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Let $k$ be a field. Let $A$ and $B$ be two $k$-algebras, ie. two rings that are also $k$-vector spaces and their multiplication is $k$-bilinear.

Any isomorphism of $k$-algebras is also a ring isomorphism, so if $A$ and $B$ are isomorphic as $k$-algebras, they are isomorphic as rings.

I would guess that the converse fails. Is there any example of $A$ and $B$ that are isomorphic as rings, but not as $k$-algebras?

The reason I came up with this question is the following. Two affine varieties are isomorphic if and only if their coordinate rings are isomorphic as $k$-algebras. I am interested in finding an example where coordinate rings are isomorphic as rings, but the varieties are not isomorphic.

spin
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  • First thought: start with a ring $R$ and find two essentially different ringhomomorphisms $\alpha,\beta:k\rightarrow R$, both with an image in the center of $R$. Then for $\lambda\in k$ and $r\in R$ the action $\lambda.r$ can be defined as $\alpha\left(\lambda\right)r$ and also as $\beta\left(\lambda\right)r$. Producing probably/hopefully two essentially different $k$-algebras. – drhab Jul 03 '14 at 09:49

1 Answers1

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Let $\sigma : k \to k$ be any homomorphism. By restriction of scalars, we obtain a $k$-algebra $A$ whose underyling ring is just $k$. But $A$ is isomorphic to $k$ as $k$-algebra if and only if $\sigma$ is an isomorphism.

More generally: Let $A$ be a commutative ring and $f,g : k \to A$ be two homomorphisms. These may be considered as two $k$-algebras. The underlying rings are equal to $A$. But the $k$-algebras are isomorphic if and only if there is an automorphism $h : A \to A$ such that $hf = g$. The first paragraph deals with the somewhat pathological special case $A=k$ and $f=\mathrm{id}, g=\sigma$. But of course there are lots of other examples, too (unless $k$ is a prime field or something similar).

For example, consider the two embeddings $\mathbb{Q}(\sqrt{2}) \rightrightarrows \mathbb{R}$ given by $\sqrt{2} \mapsto \pm \sqrt{2}$. They don't differ by an automorphism of $\mathbb{R}$, since $\mathrm{End}(\mathbb{R})=\{\mathrm{id}\}$.

Martin Brandenburg
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  • But $k$ is a field, so $\sigma$ is either $0$ or an isomorphism (if $\sigma(1)\neq 0$, then $\sigma(1)=1$, because $\sigma(1)^2=\sigma(1)$), or am I missing something? – tomasz Jul 03 '14 at 09:57
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    @tomasz: no, $\sigma$ can be a non-surjective embedding; a famous example is $x \mapsto x^p$ on $\mathbf F_p (x)$. –  Jul 03 '14 at 09:58
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    @AsalBeagDubh: good point :) – tomasz Jul 03 '14 at 10:00
  • Also, field homomorphisms preserve $1$ by definition ... – Martin Brandenburg Jul 03 '14 at 10:01
  • @MartinBrandenburg: Well, yes. But non-unital ring homomorphism don't (and $\sigma=0$ makes for a perfectly good example if you consider non-unital agebras). – tomasz Jul 03 '14 at 10:03
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    @AsalBeagDubh: Or just $k(x) \to k(x), x \mapsto x^2$ for any field $k$. – Martin Brandenburg Jul 03 '14 at 10:09
  • @MartinBrandenburg: sorry, what I meant was that every element of $\mathbf F_p(x)$ is sent to its $p$-th power. (My notation was unclear.) Does your example really extend to a field homomorphism if $k$ has char. different from 2? –  Jul 03 '14 at 10:16
  • @AsalBeagDubh: Sure! This is because $x^2$ is trancendental. Not every polynomial is squared, we have $5x-1 \mapsto 5x^2-1$ for example. – Martin Brandenburg Jul 03 '14 at 10:17
  • @MartinBrandenburg: oops, of course. As usual, my fingers were working faster than my brain! –  Jul 03 '14 at 10:20