# What is an example of two k-algebras that are isomorphic as rings, but not as k-algebras?

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.

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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 3 '14 at 9:49

## 2 Answers

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

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

For $t \in k$, let $A_t$ be $k$ viewed as the $k(x)$-algebra induced by $x \to t$.

Then for any $k(x)$-algebra isomorphism $\varphi: A_s \to A_t$, we have:

$$s = s \cdot 1 = s \cdot \varphi(1) = \varphi(s \cdot 1) = \varphi(x \cdot 1) = x \cdot \varphi(1) = t \cdot 1 = t$$

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There is no homomorphism of rings $k(x) \to k$, $x \mapsto t$. In fact, $x-t$ lies in the kernel, but homomorphisms of fields are injective. Thus, there is no $k(x)$-algebra structure on $k$ as you described. –  Martin Brandenburg Jul 3 '14 at 10:10
The assignment you've noted doesn't seem to induce a $k(x)$-algebra structure: note that $x-t\mapsto 0$. @MartinBrandenburg: beat me to it. :) –  tomasz Jul 3 '14 at 10:11
Ah right, I was thinking about $k[x]$-algebras –  Hurkyl Jul 3 '14 at 11:09