# Homomorphism of $k$-algebras induce homomorphism of maximal spectrum

For $$k$$ a algebraically closed field, we define an affine $$k$$-algebra to be a finitely generated $$k$$-algebra that is reduced (i.e. $$\sqrt{(0)} = (0)$$). For an affinie $$k$$-algebra $$A$$, we define $$\operatorname{specm} A$$ to be the set of maximal ideals. Then we have the following proposition:

If $$\alpha: A \rightarrow B$$ is a homomorphism of affine $$k$$-algebras, then $$\alpha$$ induces a continuous map of topological spaces $$\phi: \operatorname{specm} B \rightarrow \operatorname{specm} A$$ where for a maximal ideal $$m \subset B$$,

$$\phi(m) = \alpha^{-1}(m).$$

I am having trouble understanding the first half of the proof which goes as follows:

Proof:

1. For any $$h \in A$$, $$\alpha(h)$$ is invertible in $$B_{\alpha(h)}$$ (which denotes the localization of $$B$$ at $$\alpha(h)$$), so the homomorphism $$A \rightarrow B \rightarrow B_{\alpha(h)}$$ extends to a homomorphism $$\frac{g}{h^m} \rightarrow \frac{\alpha(g)}{\alpha(h)^m}: A_h \rightarrow B_{\alpha(h)}$$

2. For any maximal ideal $$n \in B$$, $$m = \alpha^{-1}(n)$$ is maximal in $$A$$ because $$A/m \rightarrow B/n$$ is an injective map of k-algebras which implies that $$A/m$$ is $$k$$.

I do not believe step 1 is used anywhere else in the proof so it seems that step 2 must be a consequence of step 1. Could someone explain how? In particular, is step 1 the reason that the map in step 2 is injective? Thank you!

• A formatting tip: please use \operatorname to format operators like $\operatorname{specm}$ which do not have predefined latex commands. This improves readability of your post. I've made the upgrade for you this time. Aug 19 '20 at 0:26
• Can you add in some info about what source you're reading the proof in, with a link if possible? It's hard to help with such a limited view of the full proof. Aug 19 '20 at 4:15
• This can be found in "Introduction to Algebraic Geometry" by Justin Smith on page 72. I'm sorry, I don't have a link. Aug 19 '20 at 13:47

I don't know where you're reading, but this seems overly complicated. Suppose that $$\alpha: A\to B$$ is a map of $$k$$-algebras where $$A$$ and $$B$$ are finite type. Let $$\mathfrak{m}$$ be a maximal ideal. We want to show that $$\alpha^{-1}(\mathfrak{m})$$ is a maximal ideal. Note though that the induced map

$$\alpha:A/\alpha^{-1}(\mathfrak{m})\to B/\mathfrak{m}$$

is injective sice if $$\alpha(a\alpha^{-1}(\mathfrak{m}))=\alpha(a)\mathfrak{m}$$ is zero, then this says that $$\alpha(a)\in\mathfrak{m}$$ so that $$a\in \alpha^{-1}(\mathfrak{m})$$ which says that $$a\alpha^{-1}(m)$$ is zero.

Now, let us note that while we might be worried that $$\alpha^{-1}(\mathfrak{m})$$ is not maximal it is certainly prime. Indeed, if $$ab\in\alpha^{-1}(\mathfrak{m})$$ then $$\alpha(ab)\in \mathfrak{m}$$. But, this implies that $$\alpha(a)\alpha(b)\in\mathfrak{m}$$ so then either $$\alpha(a)\in\mathfrak{m}$$ or $$\alpha(b)\in\mathfrak{m}$$. But, this means precisely that $$a\in\alpha^{-1}(\mathfrak{m})$$ or $$\alpha^{-1}(b)\in\mathfrak{m}$$. Since $$a$$ and $$b$$ were arbitrary we see that $$\alpha^{-1}(\mathfrak{m})$$ is prime as desired (NB: of course this didn't use that $$\mathfrak{m}$$ is maximal and works for any prime ideal).

So, we see that $$\alpha$$ induces an inclusion of the integral domain $$A/\alpha^{-1}(\mathfrak{m})$$ into the field $$B/\mathfrak{m}$$. If we were dealing with arbitrary rings then this would be the full extent of what we could really say. But, the fact that we're dealing with finite type $$k$$-algebras is what says the day.

How so? By the Nullstellensatz since $$B$$ is a finite-dimensional $$k$$-algebra we have that $$B/\mathfrak{m}$$ is a finite-dimensional $$k$$-algebra! So, in particular, since $$A/\alpha^{-1}(\mathfrak{m})$$ embeds into $$B/\mathfrak{m}$$ as a $$k$$-algebra we see that $$A/\alpha^{-1}(\mathfrak{m})$$ is an integral domain which is also a $$k$$-algebra which is finite dimensional over $$k$$. This is enough.

Namely, in absolute complete generality if $$\ell$$ is a field and $$R$$ is an integral domain which is an $$\ell$$-alebrawith $$\dim_\ell R<\infty$$ then $$R$$ is a field.

Why? We need to show that for any $$r\in R$$ which is non-zero that $$r$$ has a multiplicative inverse. But, this just means precisely that the map

$$m_r:R\to R:x\mapsto rx$$

is invertible-- evidently if $$r$$ has a multiplicative inverse then $$m_r^{-1}=m_{r^{-1}}$$ and if $$m_r$$ is invertible then $$1$$ is in the image of $$m_r$$ which means that there exists $$x$$ such that $$1=m_r(x)=rx$$.

But, note that since $$R$$ is a domain that $$m_r$$ is injective-- if $$m_r(x)=0$$ then $$rx=0$$ which implies, by the domain property, that $$x=0$$ since $$r\ne 0$$. But, note that $$m_r$$ is clearly a map of $$k$$-vector spaces, and since any injective endomorphism of a finite-dimensional vector space is an automorphism, we win!

• Thank you for the answer. It seems that Step 2 can stand on its own, so I am now very curious where Step 1 is used. This proof is from page 72 of Justin Smith's "Introduction to Algebraic Geometry" Aug 20 '20 at 16:25
• @JohnKnoxV Yeah, I'm not sure. You could, if you want, make a separate question asking this with a screenshot of the book. It seems to me that it might just be a mistake. Aug 20 '20 at 17:08