# Showing if $R$ is a PID and $0\neq r\in R$ is irreducible, then $R[X]/(r)\cong (R/(r))[X]$

I've been looking at the statements I found on these two Stack Exchange answers and I've been trying to prove them. The first claim is:

If $$R$$ is a PID and $$0\neq r\in R$$ is irreducible, then $$R[X]/(r)\cong (R/(r))[X]$$

The second claim follows from the first:

If $$k$$ is a field, and $$p(X)$$ is an irreducible in $$k[X]$$, then $$K[X,Y]/(p(X)) \cong (K[X]/(p(X)))[Y]$$

Attempt to prove claim 1: It looks like I can prove this using the First Isomorphism Theorem by finding a ring morphism $$R[X] \hookrightarrow (R/(r))[X]$$ with $$(r)$$ as the kernel. However the only one I can think of is the following:

Let $$f:R\rightarrow R/(r)$$ be the projection $$f(x)=\bar{x}$$. Then we have the morphism of rings $$\phi:R[X]\rightarrow (R/(r))[X]$$ with the kernel $$(r)[X]$$ described as $$\phi(a_nx^n+\cdots+a_1x+a_0)=\phi(a_nx^n)+\cdots+\phi(a_1x)+\phi(a_0)$$ $$=f(a_n)x^n+\cdots+f(a_1)x+f(a_0)=\bar{a_n}x^n+\cdots+\bar{a_1}x+\bar{a_0}$$

Any help to prove this is appreciated!

edit: Further comment that I am interested whether the isomorphism hold only when $$R$$ is a commutative, unitary ring with any ideal $$I$$. There is utility in pulling out the '$$[X]$$' and treating any quotient of a ring polynomial as a ring polynomial with coefficients that are quotients i.e. $$R[X]/I\cong (R/I)[X]$$. Note I'm aware $$R[X]/I[X] \cong (R/I)[X]$$ holds.

• That $R[X]/I[x] \cong (R/I) [x]$ holds implies the first claim you made, taking $I = rR$. $I[x] = I ( R[x])$ in that case. You mentioned you're aware of this fact. Did you want someone to show this, or did you want something else? – Dean Young Mar 3 at 1:59
• Well in that case, may you please show me how the implication to my first claim is made? Or is that what you've shown below? Note I'm still reading it. I presume you mean $R[X]/I[X]\cong R[X]/I$, with $I=rR$? – CloudIcarus Mar 3 at 2:08
• $rR$ is not an ideal of $R[x]$. Take $I = rR$ We have (a) $R[x] / I[x] \cong (R/I)[x]$ (you already know this), (b) $I[x]= (rR)R[x]$ (the extension of $rR$ in $R[x]$, which could also be expressed as $rR[x]$ if you find it simpler)), (c) $R/I$ is identically $R/rR$, or $R/(r)$ in your notation. – Dean Young Mar 3 at 2:14
• As for (b), an element in $(rR)R[x]$ is of the form $\sum_{i = 1}^n r f_i$ for $f_i \in R[x]$. Elements of this form are merely polynomials in $R[x]$ with coefficients in $rR$. – Dean Young Mar 3 at 2:16
• Note: we made no use of $r$ being irreducible. – Dean Young Mar 3 at 2:17

Theorem For a ring $$R$$ and an element $$r \in R$$, $$R/rR[x] \cong R[x] / r R[x]$$.

That $$r$$ is irreducible is an unnecessary assumption.

You gave one map $$R[x] \rightarrow R/rR [x]$$. For a map going the other way, take $$R \rightarrow R[x] \rightarrow R[x] / rR[x]$$. This kills $$rR$$, and so factors through $$R/rR$$ by a map $$R/rR \rightarrow R[x] / rR[x]$$. There is a canonical map $$R/rR[x] \rightarrow R[x] / rR[x]$$ sending $$x$$ to $$\overline{x}$$. The resulting map sends $$\sum_{i = 1}^n \overline{a_i} x^i$$ to $$\overline{\sum_{i = 1}^n a_i x_i}$$.

Now one must show these two maps are inverse. But the map you constructed sends $$\overline{\sum_{i = 1}^n a_i x_i }$$ to $$\sum_{i = 1}^n \overline{a_i} x^i$$, so that these maps are inverse is immediate.