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!

# 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.