# Prove that if I is a prime ideal of R then I[x] is a prime ideal of R[x]

Prove that if $I$ is a prime ideal of $R$ then $I[x]$ is a prime ideal of $R[x]$.

This is homework. I have been trying to assume that there is an $fg$ in $I[x]$ such that neither $f$ nor $g$ is in $I[x]$. Hence $f$ and $g$ have at least one coefficient not in $I$. I was trying to show that $fg$ would then have a coefficient not in $I$, obtaining a contradiction. But I don't see how to control the terms. If $f$ and $g$ had only one coefficient not in $I$, then I think I could use the properties of the ideal to complete the proof. The problem is not being able to know for certain which if any of the coefficients of $f$ and $g$ are in $I$.

Perhaps my approach is wrong to begin. Please help. Even a hint in the right direction will much appreciated.

• The ring is commutative. Sorry to have omitted that! – OLP Mar 1 '15 at 18:13

## 1 Answer

I assume that $R$ is commutative, because the ring of polynomials with a non-commutative ring of coefficients is a very rare beast.

Can you combine the following pieces?

• An ideal $I$ of a commutative ring $R$ is prime, iff the quotient ring $R/I$ is an integral domain.
• The polynomial ring $R[x]$ is an integral domain if and only if $R$ is.
• There is an isomorphism $(R/I)[x]\cong R[x]/I[x]$.

Alternatively, following your own line of attack: Assume that $f$ and $g$ are two polynomials, neither in $I[x]$. Let $f(x)=\cdots +a x^m+\cdots$ and $g(x)=\cdots+ bx^n+\cdots$. Assume that the highlighted terms $ax^m$ and $bx^n$ are the highest degree term with the property that the coefficients don't belong to the ideal $I$. What can you say about the term of degree $m+n$ in the product $fg$?

• Thanks for the hint! I will see what I can manage. – OLP Mar 1 '15 at 18:13
• Sorry about neglecting your own line of attack. I added another hint suggesting how you might follow up on that idea. – Jyrki Lahtonen Mar 1 '15 at 18:20
• So R/I is an ID, hence (R/I)[x] is an ID, hence R[x]/I[x] is an ID, hence I[x] is prime. It's a very high level argument. I need to get better at those instead of getting tangled up in coefficients. – OLP Mar 1 '15 at 18:21
• The term In question can't belong to the ideal since it is prime. Nice. But does this imply that the coefficient as a whole is not in I? I mean in terms of polynomial multiplication I can see that fg will have a term within one of its coefficients not in I but I don't see how to show that the coefficient as a whole is not in I. – OLP Mar 1 '15 at 18:23
• The isomorphism is gotten from the projection, coefficient by coefficient $p:R[x]\to (R/I)[x]$. It is a surjection, because the projection $R\to R/I$ is one, and its kernel is $I[x]$. Then apply the first homomorphism theorem. – Jyrki Lahtonen Mar 1 '15 at 18:36