# Reconstructing an Affine Variety from its Coordinate Ring

I'm trying to understand the construction often written as $V=\operatorname{Spec}(R)$, where $R$ is a finitely generated $\mathbb C$-algebra with no nonzero nilpotents.

At first glance, the notation $\operatorname{Spec}(R)$ is introduced just as the set of maximal ideals. If $R=\mathbb C[V]$ is the coordinate ring of an affine variety $V\subseteq\mathbb C^n$, we know that points of $V$ correspond to maximal ideals of $\mathbb C[V]$, so $V=\operatorname{Spec}(\mathbb C[V])$ is a legitimate equation on the level of sets. But to make $\operatorname{Spec}(R)$ a proper affine variety if we don't already know the variety that has $R$ as its coordinate ring, we need some way to equip the set of maximal ideals with the structure of an affine variety.

From what I read, the "proper" way to look at this, is learning about schemes. Since I'm only beginning algebraic geometry (to study toric varieties), I don't know anything about the theory of schemes for now.

The way I managed to equip $\operatorname{Spec}(R)$ with the structure of an affine variety is the following: Let $R$ be a finitely generated $\mathbb C$-algebra with no nonzero nilpotents. Pick generators $f_1,\dots,f_r$ and consider the surjective $\mathbb C$-algebra homomorphism $\varphi:\mathbb C[x_1,\dots,x_r]\to R$ with $x_i\mapsto f_i$. By the homomorphism theorem we have an isomorphism $R\cong\mathbb C[x_1,\dots,x_r]/I$, where $I$ is the kernel of $\varphi$. Since $R$ has no nonzero nilpotents, the ideal $I$ is radical. Let $V=\mathbf V(I)\subseteq\mathbb C^r$ be the affine variety given by the ideal $I$, then $\mathbf I(V)=\sqrt{I}=I$ by Hilbert's Nullstellensatz and therefore $\mathbb C[V] = \mathbb C[x_1,\dots,x_r]/I\cong R$. Since $V=\operatorname{Spec}(R)$ as sets, this construction equips $\operatorname{Spec}(R)$ with the structure of an affine variety.

Is this the right way to think about this, when I come across varieties defined as the $\operatorname{Spec}$ of some $\mathbb C$-algebra?

• Yes. But this construction is not intrinsic, it depends on chosen generators of $R$. The category of varieties in the "modern" sense (spaces with a nice sheaf on it) is somehow more natural. – Martin Brandenburg Oct 22 '13 at 20:48
• Besides, I would write $\mathrm{Spm}(R)$ or $X(\mathbb{C})$ instead of $X=\mathrm{Spec}(R)$ (which contains all prime ideals). – Martin Brandenburg Oct 22 '13 at 20:55

You are correct in your reasoning. Though it is not intrinsic, it is often adequate for your purposes of studying toric varieties. One of the many reasons toric varieties are nice to study is that you can learn many things about them, and algebraic geometry as a whole, without knowing what a sheaf is (as strange as that may sound to other people). For affine toric varieties there is actually a relatively standard geometric realisation.

Proposition: Suppose $\sigma$ is a rational strongly convex polyhedral cone in a lattice $N,$ and let the dual lattice be $M.$ By Gordan's lemma, the monoid $S_{\sigma}:= \sigma^{\vee}\cap M$ is finitely generated, say by $\mu_1, \ldots, \mu_k.$ Then the affine toric variety constructed from the cone can be realised as $U_{\sigma} = V(I_{\sigma}) \subseteq \mathbb{C}^k,$ where $$I_{\sigma} = \big\langle \ X_1^{a_1} X_2^{a_2} \cdots X_k^{a_k} - X_1^{b_1} X_2^{b_2} \cdots X_k^{b_k} \ \big| \ a_i, b_i \in \mathbb{Z}_{\geq 0} \ , \ \sum_{i=1}^k a_i\mu_i=\sum_{i=1}^k b_i\mu_i\ \big\rangle.$$

