$ for this situation. If every element of $l$ is the root of a polynomial in $k[x]$ we call $l/k$ an algebraic extension. Finally we say that the algebraic closure of $k$ is the algebraic extension $\bar{k}$ of $k$ which is itself algebraically closed.

(For those with a more technical background, recall that the algebraic closure is unique up to $k$-isomorphisms, provided one is willing to apply Zorn’s Lemma).

The idea of algebraic closure gives us a pleasant way to construct algebraically closed fields. However it gives us little intuition about what these fields ‘look like’. An illustrative example is provided by the algebraic closure of the finite field of order $p^d$ for $p$ prime. We’ll write $\mathbb{F}_{p^d}$ for this field, as is common practice. It’s not too hard to prove the following

Theorem 3.3 $\mathbb{F}_{p^d}=\bigcup_{n=1}^{\infty}\mathbb{F}_{p^{n!}}$

Proof Read this PlanetMath article for details.

Now we’ve got a little bit of an idea what algebraically closed fields might look like! In particular we’ve constructed such fields with characteristic $p$ for all $p$. From now on we shall boldly assume that for our purposes

every field $k$ is algebraically closed

I imagine that you may have an immediate objection. After all, I’ve been recommending that you use $\mathbb{R}^n$ to gain an intuition about $\mathbb{A}^n$. But we’ve just seen that $\mathbb{R}$ is not algebraically closed. Seems like we have an issue.

At this point I have to wave my hands a bit. Since $\mathbb{R}^n$ is a subset of $\mathbb{C}^n$ we can recover many (all?) of the geometrical properties we want to study in $\mathbb{R}^n$ by examining them in $\mathbb{C}^n$ and projecting appropriately. Moreover since $\mathbb{C}^n$ can be identified with $\mathbb{R}^{2n}$ in the Euclidean topology, our knowledge of $\mathbb{R}^n$ is still a useful intuitive guide.

However we should be aware that when we are examining affine plane curves with $k=\mathbb{C}$ they are in some sense $4$ dimensional objects – subsets of $\mathbb{C}^2$. If you can imagine $4$ dimensional space then you are a better person than I! That’s not to say that these basic varieties are completely intractable though. By looking at projections in $\mathbb{R}^3$ and $\mathbb{R}^2$ we can gain a pretty complete geometric insight. And this will soon be complemented by our burgeoning algebraic understanding.

Now that I’ve finished rambling, here’s the promised proof!

Lemma 3.4 Every nonempty Zariski open subset of $\mathbb{A}^1$ is dense.

Proof Recall that $k[x]$ is a principal ideal domain. Thus any ideal $I\subset k[x]$ may be written $I=(f)$. But $k$ algebraically closed so $f$ splits into linear factors. In other words $I = ((x-a_1)\dots (x-a_n))$. Hence the nontrivial Zariski closed subsets of $\mathbb{A}^1$ are finite, so certainly the Zariski open subsets of $\mathbb{A}^1$ are dense. $\blacksquare$

I believe that the general case is true for the complement of an irreducible variety, a concept which will be introduced next. However I haven’t been able to find a proof, so have asked here.

How do varieties split apart? This is a perfectly natural question. Indeed many objects, both in mathematics and the everyday world, are made of some fundamental building block. Understanding this ‘irreducible atom’  gives us an insight into the properties of the object itself. We’ll thus need a notion for what constitutes an ‘irreducible’ or ‘atomic’ variety.

Definition 3.5 An affine variety $X$ is called reducible if one can write $X=Y\cup Z$ with $Y,\ Z$ proper subsets of $X$. If $X$ is not reducible, we call it irreducible.

This seems like a good and intuitive way of defining irreducibility. But we don’t yet know that every variety can be constructed from irreducible building blocks. We’ll use the next few minutes to pursue such a theorem.

As an aside, I’d better tell you about some notational confusion that sometimes creeps in. Some authors use the term algebraic set for  my term affine variety. Such books will often use the term affine variety to mean irreducible algebraic set. For the time being I’ll stick to my guns, and use the word irreducible when it’s necessary!

Before we go theorem hunting, let’s get an idea about what irreducible varieties look like by examining some examples. The ‘preschool’ example is that $V(x_1 x_2)\subset \mathbb{A}^2$ is reducible, for indeed $V(x_1 x_2) = V(x_1)\cup V(x_2)$. This is neither very interesting nor really very informative, however.

A better example is the fact that $\mathbb{A}^1$ is irreducible. To see this, recall that earlier we found that the only proper subvarieties of $\mathbb{A}^1$ are finite. But $k$ is algebraically closed, so infinite. Hence we cannot write $\mathbb{A}^1$ as the union of two proper subvarities!

What about the obvious generalization of this to $\mathbb{A}^n$? Turns out that it is indeed true, as we might expect. For the sake of formality I’ll write it up as a lemma.

Lemma 3.6 $\mathbb{A}^n$ is irreducible

Proof Suppose we could write $\mathbb{A}^n=V(f)\cup V(g)$. By Lemma 2.5 we know that $V(f)\cup V(g) = V((f)\cap (g))$. But $(f)\cap(g)\supset (fg)$ so $V((f)\cap(g))\subset V(fg)$ again by Lemma 2.5. Conversely if $x\in V(fg)$ then either $f(x) = 0$ or $g(x) = 0$, so $x \in V(f)\cup V(g)$. This shows that $V(f)\cup V(g)=V(fg)$.

Now $V(fg)=\mathbb{A}^n$ immediately tells us $fg(x) = 0 \ \forall x\in k$. Suppose that $f$ is nonzero. We’ll prove that $g$ is the zero polynomial by induction on $n$. Then $V(g)=\mathbb{A}^n$ so $\mathbb{A}^n$ not irreducible, as required.

We first note that since $k$ algebraically closed $k$ infinite. For $n=1$ suppose $f,\ g \neq 0$. Then $f,\ g$ are each zero at finite sets of points. Thus since $k$ infinite, $fg$ is not the zero polynomial, a contradiction.

