Let $h:\mathbb{F}[x]\rightarrow\mathbb{F}[x]$ be a ring homomorphism where $\mathbb{F}$ is a field. Prove/disprove (give counter example) that the only such homomorphism is the $h=\text{Id}_{\mathbb{F}[x]}$

  • $\begingroup$ My first reaction was that it was false so tried to come up with a counter example. I tried a shifting map, one that multiplies by x. But it doesn't seem to preserve multiplication. $\endgroup$ – NormalsNotFar Mar 9 '16 at 17:35
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    $\begingroup$ How about this: $h(a_0 + a_1x + a_2x^2 + a_3x^3 + ... + a_nx^n) = a_0$? $\endgroup$ – user49685 Mar 9 '16 at 17:38

Hint: there are many so-called ''evaluation homomorphisms'' from $\mathbb{F}[x]$ to $\mathbb{F}$. Can you compose these with the inclusion of $\mathbb{F}$ into $\mathbb{F}[x]$ to get more than one ring homomorphism? Why are these maps not the identity on $\mathbb{F}[x]$?


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