6
$\begingroup$

Consider the following claim:

Let $X \subset k[x_1,\dots,x_n]$ and $Y\subset k[y_1,\dots,y_m]$ be algebraic sets and suppose that we have a ring isomorphism $\varphi: \mathcal{O}_Y \to \mathcal{O}_X$. Show that the algebraic sets $X$ and $Y$ are isomorphic.

Now, I know that if $\varphi$ is an isomorphism of $k$-algebras, then $X$ and $Y$ are isomorphic. And I know that any isomorphism of $k$-algebras is also an isomorphism of the underlying rings.

Question: However, isn't it the case that not every ring homomorphism $\varphi: \mathcal{O}_Y \to \mathcal{O}_X$ is a $k$-algebra homomorphism $\mathcal{O}_Y \to \mathcal{O}_X$, and that only $k$-algebra homomorphisms $ \mathcal{O}_Y \to \mathcal{O}_X$ correspond to morphisms $X \to Y$, not arbitrary ring homomorphsims $\mathcal{O}_Y \to \mathcal{O}_X$?

In other words, which of the following two statements are correct? Are they even mutually exclusive? Or are they equivalent? If they are equivalent, why?

Two algebraic sets are isomorphic if and only if their coordinate rings are isomorphic as rings.

OR

Two algebraic sets are isomorphic if and only if their coordinate rings are isomorphic as $k$-algebras.

Context: My book, in sections 4.18 and 4.19, as well as this question on Math.SE implies that the first statement is true. However, section 4.8 of the same book, and every other source I have found (e.g. here or here), implies only that the second statement is true. Moreover, I was only able to prove the second statement, not the first. Which is correct?

This is a follow-up to my previous question, an attempt to show that the two statements are mutually exclusive and not equivalent. It was unanswered though, so I don't know if my attempt was successful -- for all I know, the two statements could be equivalent. If so, why?

EDIT: What I showed in my notes, and was agreed to be true in the comments is that:

There is a one-one correspondence between morphisms $X \to Y$ and ring homomorphisms $\mathcal{O}_Y \to \mathcal{O}_X$. (FALSE)

There is a one-one correspondence between morphisms $X \to Y$ and $k$-algebra homomorphisms $\mathcal{O}_Y \to \mathcal{O}_X$. (TRUE)

$\endgroup$
  • 3
    $\begingroup$ I don't have access to the book you're reading. But the linked answer does not claim that the first statement is true. It says that an alg. map between two alg. sets is an isomorphism iff the induced map is an isomorphism of coordinate rings. But notice that the induced map is automatically a homomorphism of $k$-algebras. (The second statement you wrote makes the stronger claim that an isomorphism of $k$-algebras must actually be induced by an isomorphism of alg. sets.) But the linked question never claims that there is a 1-1 correspondence between alg. maps and ring homomorphisms. $\endgroup$ – Nefertiti Mar 24 '17 at 14:38
  • 1
    $\begingroup$ Does anyone have an example of finite type $k$-algebras that are not isomorphic, but isomorphic as rings? $\endgroup$ – MooS Mar 24 '17 at 14:45
  • 1
    $\begingroup$ @MooS: not quite, but there is a ring homomorphism from $\mathbf C$ to $\mathbf C$ that isn't a $\mathbf C$-algebra homomorphism: complex conjugation. (Oops, I just saw the previous comment.) $\endgroup$ – Nefertiti Mar 24 '17 at 14:52
  • 1
    $\begingroup$ Yes, of course. I was being stupid. @Will: The induced homomorphism of coordinate rings simply fixes $k$ by construction, i.e. they are homomorphism of $k$-algebras. I think the confusion which arises here is a great argument that it is very convenient to work with schemes instead of classical algebrsic sets. In the scheme set up, the $k$-algebra structure is part of the data of a $k$-scheme and is not implicitly hidden in the definition. $\endgroup$ – MooS Mar 24 '17 at 15:08
  • 1
    $\begingroup$ A ring map induces a $k$-algebra map (just twist the algebra structure of the target by the automorphism of $k$ given by the restriction of the ring map) and thus induces a map of algebraic sets. But then the 1-1 correspondence is lost. $\endgroup$ – MooS Mar 24 '17 at 15:49
1
$\begingroup$

Attempt: The anti-equivalence of $k$-algebras and affine varieties (1)(2 pp.3-4), gives us automatically that:

Two algebraic sets are isomorphic if and only if their coordinate rings are isomorphic as $k$-algebras. (TRUE)

