# 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.

Presumably for general $\mathbb{A}^n$ one can ask a similar question about the Zariski topology? I haven’t thought hard about whether or not it’s true yet, however. That may be the subject for another blog post, in due course!