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:


  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!

  • $\begingroup$ 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. $\endgroup$
    – KReiser
    Aug 19 '20 at 0:26
  • 1
    $\begingroup$ 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. $\endgroup$ Aug 19 '20 at 4:15
  • $\begingroup$ This can be found in "Introduction to Algebraic Geometry" by Justin Smith on page 72. I'm sorry, I don't have a link. $\endgroup$
    – JohnKnoxV
    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!

  • $\begingroup$ 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" $\endgroup$
    – JohnKnoxV
    Aug 20 '20 at 16:25
  • $\begingroup$ @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. $\endgroup$ Aug 20 '20 at 17:08

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