Show that $R[x]/xR[x]\cong R$, where $R$ is a commutative ring.

My first thought was that I would need to use one of the homomorphism theorems. I was thinking that I would need a surjective ring homomorphism, and that it should be $\varphi(p) = p(a)$ or $\psi(p) = p(0)$. I'm not sure. Then $\ker \varphi = \{p\mid p(a) = 0\}$ and $\ker\psi = \{p\mid p(0) = 0\}$. And from what I understand, $p$ is informally $p = a_0+a_1x+\dotsb+a_nx^n$, $a_i\in R$.

Then I considered $xR[x]$. I believe that this is the set $\{xp\mid p\in R[x] \}$.

But then I'm not sure where to go from there, how to fit the pieces together. I was hoping for some guidance. Thanks!

  • $\begingroup$ That is, the polynomial in R[x]/xR[x] is strictly determined by its constant term. But the other terms are "absorbed" into xR[x]. $\endgroup$ Aug 8, 2016 at 11:29
  • $\begingroup$ Sum(0 to infinity)a_ix^i + { Sum(1 to infinity) b_i x^i | b_i are in R)}= a_0 +Sum(1 to infinity)a_ix^i +{Sum(1 to infinity) b_i x^i |b_i are in R} = a_0+{Sum(1 to infinity) (b_i-a_i)x^i | b_i are in R} = a_0+{Sum(1 to infinity)c_i x^i | c_i are in R} $\endgroup$ Aug 8, 2016 at 11:44
  • $\begingroup$ The elements are literally r+xR[x] for r in R. It is easy to show these cosets are unique. $\endgroup$ Aug 8, 2016 at 12:04

1 Answer 1


You're on the right track. Put $f\colon R[x] \to R$, $f(p(x)) = p(0)$. Check that this is a ring morphism. Also, given $a \in R$, $f(a) = a$, so $f$ is an epimorphism. Finally, if $p(x) = \sum_{k=0}^na_kx^k$, we have $f(p(x)) = a_0$. You can check that $\ker f = x\,R[x]$ directly, by factoring $x$ out, if $p(x) \in \ker f$. On the other hand, if $p(x) \in x\,R[x]$, then $f(p(x)) = 0$ (why?).

Now conclude using the first isomorphism theorem.

  • $\begingroup$ Why do you bother with morphism and stuff when you can easily do it from first principles? Is it supposed to be practice for "real" problems? $\endgroup$ Aug 8, 2016 at 11:47
  • $\begingroup$ This sort of reasoning appears again in a similar way when you study field extensions and quotients of $K[x]$... might as well start getting used now. $\endgroup$
    – Ivo Terek
    Aug 8, 2016 at 11:51
  • $\begingroup$ I will think about your post some more. Thanks. $\endgroup$
    – David
    Aug 8, 2016 at 22:11
  • $\begingroup$ Ok. Feel free to ask about anything you don't understand. $\endgroup$
    – Ivo Terek
    Aug 8, 2016 at 22:13

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