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