Now let $n>1$.  Consider $f,\ g$ nonzero polynomials in $k[\mathbb{A}^n]$. Fix $x_n \in k$. Then $f,\ g$ polynomials in $k[\mathbb{A}^{n-1}]$. For some $x_n$, $f,\ g$ nonzero as polynomials in $k[\mathbb{A}^{n-1}]$. By the induction hypothesis $fg\neq 0$. This completes the induction. $\blacksquare$

I’ll quickly demonstrate that $\mathbb{A}^n$ is quite strange, when considered as a topological space with the Zariski topology! Indeed let $U$ and $V$ be two nonempty open subsets. Then $U\cap V\neq \emptyset$. Otherwise $\mathbb{A}^n\setminus U,\ \mathbb{A}^n\setminus V$ would be proper closed subsets (affine subvarieties) which covered $\mathbb{A}^n$, violating irreducibility. This is very much not what happens in the Euclidean topology! Similarly we now have a rigorous proof that an open subset $U$ of $\mathbb{A}^n$ is dense. Otherwise $\bar{U}$ and $\mathbb{A}^n\setminus U$ would be proper subvarieties covering $\mathbb{A}^n$.

It’s all very well looking for direct examples of irreducible varieties, but in doing so we’ve forgotten about algebra! In fact algebra gives us a big helping hand, as the following theorem shows. For completeness we first recall the definition of a prime ideal.

Definition 3.7 $\mathfrak{p}$ is a prime ideal in $R$ iff whenever $fg \in \mathfrak{p}$ we have $f\in \mathfrak{p}$ or $g \in \mathfrak{p}$. Equivalently $\mathfrak{p}$ is prime iff $R/\mathfrak{p}$ is an integral domain.

Theorem 3.8 Let $X$ be a nonempty affine variety. Then $X$ irreducible iff $I(X)$ a prime ideal.

Proof [“$\Rightarrow$“] Suppose $I(X)$ not prime. Then $\exists f,g \in k[\mathbb{A}^n]$ with $fg \in I(X)$ but $f,\ g \notin I(X)$. Let $J_1 = (I(X),f)$ and $J_2 = (I(X),g)$. Further define $X_1 = V(J_1), \ X_2 = V(J_2)$. Then $V(X_1), \ V(X_2) \subset X$ so proper subsets of $\mathbb{A}^n$. On the other hand $X\subset X_1 \cup X_2$. Indeed if $P\in X$ then $fg(P)=0$ so $f(P)=0$ or $g(P)=0$ so $P \in X_1\cup X_2$.

[“$\Leftarrow$“] Suppose $X$ is reducible, that is $\exists X_1,\ X_2$ proper subvarieties of $X$ with $X=X_1\cup X_2$. Since $X_1$ a proper subvariety of $X$ there must exist some element $f \in I(X_1)\setminus I(X)$. Similarly we find $g\in I(X_2)\setminus I(X)$. Hence $fg(P) = 0$ for all $P$ in $X_1\cup X_2 = X$, so certainly $fg \in I(X)$. But this means that $I(X)$ is not prime. $\blacksquare$

This easy theorem is our first real taste of the power that abstract algebra lends to the study of geometry. Let’s see it in action.

Recall that a nonzero principal ideal of the ring $k[\mathbb{A}^n]$ is prime iff it is generated by an irreducible polynomial. This is an easy consequence of the fact that $k[\mathbb{A}^n]$ is a UFD. Indeed a nonzero principal ideal is prime iff it is generated by a prime element. But in a UFD every prime is irreducible, and every irreducible is prime!

Using the theorem we can say that every irreducible polynomial $f$ gives rise to an irreducible affine hypersurface $X$ s.t. $I(X)=(f)$. Note that we cannot get a converse to this – there’s nothing to say that $I(X)$ must be principal in general.

Does this generalise to ideals generated by several irreducible polynomials? We quickly see the answer is no. Indeed take $f = x\, g = x^2 + y^2 -1$ in $k[\mathbb{A}^2]$. These are both clearly irreducible, but $(f,g)$ is not prime. We can see this in two ways. Algebraically $y^2 \in (f,g)$ but $y \notin (f,g)$. Geometrically, recall Lemma 2.5 (3). Also note that by definition $(f,g) = (f)+(g)$. Hence $V(f,g) = V(f)\cap V(g)$. But $V(f) \cap V(g)$ is clearly just two distinct points (the intersection of the line with the circle). Hence it is reducible, and by our theorem $(f,g)$ cannot be prime.

We can also use the theorem to exhibit a more informative example of a reducible variety. Consider $X = V(X^2Y - Y^2)$. Clearly $\mathfrak{a}=(X^2Y-Y^2)$ is not prime for $Y(X^2 - Y) \in \mathfrak{a}$ but $Y\notin \mathfrak{a}, \ X^2 - Y \notin \mathfrak{a}$. Noting that $\mathfrak{a}=(X^2-Y)\cap Y$ we see that geometrically $X$ is the union of the $X$-axis and the parabola $Y=X^2$, by Lemma 2.5.

Having had such success with prime ideals and irreducible varieties, we might think – what about maximal ideals? Turns out that they have a role to play too. Note that maximal ideals are automatically prime, so any varieties they generate will certainly be irreducible.

Definition 3.9 An ideal $\mathfrak{m}$ of $R$ is said to be maximal if whenever $\mathfrak{m}\subset\mathfrak{a}\subset R$ either $\mathfrak{a} = \mathfrak{m}$ or $\mathfrak{a} = R$. Equivalently $\mathfrak{m}$ is maximal iff $R/\mathfrak{m}$ is a field.

Theorem 3.10 An affine variety $X$ in $\mathbb{A}^n$ is a point iff $I(X)$ is a maximal ideal.

Proof  [“$\Rightarrow$“] Let $X = \{(a_1, \dots , a_n)\}$ be a single point. Then clearly $I(X) = (X_1-a_1,\dots ,X_n-a_n)$. But $k[\mathbb{A}^n]/I(X)$ a field. Indeed $k[\mathbb{A}^n]/I(X)$ isomorphic to $k$ itself, via the isomorphism $X_i \mapsto a_i$. Hence $I(X)$ maximal.