Proof: We have $\mathbb{C}[S_{\sigma}] = \mathbb{C}[\chi^{\mu_1}, \ldots, \chi^{\mu_k}].$ Then $\mathbb{C}$-algebra homomorphism $$\varphi: \mathbb{C}[X_1,\ldots, X_k] \to \mathbb{C}[S_{\sigma}] \ : \ X_i\mapsto \chi^{\mu_i}$$ is surjective so $\mathbb{C}[S_{\sigma}] \cong \mathbb{C}[X_1,\ldots, X_k]/(\ker \varphi).$ We need to show that $I_{\sigma} = \ker \varphi.$ It is clear that $I_{\sigma}\subseteq \ker \varphi.$ For $\mu \in S_{\sigma}$ let $\pi(\mu) = \{ a=(a_1,\ldots, a_k) \ | \ a_1 \mu_1 + \ldots a_k \mu_k = \mu \}$ and adopt the notation $X^a := X_1^{a_1} \cdots X_k^{a_k}.$ Suppose $p= \sum \lambda_a X^a$ is in $\ker \varphi$ so $\displaystyle \varphi(p) = \sum_{\mu \in S_{\sigma}} \left( \sum_{a\in \pi(\mu) } \lambda_a \right) \chi^{\mu}=0.$ Therefore $\displaystyle \sum_{a\in \pi(\mu)} \lambda_a =0$ for all $\mu \in S_{\sigma}.$ We can write $\displaystyle p=\sum_{\mu \in S_{\sigma}} p_{\mu}$ where $\displaystyle p_{\mu} = \sum_{a\in\pi(\mu)} \lambda_a X^a$ so it suffices to show that $p_{\mu} \in I_{\sigma}.$ The number of non-zero coefficients of $p_{\mu}$ is finite, so suppose there are $m$ of them indexed with superscripts. We have \begin{align*} \sum_{i=1}^m \lambda_{a^i} X^{a^i} &= \lambda_{a^1} ( X^{a^1} - X^{a^2} ) + (\lambda_{a^1} + \lambda_{a^2} ) (X^{a^2} -X^{a^3}) + \ldots\\ & +\left(\sum_{i=1}^{m-1} \lambda_{a^i}\right) ( X^{a^{m-1}} - X^{a^m} ) + \left(\sum_{i=1}^{m} \lambda_{a^i}\right)( X^{a^m} - X^{a^1}) + \left(\sum_{i=1}^{m} \lambda_{a^i}\right) X^{a^1}. \end{align*} Since $\displaystyle \sum_{i=1}^{m} \lambda_{a^i}=0$ we see that $p_{\mu}$ is the sum of elements of $I_{\sigma},$ which completes the proof.

It is true that if $\sigma$ has codimension zero in the lattice then there is actually a unique minimal set of monoid generators (that is, no element in the set is generated by the others), so for that case there are no choices to be made.

I am actually in the process of writing an introduction to Toric varieties that does not assume any familiarity with schemes or anything beyond the bare basics of a first course in algebraic geometry. If you are interested I can send it to you when I finish writing them next week.

• Thank you very much for your answer! I'm indeed interested in the introduction you are writing, so I'd be happy if you sent me copy, thanks! – Christoph Oct 23 '13 at 6:28
• A little off-topic, but I couldn't help commenting on "as strange as that may sound to other people". It doesn't sound strange at all to me: it's important to keep in mind that the list of people who didn't need to know what a sheaf is includes Newton, Euler, Cayley, Salmon, M. Noether, Schubert, Castelnuovo, Enriques, Fano, Lefschetz, Hodge... – user64687 Oct 23 '13 at 8:38
• My point being that I think some students get the wrong impression about algebraic geometry, namely that one has to learn all the intricacies of sheaves and schemes before getting started. Those things are important to learn in time, but they shouldn't stop one from diving in! – user64687 Oct 23 '13 at 8:41