Find all surjective ring homomorphisms $\phi:\mathbb{R}[x] \rightarrow \mathbb{R}.$
Attempt. Functions $\phi(f(x))=f(a)$ are all surjective ring homomorphisms. It seems to me that these are exactly the surjective ring homomorphisms from $\mathbb{R}[x]$ to $\mathbb{R}$. One thought is that $R[x]/Ker\phi\cong Im\phi=\mathbb{R}$ would imply that the principal ideal $Ker\phi=<p(x)>$ is such that $p(x)$ is irreducible, but i am not sure if i am on the right path. Another thought is that $$\phi(1)=\phi(1\cdot 1)=\phi(1)\cdot\phi(1)$$ so $\phi(1)=1$ (since $\phi$ is onto the reals).
Thanks in advance!
I think the hard part of this is showing that any ring homomorphism $\phi:\Bbb R[x] \to \Bbb R$ is $\Bbb R$-linear.
I'll denote the restriction of $\phi$ to the subring of constant polynomials with $\psi: \Bbb R \to \Bbb R$. It is a fact that the only ring homomorphism from $\Bbb R \to \Bbb R$ is the identity, see this question. Thus we conclude that $\psi$ is the identity, so $\phi$ is $\Bbb R$-linear. From there it's easy to see that $\phi(f(x))=f(\phi(x))$, so if we let $\phi(x)=a$, we have $\phi(f(x))=f(a)$.