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.

  • $\begingroup$ 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. $\endgroup$ – drhab Jul 3 '14 at 9:49

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 $\mathrm{End}(\mathbb{R})=\{\mathrm{id}\}$.

  • $\begingroup$ 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? $\endgroup$ – tomasz Jul 3 '14 at 9:57
  • 2
    $\begingroup$ @tomasz: no, $\sigma$ can be a non-surjective embedding; a famous example is $x \mapsto x^p$ on $\mathbf F_p (x)$. $\endgroup$ – user64687 Jul 3 '14 at 9:58
  • 1
    $\begingroup$ @AsalBeagDubh: good point :) $\endgroup$ – tomasz Jul 3 '14 at 10:00
  • $\begingroup$ Also, field homomorphisms preserve $1$ by definition ... $\endgroup$ – Martin Brandenburg Jul 3 '14 at 10:01
  • $\begingroup$ @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). $\endgroup$ – tomasz Jul 3 '14 at 10:03

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.