[“$\Leftarrow$“] We’ll see this next time. In fact all we need to show is that $(X_1-a_1,\dots,X_n-a_n)$ are the only maximal ideals. $\blacksquare$

Theorems 3.8 and 3.10 are a promising start to our search for a dictionary between algebra and geometry. But they are unsatisfying in two ways. Firstly they tell us nothing about the behaviour of reducible affine varieties – a very large class! Secondly it is not obvious how to use 3.8 to construct irreducibly varieties in general. Indeed there is an inherent asymmetry in our knowledge at present, as I shall now demonstrate.

Given an irreducible variety $X$ we can construct it’s ideal $I(X)$ and be sure it is prime, by Theorem 3.8. Moreover we know by Lemma 2.5 that $V(I(X))=X$, a pleasing correspondence. However, given a prime ideal $J$ we cannot immediately say that $V(J)$ is prime. For in Lemma 2.5 there was nothing to say that $I(V(J))=J$, so Theorem 3.8 is useless. We clearly need to find a set of ideals for which $I(V(J))=J$ holds, and hope that prime ideals are a subset of this.

It turns out that such a condition is satisfied by a class called radical ideals. Next time we shall prove this, and demonstrate that radical ideals correspond exactly to algebraic varieties. This will provide us with the basic dictionary of algebraic geometry, allowing us to proceed to deeper results. The remainder of this post shall be devoted to radical ideals, and the promised proof of an irreducible decomposition.

Definition 3.11 Let $J$ be an ideal in a ring $R$. We define the radical of $J$ to be the ideal $\sqrt{J}=\{f\in R : f^m\in J \ \textrm{some} \ m\in \mathbb{N}\}$. We say that $J$ is a radical ideal if $J=\sqrt{J}$.

(That $\sqrt{J}$ is a genuine ideal needs proof, but this is merely a trivial check of the axioms).

At first glance this appears to be a somewhat arbitrary definition, though the nomenclature should seem sensible enough. To get a more rounded perspective let’s introduce some other concepts that will become important later.

Definition 3.12polynomial function or regular function on an affine variety $X$ is a map $X\rightarrow k$ which is defined by the restriction of a polynomial in $k[\mathbb{A}^n]$ to $X$. More explicitly it is a map $f:X\rightarrow k$ with $f(P)=F(P)$ for all $P\in X$ where $F\in k[\mathbb{A}^n]$ some polynomial.

These are eminently reasonable quantities to be interested in. In many ways they are the most obvious functions to define on affine varieties. Regular functions are the analogues of smooth functions in differential geometry, or continuous functions in topology. They are the canonical maps.

It is obvious that a regular function $f$ cannot in general uniquely define the polynomial $F$ giving rise to it. In fact suppose $f(P)=F(P)=G(P) \ \forall P \in X$. Then $F-G = 0$ on $X$ so $F-G\in I(X)$. This simple observation explains the implicit claim in the following definition.

Definition 3.13 Let $X$ be an affine variety. The coordinate ring $k[X]$ is the ring $k[\mathbb{A}^n]|_X=k[\mathbb{A}^n]/I(X)$. In other words the coordinate ring is the ring of all regular functions on $X$.

This definition should also appear logical. Indeed we define the space of continuous functions in topology and the space of smooth functions in differential geometry. The coordinate ring is merely the same notion in algebraic geometry.  The name  ‘coordinate ring’ arises since clearly $k[X]$ is generated by the coordinate functions $x_1,\dots ,x_n$ restricted to $X$. The reason for our notation $k[x_1,\dots ,x_n]=k[\mathbb{A}^n]$ should now be obvious. Note that the coordinate ring is trivially a finitely generated $k$-algebra.

The coordinate ring might seem a little useless at present. We’ll see in a later post that it has a vital role in allowing us to apply our dictionary of algebra and geometry to subvarieties. To avoid confusion we’ll stick to $k[\mathbb{A}^n]$ for the near future. The reason for introducing coordinate rings was to link them to radical ideals. We’ll do this via two final definitions.

Definition 3.14 An element $x$ of a ring $R$ is called nilpotent if $\exists$ some positive integer $n$ s.t. $x^n=0$.

Definition 3.15 A ring $R$ is reduced if $0$ is its only nilpotent element.

Lemma 3.16 $R/I$ is reduced iff $I$ is radical.

Proof Let $x+I$ be a nilpotent element of $R/I$ i.e. $(x^n + I) = 0$. Hence $x^n \in I$ so by definition $x\in \sqrt{I}=I$. Conversely let $x\in R s.t. x^m \in I$. Then $x^m + I = 0$ in $R/I$ so $x+I = 0+I$ i.e. $x \in I$. $/blacksquare$

Putting this all together we immediately see that the coordinate ring $k[X]$ is a reduced, finitely generated $k$-algebra. That is, provided we assume that for an affine variety $X$, $I(X)$ is radical, which we’ll prove next time. It’s useful to quickly see that these properties characterise coordinate rings of varieties. In fact given any reduced, finitely generated $k$-algebra $A$ we can construct a variety $X$ with $k[X]=A$ as follows.

Write $A=k[a_1,\dots ,a_n]$ and define a surjective homomorphism $\pi:k[\mathbb{A}^n]\rightarrow A, \ x_i\mapsto a_i$. Let $I=\textrm{ker}(\pi)$ and $X=V(I)$. By the isomorphism theorem $A = k[\mathbb{A}^n]/I$ so $I$ is radical since $A$ reduced. But then by our theorem next time $X$ an affine variety, with coordinate ring $A$.

We’ve come a long way in this post, and congratulations if you’ve stayed with me through all of it! Let’s survey the landscape. In the background we have abstract algebra – systems of equations whose solutions we want to study. In the foreground are our geometrical ideas – affine varieties which represent solutions to the equations. These varieties are built out of irreducible blocks, like Lego. We can match up ideals and varieties according to various criteria. We can also study maps from geometrical varieties down to the ground field using the coordinate ring.

Before I go here’s the promised proof that irreducible varieties really are the building blocks we’ve been talking about.

