Is $\mathbb{Z}[x] / (x - a) \cong \mathbb{Z}$ where $a \in \mathbb{Z}$?

I'm trying to get more comfortable with these questions so please critique my attempt, or if you can, suggest other (better) ways of doing this if you can, thanks.

The two rings are isomorphic. Consider the evaluation map $\phi : \mathbb{Z}[x] \to \mathbb{Z}, x \mapsto a$. Since this is an evaluation map it is a homomorphism and it is surjective since $\phi(n) = n$ for any $n \in \mathbb{Z}$.

Now we show $\ker(\phi) = (x - a)$. If $f \in \ker(\phi)$, then $f(a) = 0$, thus $x - a \mid f$ hence $f \in (x - a)$. If $f \in (x-a)$, then $f(x) = g(x)(x-a)$ hence $f(a) = 0$, so $f \in \ker(\phi)$. Thus by the first isomorphism theorem $\mathbb{Z}[x] / (x-a) \cong \mathbb{Z}$.

  • 1
    $\begingroup$ Looks good to me. $\endgroup$ – Patrick Stevens Feb 25 '18 at 21:24
  • 1
    $\begingroup$ Indeed. Well done! $\endgroup$ – Quoka Feb 25 '18 at 21:24
  • 1
    $\begingroup$ Good job${}{}{}$ $\endgroup$ – Andres Mejia Feb 25 '18 at 21:24
  • $\begingroup$ Thank you all for your taking the time to verify. It's very much appreciated and helpful. $\endgroup$ – user330531 Feb 25 '18 at 21:29
  • 1
    $\begingroup$ You should just mention the fact that $f(a)=0$ implies $(x-a)\mid f$ because $x-a$ is monic. $\endgroup$ – egreg Feb 25 '18 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.