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Define $P[x]$ as the set of polynomials with coefficients in $P$.

Let $$f(x)=\sum_{i=0}^m a_i x^i,\quad g(x)=\sum_{j=0}^n b_j x^j$$ where $a_i, b_j \in R$ (so $f,g\in R[x]$)

Let $f*g \in P[x]$.

I want to show that either $f$ or $g$ has to be in $P[x]$.

$$f*g =(a_0 b_0 + a_0 b_1 x + a_0 b_2 x^2 + \cdots + a_0 b_n x^n) +(a_1 b_0 x + a_1 b_1 x^2 + \cdots + a_1 b_n x^{n+1}) +\cdots + (a_m b_0 x^m + a_m b_1 x^{m+1} + \cdots a_m b_n x^{m+n})$$

Obviously if $f$ or $g$ was in $P[x]$, $f*g$ would be in $P[x]$. But what if the odd coefficients of a and the even coefficients of $b$ were in $P$, while no other coefficients are in $P$. Would that be a counterexample? I'm struggling with showing that if $a_i b_j \in P$ $\forall i,j$ then $a_i \in P$ $\forall i$ or $b_j \in P$ $\forall j$.

I'm also concerned about the coefficients that aren't just a product of coefficients from $f$ and $g$ (e.g. $a_0b_1+b_0a_1$ or $a_0b_2 +a_1b_1+a_2b_0$). If all $a_ib_j \in P$ then these coefficients are still in $P$, but I don't see why it couldn't be that none of the terms in one of these coefficients are in $P$, but still add to an element of $P$.

PNT
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Ben Betts
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  • I assume $R$ is commutative with identity. If so, prove that $R[x]/P[x]\cong (R/P)[x]$ and use the fact that an ideal is prime if and only if the quotient is an integral domain. – Arturo Magidin Apr 13 '22 at 20:55
  • "If the odd coefficients of [$f$] and the even coefficients of [$g$] are in $P$"... What would be the coefficient of $x$ in such a product? It would be $a_0b_1+a_1b_0$. Note, $a_1b_0$ is in $P$, so the sum is in $P$ if and only if $a_0b_1$ is in $P$. Do you have a counterexample, then? – Arturo Magidin Apr 13 '22 at 21:15
  • The standard proofs already occur here many times, viz. either reduce it to the trivial domain case by factoring out $P[x]$ as in the first linked dupe, or else effectively do the same element-wise with polynomials (one form of Gauss's Lemma), as in the 2nd linked dupe (which immediately extends from principal ideals to general ideals - just replace $,p\mid a,$ by $,a\in P,$ etc). – Bill Dubuque Apr 13 '22 at 23:15
  • In particular, this answer in the 2nd linked dupe shows how to give a very simple direct element-wise proof in the form that you attempted. – Bill Dubuque Apr 13 '22 at 23:28

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Assuming $R$ is commutative with identity, the easiest way of doing this is to use the characterization of prime ideals in terms of quotients:

Let $R$ be a commutative ring with identity, and $P$ an ideal of $R$. Then $P$ is a prime ideal if and only if $R/P$ is an integral domain.

Let $P$ be a prime ideal of $R$. Consider the morphism $R[x]\to (R/P)[x]$ given by $a_0+a_1x+\cdots+a_nx^n \mapsto \overline{a_0}+\overline{a_1}x+\cdots + \overline{a_n}x^n$, where $\overline{b}$ is the image of $b\in R$ in $R/P$ under the canonical projection.

Prove that this is a surjective homomorphism, that the kernel is precisely $P[x]$, and that $(R/P)$ is an integral domain because $P$ is a prime ideal.

If you want to go through the definition by elementwise multiplication, you should verify that $P[x]$ is not all of $R[x]$ (which you omitted). And you should write the product as $$fg = c_0 + c_1x + \cdots + c_{n+m}x^{n+m}$$ where $$c_r = \sum_{j=0}^r a_jb_{r-j}$$ (setting $a_k=0$ if $k\gt m$, and $b_k=0$ if $k\gt n$). Assume that $f\notin P[x]$ and $g\notin P[x]$, and show that $fg\notin P[x]$. Then there exist smallest indices $i_0$ and $j_0$ such that $a_{i_0}\notin P$ and $b_{j_0}\notin P$. Consider $c_{i_0+j_0}$.

Arturo Magidin
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