Theorem 3.17 Every affine variety $X$ has a unique decomposition as $X_1\cup\dots\cup X_n$ up to ordering, where the $X_i$ are irreducible components and $X_i\not\subset X_j$ for $i\neq j$.

Proof (Existence) An affine variety $X$ is either irreducible or $X=Y\cup Z$ with $Y,Z$ proper subset of $X$. We similarly may decompose $Y$ and $Z$ if they are reducible, and so on. We claim that this process stops after finitely many steps. Suppose otherwise, then $X$ contains an infinite sequence of subvarieties $X\supsetneq X_1 \supsetneq X_2 \supsetneq \dots$. By Lemma 2.5 (5) & (7) we have $I(X)\subsetneq I(X_1) \subsetneq I(X_2) \subsetneq \dots$. But $k[\mathbb{A}^n]$ a Noetherian ring by Hilbert’s Basis Theorem, and this contradicts the ascending chain condition! To satisfy the $X_i \not\subset X_j$ condition we simply remove any such $X_i$ that exist in the decomposition we’ve found.

(Uniqueness) Suppose we have another decomposition $X=Y_1\cup Y_m$ with $Y_i\not\subset Y_j$ for $i\neq j$. Then $X_i = X_i\cap X = \bigcup_{j=1}^{m}( X_i\cap Y_j)$. Since $X_i$ is irreducible we must have $X_i\cap Y_j = X_i$ for some $j$. In particular $X_i \subset Y_j$. But now by doing the same with the $X$ and $Y$ reversed we find $X_k$ width $X_i \subset Y_j \subset X_k$. But this forces $i=k$ and $Y_j = X_i$. But $i$ was arbitrary, so we are done. $\blacksquare$

If you’re interested in calculating some specific examples of ideals and their associated varieties have a read about Groebner Bases. This will probably become a topic for a post at some point, loosely based on the ideas in Hassett’s excellent book. This question is also worth a skim.

I leave you with this enlightening MathOverflow discussion , tackling the irreducibility of polynomials in two variables. Although some of the material is a tad dense, it’s nevertheless interesting, and may be a useful future reference!

# Invariant Theory and David Hilbert

Health warning: this post is part of a more advanced series on commutative algebra. It may be a little tricky for the layman to understand!

David Hilbert was perhaps the greatest mathematicians of the late 19th century. Much of his work laid the foundations for our modern study of commutative algebra. In doing so, he was sometimes said to have killed the study of invariants by solving the central problem in the field. In this post I’ll give a sketch of how he did so.

Motivated by Galois Theory we ask the following question. Given a polynomial ring $S = k[x_1,\dots,x_n]$ and a group $G$ acting on $S$ as a $k$-automorphism, what are the elements of $S$ that are invariant under the action of $G$? Following familiar notation we denote this set $S^G$ and note that it certainly forms a subalgebra of $S$.

In the late 19th century it was found that $S^G$ could be described fully by a finite set of generators for several suggestive special cases of $G$. It soon became clear that the fundamental problem of invariant theory was to find necessary and sufficient conditions for $S^G$ to be finitely generated. Hilbert’s contribution was an incredibly general sufficient condition, as we shall soon see.

To begin with we shall recall the alternative definition of a Noetherian ring. It is a standard proof that this definition is equivalent to that which invokes the ascending chain condition on ideals. As an aside, also recall that the ascending chain condition can be restated by saying that every nonempty collection of ideals has a maximal element.

Definition A.1 A ring $R$ is Noetherian if every ideal of $R$ is finitely generated.

We shall also recall without proof Hilbert’s Basis Theorem, and draw an easy corollary.

Theorem A.2 If $R$ Noetherian then $R[x]$ Noetherian.

Corollary A.3 If $S$ is a finitely generated algebra over $R$, with $R$ Noetherian, then $S$ Noetherian.

Proof We’ll first show that any homomorphic image of $R$ is Noetherian. Let $I$ be an ideal in the image under than homomorphism $f$. Then $f^{-1}(I)$ an ideal in $R$. Indeed if $k\in f^{-1}(I)$ and $r\in R$ then $f(rk)=f(r)f(k)\in I$ so $rk \in f^{-1}(I)$. Hence $f^{-1}(I)$ finitely generated, so certainly $I$ finitely generated, by the images of the generators of $f^{-1}(I)$.

Now we’ll prove the corollary. Since $S$ is a finitely generated algebra over $R$, $S$ is a homomorphic image of $R[x_1,\dots,x_n]$ for some $n$, by the obvious homomorphism that takes each $x_i$ to a generator of $S$. By Theorem A.2 and induction we know that $R[x_1,\dots,x_n]$ is Noetherian. But then by the above, $S$ is Noetherian. $\blacksquare$

Since we’re on the subject of Noetherian things, it’s probably worthwhile introducing the concept of a Noetherian module. The subsequent theorem is analogous to A.3 for general modules. This question ensures that the theorem has content.

Definition A.4 An $R$-module $M$ is Noetherian if every submodule $N$ is finitely generated, that is, if every element of $N$ can be written as a polynomial in some generators $\{f_1,\dots,f_n\}\subset N$ with coefficients in $R$.

Theorem A.5 If $R$ Noetherian and $M$ a finitely generated $R$-module then $M$ Noetherian.

Proof Suppose $M$ generated by $f_1,\dots,f_t$, and let $N$ be a submodule. We show $N$ finitely generated by induction on $t$.

If $t=1$ then clearly the map $h:R\rightarrow M$ defined by $1\mapsto f_1$ is surjective. Then the preimage of $N$ is an ideal, just as in A.3, so is finitely generated. Hence $N$ is finitely generated by the images of the generators of $h^{-1}(N)$.  (*)

Now suppose $t>1$. Consider the quotient map $h:M \to M/Rf_1$. Let $\tilde{N}$ be the image of $N$ under this map. Then by the induction hypothesis $\tilde{N}$ is finitely generated as it is a submodule of $M/Rf_1$. Let $g_1,\dots,g_s$ be elements of $N$ whose images generate $\tilde{N}$. Since $Rf_1$ is a submodule of $M$ generated by a single element, we have by (*) that it’s submodule $Rf_1\cap N$ is finitely generated, by $h_1,\dots,h_r$ say.

