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!