Studying basic algebraic geometry with professor Milne's notes, the following question came to me: given a (commutative) field $k$ and two $k-$algebras $R$ and $R'$, is every ring map $\phi: R \to R'$ a map of $k-$algebras? My working definition of $k-$algebra is: a commutative ring $S$ is a $k-$algebra if there exists a ring map $k \to S$ called the "structure map"... and to me, a map of $k-$algebras is a ring map that commutes with the structure maps; so for instance, if $\sigma, \sigma'$ are the structure maps of $R$ and $R'$ respectively, the ring map $\phi$ is a $k-$algebra map if $\phi\circ\sigma = \sigma'$ (I require ring maps to satisfy $1\mapsto 1$).
So I tried to prove this is always true owing to the fact that $k$ is a field, but couldn't get anywhere. All I could say is that $(\phi\circ\sigma)(k-\{0\}) \subset R'^{\times}$ and that if $q$ is the ground field of $k$, $(\phi\circ\sigma)(q) = \sigma'(q)$. However, since $\phi\circ\sigma: k\to R'$ is a ring map, it is injective, so $(\phi\circ\sigma)(k)$ is a subring of $R'$ that is a field isomorphic to $k$, as is the case for $\sigma'(k)$, but a priori they could be different subrings.
Therefore, I wonder if field/Galois theory can say something about how a field $k$ can fit inside a ring in different but isomorphic ways. Also, I think perhaps if $R, R'$ are assumed to be finitely-generated $k-$algebras the question may be easier to answer. I prefer to think about $k$ as of characteristic zero, just in case... Appreciate any help/comments :)
Take $k=\Bbb Q(i)$, and take the two field embeddings $\sigma,\sigma':\Bbb Q(i)\to\Bbb C$, i.e. such that $\sigma(i)=i$ and $\sigma'(i)=-i$. Let $R$ be $\Bbb C$ with the $k$-algebra structure determined by $\sigma$, and $R'$ be the same but for $\sigma'$. Then the identity map $\Bbb C\to\Bbb C$ gives you a ring homomorphism $R\to R'$ which is not a $k$-algebra homomorphism.
Thinking about fields can lead you to a large class of counter-examples in general; if $L/k$ is a field extension, then the inclusion $k\to L$ gives $L$ a $k$-algebra structure in your terms, and then any field automorphism $L\to L$ is a ring homomorphism and saying it is a $k$-algebra homomorphism is equivalent to saying it fixes elements of $k$, which obviously need not always hold.