We claim that $\{g_1,\dots,g_s,h_1,\dots,h_r\}$ generate $N$. Indeed given $n \in N$ the image of $n \in N$ is a linear combination of the images of the $g_i$. Hence subtracting the relevant linear combination of the $g_i$ from $n$ produces an element of $N \cap Rf_1$ which is precisely a linear combination of the $h_i$ by construction. This completes the induction. $\blacksquare$

We’re now ready to talk about the concrete problem that Hilbert solved using these ideas, namely the existence of finite bases for invariants. We’ll take $k$ to be a field of characteristic $0$ and $G$ to be a finite group, or one of the linear groups $\textrm{ GL}_n(k),\ \textrm{SL}_n(k)$. As in our notation above, we take $S=k[x_1,\dots,x_n]$.

Suppose also we are given a group homomorphism $\phi:G \to \textrm{GL}_r(k)$, which of course can naturally be seen as the group of invertible linear transformations of the vector space $V$ over $k$ with basis $x_1,\dots,x_r$. This is in fact the definition of a representation of $G$ on the vector space $V$. As is common practice in representation theory, we view $G$ as acting on $V$ via $(g,v)\mapsto \phi(g)v$.

If $G$ is $\textrm{SL}_n(k)$ or $\textrm{GL}_n(k)$ we shall further suppose that our representation of $G$ is rational. That is, the matrices $g \in G$ act on $V$ as matrices whose entries are rational functions in the entries of $g$. (If you’re new to representation theory like me, you might want to read that sentence twice)!

We now extend the action of $g\in G$ from $V$ to the whole of $S$ by defining $(g,f)\mapsto f(g^{-1}(x_1),\dots,g^{-1}(x_r),x_{r+1},\dots,x_n)$. Thus we may view $G$ as an automorphism group of $S$. The invariants under $G$ are those polynomials left unchanged by the action of every $g \in G$, and these form a subring of $S$ which we’ll denote $S^G$.

Enough set up. To proceed to more interesting territory we’ll need to make another definition.

Definition A.6 A polynomial is called homogeneous, homogeneous form, or merely a form, if each of its monomials with nonzero coefficient has the same total degree.

Hilbert noticed that the following totally obvious fact about $S^G$ was critically important to the theory of invariants. We may write $S^G$ as a direct sum of the vector spaces $R_i$ of homogeneous forms of degree $i$ that are invariant under $G$. We say that $S^G$ may be graded by degree and use this to motivate our next definition.

Definition A.7 A graded ring is a ring $R$ together with a direct sum decomposition as abelian groups $R = R_0 \oplus R_1 \oplus \dots$, such that $R_i R_j \subset R_{i+j}$.

This allows us to generalise our notion of homogeneous also.

Definition A.8 A homogeneous element of a graded ring $R$ is an element of one of the groups $R_i$. A homogeneous ideal of $R$ is an ideal generated by homogeneous elements.

Be warned that clearly homogeneous ideals may contain many inhomogeneous elements! It’s worth mentioning that there was no special reason for taking $\mathbb{N}$ as our indexing set for the $R_i$. We can generalise this easily to $\mathbb{Z}$, and such graded rings are often called $\mathbb{Z}$-graded rings. We won’t need this today, however.

Note that if $f \in R$ we have a unique expression for $f$ of the form $f = f_0 + f_1 + \dots + f_n$ with $f_i \in R_i$. (I have yet to convince myself why this terminates generally, any thoughts? I’ve also asked here.) We call the $f_i$homogeneous component of $f$.

The next definition is motivated by algebraic geometry, specifically the study of projective varieties. When we arrive at these in the main blog (probably towards the end of this month) it shall make a good deal more sense!

Definition A.9 The ideal in a graded ring $R$ generated by all forms of degree greater than $0$ is called the irrelevant ideal and notated $R_+$.

Now we return to our earlier example. We may grade the polynomial ring $S=k[x_1,\dots,x_n]$ by degree. In other words we write $S=S_0\oplus S_1 \oplus \dots$ with $S_i$ containing all the forms (homogeneous polynomials) of degree $i$.

To see how graded rings are subtly useful, we’ll draw a surprisingly powerful lemma.

Lemma A.10 Let $I$ be a homogeneous ideal of a graded ring $R$, with $I$ generated by $f_1,\dots,f_r$. Let $f\in I$ be a homogeneous element. Then we may write $f = \sum f_i g_i$ with $g_i$ homogeneous of degree $\textrm{deg}(f)-\textrm{deg}(f_i)$.

Proof We can certainly write $f = \sum f_i G_i$ with $G_i \in R$. Take $g_i$ to be the homogeneous components of $G_i$ of degree $\textrm{deg}(f)-\textrm{deg}(f_i)$. Then all other terms in the sum must cancel, for $f$ is homogeneous by assumption. $\blacksquare$

Now we return to our attempt to emulate Hilbert. We saw earlier that he spotted that grading $S^G$ by degree may be useful. His second observation was this. $\exists$ maps $\phi:S\to S^G$ of $S^G$-modules s.t. (1) $\phi$ preserves degrees and (2) $\phi$ fixes every element of $S^G$. It is easy to see that this abstract concept corresponds intuitively to the condition that $S^G$ be a summand of the graded ring $S$.

This is trivial to see in the case that $G$ is a finite group. Indeed let $\phi (f) = \frac{1}{|G|}\sum_{g\in G} g(f)$. Note that we have implicitly used that $k$ has characteristic zero to ensure that the multiplicative inverse to $|G|$ exists. In the case that $G$ is a linear group acting rationally, then the technique is to replace the sum by an integral. The particulars of this are well beyond the scope of this post however!

We finally prove the following general theorem. We immediately get Hilbert’s result on the finite generation of classes of invariants by taking $R=S^G$.

Theorem A.11 Take$k$ a field and $S=k[x_1,\dots,x_n]$ a polynomial ring graded by degree. Let $R$ be a $k$-subalgebra of $S$. Suppose $R$ is a summand of $S$, in the sense described above. Then $R$ is finitely generated as a $k$-algebra.

