If $R$ is a commutative ring with unity and $I$ is an ideal of $R$ then show that $(R/I)[x]\cong R[x]/I[x]$.

My effort: Define $\phi :R[x]\to (R/I)[x]$

$\phi(a_0+a_1x+a_2x^2+\cdots +a_nx^n)=(a_0+I)+(a_1+I)x+(a_2+I)x^2+\cdots +(a_n+I)x^n$

Obviously $\phi $ is a ring homomorphism and surjective.

$\ker \phi =\{a_0+a_1x+a_2x^2+\cdots +a_nx^n:\phi(a_0+a_1x+a_2x^2+\cdots +a_nx^n)=I\}$

So $(a_0+I)+(a_1+I)x+(a_2+I)x^2+\cdots +(a_n+I)x^n=I\implies a_i\in I\forall i$

So $\ker \phi=I[x]$

Is the proof correct? Please help.

  • 7
    $\begingroup$ Correct. ${}{}$ $\endgroup$
    – user26857
    Sep 2, 2016 at 15:04
  • $\begingroup$ Thank you very much @user26857 $\endgroup$
    – Learnmore
    Sep 2, 2016 at 15:15
  • 1
    $\begingroup$ Possible duplicate of $(R/I)[x]=R[x]/I$ $\endgroup$
    – user557
    Nov 13, 2018 at 23:06
  • $\begingroup$ @user531587 I voted to leave open because it's not clear if the older post you linked to is better (in terms of the question content and the answers content). But it's really a judgment call, and I'm sorry if you have a bigger plan. $\endgroup$ Nov 14, 2018 at 0:52

1 Answer 1


You definitely have the right idea, but what you've written isn't quite right. If we let $\pi:R \longrightarrow R/I$ be the canonical quotient map, then your map $\phi$ can be expressed as $\phi(a_0 + a_1 x + \cdots + a_n x^n) = \pi(a_0) + \pi(a_1)x + \cdots \pi(a_n)x^n$. As you point out, $\phi$ is a surjective ring homomorphism. By the First Isomorphism Theorem, we're done if we can show that the kernel of $\phi$ is $I[x]$.

Suppose that $\phi$ maps $a_0 + a_1x + \cdots + a_nx^n$ to zero. Recall that a polynomial is zero if and only if each of its coefficients is zero. This means that $\pi(a_i) = 0$ for each $i$. By the definition of $\pi$, this implies that $a_i \in I$ for each $i$, and we conclude that our polynomial is in $I[x]$.

  • 2
    $\begingroup$ If we replace $a_i+I$ in the OP's solution by $\pi(a_i)$ I think we get your solution. So, it's not clear what "isn't quite right" in the OP's solution. $\endgroup$
    – user26857
    Sep 3, 2016 at 21:38
  • 1
    $\begingroup$ @user26857 I stand corrected. I mistook the OP's equation $\phi(a_0 + a_1x + \cdots + a_nx^n) = I$ to be a notational mistake, but of course, it isn't. The ideal $I$ is the zero element of $(R/I)[X]$. I write this as $0 + I$, to emphasize that I'm thinking of a coset (i.e., an element of $R/I$) and not merely an ideal (i.e., a subset $I \subset R$.) But that's a matter of preference alone. $\endgroup$
    – PeterJL
    Sep 12, 2016 at 7:06

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