Dimension of k-algebras which are isomorphic as rings

Question

Let $k$ be a field and let $A$ and $B$ be $k$-algebras.

Assume that $A$ and $B$ are isomorphic as rings.

Can we conclude that $\dim_k(A)=\dim_k(B)$?

Answer 1

Suppose we have a homomorphism of fields $\phi : k \to k$ which is not surjective. (For one example to show this is possible, we can let $k := \mathbb{C}(t)$, and $\phi : f(t) \mapsto f(t^2)$.) Then $\phi$ induces a $k$-algebra structure on $k$ given by $\lambda * x := \phi(\lambda) x$.

Now, let $A := k$ with the usual $k$-algebra structure and $B := k$ with the $k$-algebra structure induced by $\phi$. Then $A$ and $B$ are certainly isomorphic as rings, yet $\dim_k(A) = 1$ and $\dim_k(B) > 1$. (The second follows from the fact that the nonzero vector $1$ has span $\operatorname{im}(\phi) \ne B$ when you consider the corresponding $k$-vector space structure on $B$.)