Proof Let $I\subset R$ be the ideal of $R$ generated by all homogeneous elements of degree $> 0$. By the Basis Theorem $S$ is Noetherian, and $IS$ an ideal of $S$, so finitely generated. By splitting each generator into its homogeneous components we may assume that $IS$ is generated by some homogeneous elements $f_1,\dots,f_s$ which we may wlog assume lie in $I$. We’ll prove that these elements precisely generate $R$ as a $k$-algebra.

Now let $R'$ be the $k$-subalgebra of $S$ generated by $f_1,\dots,f_s$ and take $f\in R$ a general homogeneous polynomial. Suppose we have shown $f\in R'$. Let $g$ be a general element of $R$. Then certainly $g\in S$ a sum of homogeneous components. But $R$ a summand of $S$, so applying the given map $\phi$ we have that the homogeneous components are wlog in $R$. Thus $g\in R'$ also, and we are done.

It only remains to prove $f \in R'$ which we’ll do by induction on the degree of $f$. If $\textrm{deg}(f)=0$ then $f\in K\subset R'$. Suppose $\textrm{deg}(f)>0$ so $f\in I$. Since the $f_i$ generate $IS$ as a homogeneous ideal of $S$ we may write $f = \sum f_i g_i$ with $g_i$ homogeneous of degree $\textrm{deg}(f)-\textrm{deg}(f_i)<\textrm{deg}(f)$ by Lemma A.10. But again we may use the map $\phi$ obtained from our observation that $R$ a summand of $S$. Indeed then $f=\sum \phi(g_i)f_i$ for $f,\ f_i \in R$. But $\phi$ preserves degrees so $\phi(g_i)$ 0f lower degree than $f$. Thus by the induction hypothesis $\phi(g_i) \in R'$ and hence $f\in R'$ as required. $\blacksquare$

It’s worth noting that such an indirect proof caused quite a furore when it was originally published in the late 19th century. However the passage of time has provided us with a broader view of commutative algebra, and techniques such as this are much more acceptable to modern tastes! Nevertheless I shall finish by making explicit two facts that help to explain the success of our argument. We’ll first remind ourselves of a useful definition of an algebra.

Definition A.12 An $R$-algebra $S$ is a ring $S$ which has the compatible structure of a module over $R$ in such a way that ring multiplication is $R$-bilinear.

It’s worth checking that this intuitive definition completely agrees with that we provided in the Background section, as is clearly outlined on the Wikipedia page.  The following provide an extension and converse to Corollary A.3 (that finitely generated algebras over fields are Noetherian) in the special case that $R$ a graded ring.

Lemma A.13 $S=R_0\oplus R_1 \oplus \dots$ a Noetherian graded ring iff $R_0$ Noetherian and $S$ a finitely generated $R_0$ algebra.

Lemma A.14 Let $S$ be a Noetherian graded ring, $R$ a summand of $S$. Then $R$ Noetherian.

We’ll prove these both next time. Note that they certainly aren’t true in general when $S$ isn’t graded!

# Algebra, Geometry and Topology: An Excellent Cocktail

Yes and I’ll have another one of those please waiter. One shot Geometry, topped up with Algebra and then a squeeze of Topology. Shaken, not stirred.

Okay, I admit that was both clichéd and contrived. But nonetheless it does accurately sum up the content of this post. We’ll shortly see that studying affine varieties on their own is like having a straight shot of gin – a little unpleasant, somewhat wasteful, and not an experience you’d be keen to repeat.

Part of the problem is the large number of affine varieties out there! We took a look at some last time, but it’s useful to have just a couple more examples. An affine plane curve is the zero set of any polynomial in $\mathbb{A}^2$. These crop up all the time in maths and there’s a lot of them. Go onto Wolfram Alpha and type plot f(x,y) = 0 replacing the term f(x,y) with any polynomial you wish. Here are a few that look nice

There’s a more general notion than an affine plane curve that works in $\mathbb{A}^n$. We say a hypersurface is the zero of a single polynomial in $\mathbb{A}^n$. The cone in $\mathbb{R}^3$ that we say last time is a good example of a hypersurface. Finally we say a hyperplane is the zero of a single polynomial of degree $1$ in $\mathbb{A}^n$.

Hopefully all that blathering has convinced you that there really are a lot of varieties, and so it looks like it’s going to be hard to say anything general about them. Indeed we could look at each one individually, study it hard and come up with some conclusions. But to do this for every single variety would be madness!

We could also try to group them into different types, then analyse them geometrically. This way is a bit more efficient, and indeed was the method of the Ancients when they learnt about conic sections. But it is predictably difficult to generalise this to higher dimensions. Moreover, most geometrical groupings are just the tip of the iceberg!

What with all this negativity, I imagine that a shot of gin sounds quite appealing now. But bear with me one second, and I’ll turn it into a Long Island Iced Tea! By broadening our horizons a bit with algebraic and topological ideas, we’ll see that all is not lost. In fact there are deep connections that make our (mathematical) life much easier and richer, thank goodness.

First though, I must come good on my promise to tell you about some subset’s of $\mathbb{C}^n$ that aren’t algebraic varieties. A simple observation allows us to come up with a huge class of such subsets. Recall that polynomials are continuous functions from $\mathbb{C}^n$ to $\mathbb{C}$, and therefore their zero sets must be closed in the Euclidean topology. Hence in particular, no open ball in $\mathbb{C}^n$ can be thought of as a variety. (If you didn’t understand this, it’s probably time to brush up on your topology).

There are two further ‘obvious’ classes. Firstly graphs of transcendental functions are not algebraic varieties. For example the zero set of the function $f(x,y) = e^{xy}-x^2$ is not an affine variety. Secondly the closed square $\{(x,y)\in \mathbb{C}^2:|x|,|y|\leq 1\}$ is an example of a closed set which is not an affine variety. This is because it clearly contains interior points, while no affine variety in $\mathbb{C}^2$ can contain such points. I’m not entirely sure at present why this is, so I’ve asked on math.stackexchange for a clarification!

