Let $R$ be a commutative ring, $I\subset R$ an ideal. Then $I[x]$ (all polynomials with coefficients in $I$) is an ideal of $R[x]$.
Prove or disprove:
- $I$ maximal $\implies I[x]$ maximal;
- $I$ prime $\implies I[x]$ prime.
Here is a counterexample for the first part: the ideal $(2)$ is maximal in $\mathbb Z$, but the set of all polynomials in $\mathbb Z[x]$ divisible by $2$ is not a maximal ideal since it is contained in $(2)\subset \mathbb Z[x]$. Is this correct?
I think the second implication is true. The way I tried to prove it is this. Let $I$ be prime, let $p\in I[x]$. We need to show that $p\mid fg\implies p\mid f$ or $p\mid g$. Assume the converse: $p$ does not divide both $f$ and $g$. Then $f=pq_1+r_1,\ g=pq_2+r_2,\deg r_i < \deg p, r_i\ne 0$. Need to show $$(pq_1+r_1)(pq_2+r_2)=p^2q_1q_2+pq_2r_1+pq_1r_2+r_1r_2$$ is not divisible by $p$. Write $r_1r_2=pq_3+r_3$ with $\deg r_3 < \deg p$. Then the first display becomes $$(pq_1+r_1)(pq_2+r_2)=p^2q_1q_2+pq_2r_1+pq_1r_2+pq_3+r_3$$ If $r_3\ne 0$, then this is not divisible by $p$ because $\deg r_3 < \deg p$. But I don't know whether $r_3\ne 0$. Moreover, such approach doesn't use that the coefficient of $p$ lie in a prime ideal... Is this a correct way to prove this implication at all?
(This is as a response to OP's comment on Hello_World's answer)
First of all, your approach assumes that $R[x]$ is a Euclidean domain which is not always the case (For example, take $R=\mathbb{R}[y]$. Then $R[x]=\mathbb{R}[x,y]$ is not Euclidean). So I would say it is doomed to failure.
It is indeed possible to prove it directly (Although it is more elegant using the fact that $(R/I)[x] \simeq R[x]/I[x]$).
Let $f,g \notin I[x]$. We need to show that $fg \notin I[x]$. Write $f=a_nx^n+...+a_0$ , $g=b_mx^m+...+b_0$ and $fg=c_{m+n}x^{m+n}+...+c_0$. By definition, there is some coefficient $a_i$ of $f$ and $b_j$ of $g$ with $a_i,b_j \notin I$. Take $i$ and $j$ to be minimal and put $k=i+j$. Then $c_k \notin I$ (Prove it!) and therefore $fg \notin I[x]$.