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- #1

- Jun 22, 2012

- 2,918

On page 710, D&F make the following statement:

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"In general, suppose R is a commutative ring. If P is a prime ideal in R[x] then \(\displaystyle P \cap R \) is a prime ideal in R ... ..."

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In checking out the validity of this assertion I wish to reason the following way:

Elements of \(\displaystyle P \cap R \) must be constant terms in P of the form \(\displaystyle a,b \in R\).

Given this, we reason as follows:

\(\displaystyle ab \in P \cap R \Longrightarrow ab \in P \) and \(\displaystyle ab \in R \)

Now \(\displaystyle ab \in P \Longrightarrow a \in P \) or \(\displaystyle b \in P \) since P is a prime ideal ... ... ... (1)

But we also have that:

\(\displaystyle ab \in R \Longrightarrow a \in R \) and \(\displaystyle b \in R \) ... ... ... (2) [... ? but why exactly ... does this actually follow from the closure of multiplication ... or not ?]

Then I wish to claim that (1) and (2) give me the following:

\(\displaystyle ab \in P \cap R \Longrightarrow a \in P \cap R \) or \(\displaystyle b \in P \cap R \)

Can someone please confirm that this reasoning is correct?

My problem or uncertainty is with my claim in the above that:

\(\displaystyle ab \in R \Longrightarrow a \in R \) or \(\displaystyle b \in R \)

I am not sure how to justify this reasoning.

I would appreciate a clarification.

Peter