How does algebra come into the mix then? To see that, we’ll need to recall a definition about a particular type of ring.

Definition 2.1 A Noetherian ring is a ring which satisfies the ascending chain condition on ideals. In other words given any chain $I_1 \subseteq I_2 \subseteq \dots \ \exists n$ s.t. $I_{n+k}=I_n$ for all $k\in\mathbb{N}$.

It’s easy to see that all fields are trivially Noetherian, for the only ideals in $k$ are $latex 0$ and $k$ itself. Moreover we have the following theorem due to Hilbert, which I won’t prove. You can find the (quite nifty) proof here.

Theorem 2.2 (Hilbert Basis) Let $N$ be Noetherian. Then $N[x_1]$ is Noetherian also, and by induction so is $N[x_1,\dots,x_n]$ for any positive integer $n$.

This means that our polynomial rings $k[\mathbb{A}^n]$ will always be Noetherian. In particular, we can write any ideal $I\subset k[\mathbb{A}^n]$ as $I=(f_1, \dots, f_r)$ for some finite $r$, using the ascending chain condition. Why is this useful? For that we’ll need a lemma.

Lemma 2.3 Let $Y$ be an affine variety, so $Y=V(T)$ some $T\subset K[\mathbb{A}^n]$. Let $J=(T)$, the ideal generated by $T$. Then $Y=V(J)$.

Proof By definition $T\subset J$ so $V(J)\subset V(T)$. We now need to show the reverse inclusion. For any $g\in J$ there exist polynomials $t_1,\dots, t_n$ in $T$ and $q_1,\dots,q_n$ in $K[\mathbb{A}^n]$ s.t. $g=\sum q_i t_i$. Hence if $p\in V(T)$ then $t_i(p)=0 \ \forall i$ so $p\in V(J)$. $\blacksquare$

Let’s put all these ideas together. After a bit of thought, we see that every affine variety $Y$ can be written as the zero set of a finite number of polynomials $t_1, \dots,t_n$. If you don’t get this straight away look back carefully at the theorem and the lemma. Can you see how to marry their conclusions to get this fact?

This is an important and already somewhat surprising result. If you give me any subset of $\mathbb{A}^n$ obtained from the solutions (possibly infinite) number of polynomial equations, I can always find a finite number of equations whose solutions give your geometrical shape! (At least in theory I can – doing so in practice is not always easy).

You can already see that a tiny bit of algebra has sweetened the cocktail! We’ve been able to deduce a fact about every affine variety with relative ease. Let’s pursue this link with algebra and see where it takes us.

Definition 2.4 For any subset $X \subset \mathbb{A}^n$ we say the ideal of $X$ is the set $I(X):=\{f \in k[\mathbb{A}^n] : f(x)=0\forall x\in X\}$.

In other words the ideal of $X$ is all the polynomials which vanish on the set $X$. A trivial example is of course $I(\mathbb{A}^n)=(0)$. Try to think of some other obvious examples before we move on.

Let’s recap. We’ve now defined two maps $V: \{\textrm{ideals in }k[\mathbb{A}^n]\}\rightarrow \{\textrm{affine varieties in }\mathbb{A}^n\}$ and $I:\{\textrm{subsets of }\mathbb{A}^n\}\rightarrow \{\textrm{ideals in }k[\mathbb{A}^n]\}$. Intuitively these maps are somehow ‘opposite’ to each other. We’d like to be able to formalise that mathematically. More specifically we want to find certain classes of affine varieties and ideals where $V$ and $I$ are mutually inverse bijections.

Why did I say certain classes? Well, clearly it’s not the case that $V$ and $I$ are bijections on their domains of definition. Indeed $V(x^n)=V(x)$, but $(x)=\neq(x^n)$ so $V$ isn’t injective. Furthermore working in $\mathbb{A}^1_{\mathbb{C}}$ we see that $I(\mathbb{Z})=(0)=I(\mathbb{A}^1)$ so $I$ is not injective. Finally for $n\geq 2 \ (x^n)\notin \textrm{Im}(I)$ so $I$ is not surjective.

It’ll turn out in the next post that a special type of ideal called a radical ideal will play an important role. To help motivate its definition, think of some more examples where $V$ fails to be injective. Can you spot a pattern? We’ll return to this next time.

Now that we’ve got our maps $V$ and $I$ it’s instructive to examine their properties. This will give us a feeling for the basic manipulations of algebraic geometry. No need to read it very thoroughly, just skim it to pick up some of the ideas.

Lemma 2.5 The maps $I$ and $V$ satisfy the following, where $J_i$ ideals and $X_i$ subsets of $\mathbb{A}^n$:
(1) $V(o)=\mathbb{A}^n,\ V(\mathbb{A}^n)=0$
(2) $V(J_1)\cup V(J_2)=V(J_1\cap J_2)$
(3) $\bigcap_{\lambda\in\Lambda}V(J_{\lambda}=V(\sum_{\lambda\in\Lambda}J_{\lambda})$
(4) $J_1\subset J_2 \Rightarrow V(J_2) \subset V(J_1)$
(5) $X_1\subset X_2 \Rightarrow I(X_2)\subset I(X_1)$
(6) $J_1 \subset I(V(J_1))$
(7) $X_1 \subset V(I(X_1))$ with equality iff $X_1$ is an affine variety

