If $P$ is a prime ideal of a commutative ring $R$, then $R[x]/P[x]$ is isomorphic to $(R/P)[x]$

Question

Suppose I have a commutative ring $R$ with 1, and a prime ideal $P$. Then I have read that $R[x]/P[x]$ is isomorphic to $(R/P)[x]$.

I am not totally sure what either of these two objects are, I am imagining that

$R[x]/P[x] = \{ r(x) + P[x] | r(x) \in R[x] \}$,

and $(R/P)[x]\ = \{ (r + P)[x] | r \in R \}$

but I am not sure. Further I am not sure what the map between these two sets should be before I prove it is a bijective homomorphism.

Additionally, since ideals are not necessarily subrings, is $P[x]$ even necessarily defined?

Answer 1

Consider the map $\theta:R[x] \to (R/P)[x]$ defined by $$ \theta:(a_0+\cdots+a_n x^n)\rightarrow (a_0 + P) + \cdots + (a_n +P)x^n.$$ Then the kernel of $\theta$ is $P[x]$. The map is clearly onto.

Afterwards applying the first isomorphism theorem will yield the result.