Let $A$ be a ring, $I \subset A$ a prime ideal and $\Phi : A \rightarrow A[x]$ the canonical inclusion homomorphism. Is $I[x]= \{ \sum_{k=0}^n a_k x^k : n\in \Bbb{N}, (a_k)_{k=0}^n \in I^n \}$ also prime in $A[x]$ ?
I am wondering this in order to apply it on other proof. I have search on my algebra books and notes and it does not appear theory about it. I have tried some ways to prove it, thinking about properties like $A/I$ being a integral domain since $I$ is prime, the relation between prime ideals and nilpotent elements among other things.
Any possible help or reference to already existing theory would be appreciated.
I assume, from your comment about integral domains, that you are considering commutative rings.
Consider the reduction map $A[x]\to (A/I)[x]$ given by $$\sum_{i=0}^n a_ix^i \longmapsto \sum_{i=0}^n(a_i+I)x^i.$$ This is a morphism, induced by the canonical projection $A\mapsto A/I$ and mapping $x$ to $x$, using the universal property of the polynomial ring.
The kernel is precisely $I[x]$. So by the Isomorphism Theorems, $A[x]/I[x]\cong (A/I)[x]$.
If $A$ is commutative with unity and $I$ is prime, then $A/I$ is an integral domain, hence $(A/I)[x]$ is an integral domain, proving that $I[x]$ is a prime ideal of $A[x]$.