Proof We prove each in turn.
(1) Trivial.
(2) We first prove “$\subset$“. Let $q\in V(J_1)\cup V(J_2)$. Wlog assume $q \in V(J_1)$. Then $f(q)=0 \ \forall f \in J_1$. So certainly $f(q)=0 \ \forall f\in J_1\cap J_2$, which is what we needed to prove. Now we show “$\supset$“. Let $q\not\in {V(J_1)\cup V(J_2)}$. Then $q \not\in V(J_1)$ and $q \not\in V(J_2)$. So there exists $f \in J_1, \ g\in J_2$ s.t. $f(q) \neq 0,\ g(q)\neq 0$. Hence $fg(q)\neq 0$. But $fg\in J_1\cap J_2$ s0 $q \not\in {V(J_1\cap J_2)}$.
(3) “$\subset$” is trivial. For “$\supset$” note that $0 \in J_{\lambda}\ \forall \lambda$, and then it’s trivial.
(4) Trivial.
(5) Trivial.
(6) If $p \in J_1$ then $p(q)=0\ \forall q \in V(J_1)$ by definition, so $p \in I(V(J_1))$.
(7) The relation $X_1 \subset V(I(X_1))$ follows from definitions exactly as (6) did. For the “if” statement, suppose $X_1=V(J_1)$, some ideal $J_1$. Then by (5) $J_1 \subset I(V(J_1))$ so by (4) $V(I(X_1)=V(I(V(J_1)) \subset V(J_1)=X_1$. Conversely, suppose $V(I(X_1)= X_1$. Then $X_1$ is the zero set of $I(X_1)$ so an affine variety by definition. $\blacksquare$

That was rather a lot of tedious set up! If you’re starting to get weary with this formalism, I can’t blame you. You may be losing sight of the purpose of all of this. What are these maps $V$ and $I$ and why do we care how they behave? A fair question indeed.

The answer is simple. Our $V,\ I$ bijections will give us a dictionary between algebra and geometry. With minimal effort we can translate problems into an easier language. In particular, we’ll be allowed to use a generous dose of algebra to sweeten the geometric cocktail! You’ll have to wait until next time to see that in all its glory.

Finally, how does topology fit into all of this? Well, Lemma 2.5 (1)-(3) should give you an inkling. Indeed it instantly shows that the following definition makes sense.

Definition 2.6 We define the Zariski topology on $\mathbb{A}^n$ by taking as closed sets all the affine varieties.

In some sense this is the natural topology on $\mathbb{A}^n$ when we are concerned with solving equations. Letting $k=\mathbb{C}$ we can make some comparisons with the usual Euclidean topology.

First note that since every affine variety is closed in the Euclidean topology, every Zariski closed set is Euclidean closed. However we saw in the last post that not all Euclidean closed sets are affine varieties. In fact there are many more Euclidean closed sets than Zariski ones. We say that the Euclidean topology is finer than the Zariski topology. Indeed the Euclidean topology has open balls of arbitrarily small radius. The general Zariski open set is somehow very large, since it’s the complement of a line or surface in $\mathbb{A}^n$.

Next time we’ll prove that for algebraically closed $k$ every Zariski open set is dense in the Zariski topology, and hence (if $k =\mathbb{C}$) in the Euclidean topology. In particular, no nonempty Zariski open set is bounded in the Euclidean topology. Hence we immediately see that the intersection of two nonempty Zariski open sets of $\mathbb{A}$^n is never empty. This important observation tells us the the Zariski topology is not Hausdorff. We really are working with a very strange topological space!

And how is this useful? You know what I am going to say. It gives us yet another perspective on the world of affine varieties! Rather than just viewing them as geometrical objects in abstract $\mathbb{A}^n$ we can imagine them as a fundamental world structure. We’ll now be able to use the tools of topology to help us learn things about geometry. And there’s the slice of lemon to garnish the perfect cocktail.

I leave you with this enlightening question I recently stumbled upon. Both the question, and the proposed solutions struck me as extremely elegant.

# A Slice of Algebra

and a nice cup of tea. I always find that helps. Before we get down to business, you might want to put this delightful recording on. It’s always nice to have a bit of background music, and Strauss just seems to fit with Algebra somehow.

A broad definition of Algebra could be the study of equations and their solutions. This is perhaps the type of algebra we’re all familiar with from school. Here’s a typical problem

Find $x\in \mathbb{R}$ given that $x^2-2x+1=0$

That was easy, of course. Let’s try another one

Find all $x,y \in \mathbb{R}$ such that $y-x^2 = 0$

Perhaps you had to think for a moment before realising that this just defines a parabola in 2D space, pictured below.

These example illustrate that the solutions to equations can come in the form of points, or curves, and it’s not hard to see that solutions to equations in sufficiently many variables can define surfaces of any dimension you like. For example the equation $z=0$ defines a plane in 3D space.

So we can easily see that Algebra gives rise to geometrical structures of the type we discussed in the last post. It should now seem natural to study geometrical structures from an algebraic point of view. Voila – we have the motivation for Algebraic Geometry.

There’s nothing to restrict us to studying the solutions (often referred to as zeroes) of a single equation. In fact many interesting and useful geometric constructions arise as the simultaneous zeroes of several equations. Can you see two equations in (some or all of) the variables $x,y,z$ whose simultaneous solutions give rise to the $y$-axis in 3D [2]?

The technical terminology for the collection of simultaneous zeroes of several equations is an algebraic set. It is the most fundamental object of study which we will focus on.

Here we reach a slight technical impasse. For what follows I’ll assume a familiarity with elementary abstract algebra as outlined on the Background page. This may be viewed as a technical toolkit for our forthcoming studies. I’ll also assume some very basic knowledge about Topology, though not much more than can be gleaned by a thorough reading of the Wikipedia page. If you’ve never come across abstract algebra before, now is the time to do some serious thinking! I can’t promise it’ll be easy, and it might take a couple of days to get your head around the concepts, but I promise you it’s worth it. I’ll be happy to answer any questions commented on the Background page, and will flesh out the currently sparse details in the near future.

Good luck!

[1] We only every consider polynomial equations, which are those of the form $f(x_1,\dots,x_n)=0$ where $f(x_1,\dots,x_n)$ is a finite sum of nonnegative integer powers  of products of the variables $x_1,\dots, x_n$. Thus $f(x,y)=x^2+y^2=0$, the circle, is admissible for study but $f(x,y)=x^y=0$ is not. It turns out that not much is lost by restricting our study to polynomials only. In some sense any mathematically interesting curve can be approximated arbitrarily closely by the set of solutions to polynomial equations. (This entirely depends on your definition of mathematically interesting though)!

[2] The equations are of course $x=0$ and $z=0$. Geometrically this is true since the $y$-axis is the intersection of the two planes defined by $x=0$ and $z=0$.