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Let $R$ be a commutative ring (not necessarily unital). Let I be an ideal of R. Show that $I[X]$ is a prime ideal in $R[X]$ iff I is a prime ideal in $R$.

I have attempted to use the facts:

  1. I is prime iff $R/I$ is an ID.
  2. I is maximal iff $R/I$ is a field.

But both require $R$ to be a commutative unital ring. I thought constructing an isomorphism might work but not entirely sure. A more detailed explanation is very much appreciated. Thanks.

user1868607
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user71346
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  • In my answer I assume that prime ideals are defined to be proper ideals $I$ with $ab \in I \Rightarrow a \in I \vee b \in I$. There are also other definitions in the non-unital case, which really differ from this one (for example $AB \subseteq I \Rightarrow A \subseteq I \vee B \subseteq I$ for ideals $A,B$). – Martin Brandenburg Apr 07 '13 at 02:51

1 Answers1

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$I \subseteq R$ is prime iff $R/I$ is a nontrivial ring without zero divisors. Now observe $R[X]/I[X] = (R/I)[X]$ (both sides satisfy the same universal property). Then the claim comes down to: $R$ is a nontrivial ring without zero divisors iff $R[X]$ is a nontrivial ring without zero divisors. Well, $\Leftarrow$ is clear, and $\Rightarrow$ follows from degree considerations. We don't need a unit for that.

Martin Brandenburg
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  • Thank you very much. Couldn't think of that isomorphism before. That is actually a very good way to solve the problem. – user71346 Apr 07 '13 at 02:11