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

This is homework. I have been trying to assume that there is an $fg$ in $P[x]$ such that neither $f$ nor $g$ is in $P[x]$. Hence $f$ and $g$ have at least one coefficient not in $P$. I was trying to show that $fg$ would then have a coefficient not in $P$, obtaining a contradiction. But I don't see how to control the terms. If $f$ and $g$ had only one coefficient not in $P$, 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 $P$.

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

Bill Dubuque
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OLP
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1 Answers1

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

Alternatively, following your own line of attack: Assume that $f$ and $g$ are two polynomials, neither in $P[x]$. 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 $P$. What can you say about the term of degree $m+n$ in the product $fg$?

Note for comments: notation has been changed: ideal $I$ is now $P$.

Bill Dubuque
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Jyrki Lahtonen
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  • Thanks for the hint! I will see what I can manage. – OLP Mar 01 '15 at 18:13
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    Sorry about neglecting your own line of attack. I added another hint suggesting how you might follow up on that idea. – Jyrki Lahtonen Mar 01 '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 01 '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 01 '15 at 18:23
  • One more question, if you have time. How do we know that the homomorphism in the first hint is an isomorphism? – OLP Mar 01 '15 at 18:29
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    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 01 '15 at 18:36
  • Thanks so much for your help. I was just plugging away at that with the text and I think I understand. If you have the time, I would like to understand the other attack too, but I am grateful for all the help so far. – OLP Mar 01 '15 at 18:48
  • I think I understand the other attack now. All of the other terms are in the ideal. Hence their sum is and hence the difference of this coefficient and this sum and hence the term we know cannot be in the ideal. Very nice. – OLP Mar 01 '15 at 19:07
  • @OLP: Well done in completing that other attack, too! – Jyrki Lahtonen Mar 01 '15 at 20:03
  • I think I see an issue with the original line of attack. What if the first coefficient of one and the last coefficient from the other are not in P? The lowest degree term and the highest degree term in the polynomial product each have an element from P in the product of coefficients making up the coefficient of each of said degree terms. But, each of the two polynomials have a coefficient not in P meaning neither f or g is in P[x]. – user3146 Mar 14 '24 at 03:16
  • @JyrkiLahtonen But you necessarily assumed that $R$ is an integral domain, isn't it? Please correct me, if I have missed something. – Thomas Finley May 14 '24 at 04:51
  • @ThomasFinley I did not assume that. The second bullet is applied to $R/P$ rather than $R$. In other words, the $R$ in the second bullet is a variable taking values in the realm of commutative rings. I admit that it is potentially confusing to use the same symbol for the fixed (?) commutative ring given in the problem, and a variable ring. To make it clear, mentally reformulate it as a separate theorem: For every commutative ring $R$, the polynomial ring $R[x]$ is an integral domain if and only if the ring $R$ is. Or something like that. – Jyrki Lahtonen May 14 '24 at 04:58
  • @JyrkiLahtonen Thanks for the clarification! I think, more specifically we have: Let $R$ is a commutative ring and $I$ be an ideal of $R.$ $I$ is a prime ideal (of $R$) iff* I[x] is a prime ideal (of $R[x]$).* Isn't it? – Thomas Finley May 14 '24 at 12:15