Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove: Any endomorphism of $K[x]$ which is the identity on $K$ is $E_g$ for some $g$ in $K[x]$.

Note: In my book it defines $E_g\colon K[x] \to K[x]$ by sending $x$ to $g$.

This seems like it just follows from definitions but I guess I am just looking for some reassurance in my thought.

share|cite|improve this question

Hint: the map is already additive, and is multiplicative on powers of $x$, and fixing elements of $K$ amounts to it being linear. If $\phi(x)=g\in K[x]$, verify that the mentioned properties determine the image of an arbitrary polynomial.

P.S.: The notation $E_g$ is supposed to suggest "evaluation".

share|cite|improve this answer
+1, For the explanation of the notation. – wxu Apr 30 '12 at 17:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.