Now, note that, since any $k$-algebra homomorphism is a ring homomorphism when we ignore scalar multiplication, any $k$-algebra isomorphism is also a ring isomorphism, so the above implies that:

$$\text{[two algebraic sets isomorphic }\iff\text{ coordinate rings $k$-algebra isomorphic]}\\ \implies \text{[coordinate rings are ring isomorphic]}$$

So the only real remaining question is the following:

[coordinate rings are ring isomorphic] $\overset{?????}{\implies}$[coordinate rings are $k$-algebra isomorphic]

According to the comments, this might be true, since given a ring homomorphism, we might always be able to "twist" it (I guess under the action of the Galois group of $k$) into being a $k$-algebra homomorphism. Any references confirming or denying this would be appreciated. (See also.) So would comments on my faux proof of this claim below.

"Proof" of ?????: Generalizing from coordinate rings to arbitrary associative unital algebras, we can write (see for notation) the condition for a ring homomorphism $A \to B$ to also be a $k$-algebra homomorphism as $\varphi \circ \eta_A = \eta_B$. Now assume that $\varphi: A \to B$ isn't a $k$-algebra homomorphism, but still a ring isomorphism, i.e. $\varphi \circ \eta_A \not= \eta_B$. Since $\eta_A(k)$ is a field, and $\varphi |_{\eta_A(k)}$ is a ring homorphism with a field as its domain, it is injective, and thus $\varphi(\eta_A(k))$ is also a field isomorphic to $k$, albeit a different from $\eta_B(k)$ (they would be the same if $\varphi$ were a $k$-algebra homomorphism).

To summarize, $k \cong \eta_A(k) \cong \varphi(\eta_A(k)) \cong \eta_B(k)$, with none of them equal (just isomorphic as fields/rings). Define: $$ \DeclareMathOperator*{\rl}{\rightleftharpoons} k \rl\limits_{\Upsilon^{-1}}^{\Upsilon} \eta_A(k)\,, \quad k\rl\limits_{\theta^{-1}}^{\theta} (\varphi\circ\eta_A)(k)\,, \quad k \rl\limits_{\zeta^{-1}}^{\zeta} \eta_B(k)\,, \quad \eta_A(k) \rl\limits_{\omega^{-1}}^{\omega} \eta_B(k)$$

It follows that $\theta^{-1} \circ \varphi \circ \Upsilon$ and $\zeta^{-1} \circ \omega \circ \Upsilon$ are two different automorphisms of $k$, since $\varphi \circ \eta_A \not=\eta_B$. (In particular, if $k$ had trivial Galois group, we would have a contradiction, thus if $k$ has trivial Galois group, every ring homomorphism $A \to B$ is also a $k$-algebra homomorphism.)

(I'm not really sure if $\varphi \circ \eta_A \not= \eta_B$ actually implies that the two automorphisms above are not equal, if it does, then I am skipping the step that justifies the claim, and if it doesn't, then the above is the point where the entire argument falls apart.)

For simplicity, we will define $\alpha = \theta^{-1} \circ \varphi \circ \Upsilon : k \to k$ and $\beta= \zeta^{-1}\circ \omega \circ \Upsilon : k \to k$.

Speculative claim: Define $\tilde{\varphi} = \zeta\circ\beta\circ\alpha^{-1}\circ\theta^{-1}\circ\varphi|_{\eta_A(k)} : \eta_A(k) \to \eta_B(k) $.

Then $\tilde{\varphi} \circ \eta_A = \eta_B$. (Really speculative) $\tilde{\varphi}$ can have its domain extended to all of $A$, and denoting the extension by the same symbol, we have that $\tilde{\varphi}: A \to B$ is not just a ring homomorphism, but even a $k$-algebra homomorphism (since $\tilde{\varphi} \circ \eta_A = \eta_B$). Moreover I claim (without justification) that if $\varphi$ is a ring isomorphism, then $\tilde{\varphi}$ is a ring isomorphism as well, thus a $k$-algebra isomorphism.

Thus we are able to "twist" any arbitrary ring homomorhpism $\varphi: A \to B$ via the action of the automorphism/Galois group of $k$ into a $k$-algebra homomorphism $\tilde{\varphi}:A \to B$, in such a way that the existence of a ring isomorphism $A \to B$ implies the existence of a $k$-algebra isomorphism $A \to B$.

Therefore, [coordinate rings ring isomorphic] $\implies$ [coordinate rings $k$-algebra isomorphic].

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.