Show that $R[x]/J[x]\cong (R/J)[x]$, where $J$ is an ideal of $R$.

Question

If $J$ is an ideal of the ring $R$, show that $J[x]$ is an ideal in $R[x]$ and show that $R[x]/J[x]\cong (R/J)[x]$.
Hint: Find a natural homomorphism from $R[x]$ onto $(R/J)[x]$ with kernel $J[x]$.

I think I can do the first part using the correspondence theorem.

For the second part, I'm certain we're going to need to use the/a homomorphism theorem. I thought of a few possible functions. One function was $\varphi: R[x]\to (R/J)[x]$ by $p\mapsto p-q$ where the coefficients of $q$ are in $R/J$. But after trying to show that $\ker \varphi = J[x]$, I realized that this might not be a homomorphism.

Another one was $\psi$, where $p\mapsto cp$, where $c\in R/J$. I sketched it out, and I believe $\psi$ is actually a homomorphism.

I think it's true that $\ker\psi\subseteq J[x]$. In the other direction, if $p\in J[x]$, then $\psi(p) = cp\in J[x]$. But that's where I got stuck. I don't think this shows equality. Regardless, I'm not sure if I'm even on the right track, and I've run out of potential homomorphisms. I was hoping for some guidance. Thanks.

Answer 1

Hint: Let $p:R\rightarrow R/J$ be the quotient map, consider $f:R[x]\rightarrow (R/J)(x)$ defined by $f(a_0+a_1x+...+a_nx^n)=p(a_0)+p(a_1)x+..+p(a_n)x^n$, $f$ is a morphism of rings which is surjective: Let $\sum_ib_ix^i\in R/J[x]$, there exists $a_i\in R$ such that $p(a_i)=b_i$, thus $f(\sum a_ix^i)=\sum_ib_ix^i$. $f$ is also injective since $p(a_0)+p(a_1)x+..+p(a_n)x^n=0$ implies that $p(a_i)=0, i=1,..,n$ which is equivalent to say that $a_0+a_1x+..+a_nx^n\in J[x]$ thus the kernel of $f$ is $J[x]$.

Answer 2

Define a homomorphism $\phi:R[X] \rightarrow R/J[X]$ by the formula $$\phi(a_0 +a_1X + \cdots + a_n X^n) = (a_0 + J) + (a_1 +J)X + \cdots + (a_n + J)X^n$$ Once you check that it is a homomorphism (it sends the identity to the identity, and distributes over addition and multiplication), it is immediate that it is surjective.

Injectivity follows from the fact that a polynomial is zero if and only if all its coefficients are zero. To say that $f(X) =a_0 + \cdots + a_nX^n$ gets mapped to zero is the same thing as saying that all the entries $a_i + J$ are the zero element in $R/J$. And $a_i + J$ is the zero element in $R/J$ if and only if $a_i$ is an element of $J$. Thus $f(X)$ is mapped to zero if and only if all its entries lie in $J$, i.e. $f(X) \in J[X]$.

So $J[X]$ is the kernel of the homomorphism $\phi$. This implies that $J[X]$ is an ideal; an ideal of a ring is the same thing as the kernel of a homomorphism.