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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!

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  • $\begingroup$ The dimension of $\mathbb{R}[x]/(p)$ over $\mathbb{R}$ is the degree of $p$. Therefore, $p$ has degree $1$. $p=ax+b$. Therefore, the homomorphism is evaluation at $-b/a$. $\endgroup$ – user525761 Jan 30 '18 at 13:53
  • $\begingroup$ Some authors require $\phi(1)=1$ as part of the definition of a ring homomorphism. I don't know how it is in your book. $\endgroup$ – Arthur Jan 30 '18 at 13:56
  • $\begingroup$ Let $a$ be the image of $x$ by $\phi$, then $(x-a)\subset\ker\phi$. This implies that $\phi(f)=f(a)$. $\endgroup$ – Roland Jan 30 '18 at 14:09
  • $\begingroup$ Do you mean $\mathbb R$-algebra homomorphisms instead of ring homomorphisms? The ring homomorphisms $\mathbb R[x]\to \mathbb R$ correspond to pairs $(f,a)$, where $a\in \mathbb R$ (the image of $x$) and $f\colon \mathbb R\to \mathbb R$ is a ring homomorphism. But what are all the ring homomorphisms $\mathbb R\to \mathbb R$?! $\endgroup$ – Claudius Jan 30 '18 at 15:38
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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 $\phi(f(x))=f(a)$.

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