Suppose $M/K$ is a field extension (WLOG we think of $K \subset M$ as sets) and let $a \in M$ be algebraic over $K$. If $\phi:K[x] \rightarrow K[a]$ is a homomorphism such that $x \mapsto a$ then must it be the case that $f(x) \mapsto f(a)$? I suppose this is equivalent to asking: is $\phi(k) = k$ for $k \in K$?
I may be missing something obvious here, perhaps something to do with $K[x]$ being the set of polynomials in $x$ as "formal objects". Thanks for the help.
It is true that taking $x\mapsto a$ uniquely defines a $K$-homomorphism $K[x]\to K[a]$ (i.e. having the property that $k\mapsto k$ for all $k\in K$), but you could have a more general ring homomorphism (i.e., not a $K$-homomorphism) $K[x]\to K[a]$ taking $x\mapsto a$ which doesn't fix elements of $K$.
For instance, you can take $K=\Bbb Q[i]$ and $a=i$. There is an automorphism $\sigma:K\to K$ sending $i\mapsto -i$, and you can take the (unique) $K$-homomorphism $\rho:K[x]\to K$ which sends $x\mapsto -i$, then take $\phi=\sigma\circ\rho$ to get a map $K[x]\to K$ which sends $x\mapsto i$ but has $i\mapsto -i$, so the elements of $K$ aren't fixed.