Polynomial Rings - Gauss's Lemma - Point 2 of a previous post

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• Apr 13th 2013, 09:48 PM
Bernhard
Polynomial Rings - Gauss's Lemma - Point 2 of a previous post
Point 2 of my previous post on Gauss's Lemma got well lost in the 10 replies - so I am re-posting the remaining problem - which was problem (2)

Basically I am trying to understand the proof of Gauss's Lemma as given in Dummit and Foote Section 9.3 pages 303-304 (see attached)

On page 304, part way through the proof, D&F write:

"Assume d is not a unit (in R) and write d as a product of irreducibles in R, say http://latex.codecogs.com/png.latex?..._2%20...%20p_n . Since http://latex.codecogs.com/png.latex?%20p_1 is irreducible in R, the ideal http://latex.codecogs.com/png.latex?%20%28p_1%29 is prime (cf Proposition 12, Section 8.3 - see attached) so by Proposition 2 above (see attached) the ideal http://latex.codecogs.com/png.latex?%20p_1R[x] is prime in R[x] and http://latex.codecogs.com/png.latex?%20%28R/p_1R%29[x] is an integral domain. ..."

My remaining problem with the D&F statement above are as follows:

(2) Despite reading Proposition 12 in Section 8.3 I cannot see why the ideal http://latex.codecogs.com/png.latex?%20p_1R[x] is prime in R[x] and http://latex.codecogs.com/png.latex?%20%28R/p_1R%29[x] is an integral domain. ...". (Indeed, I am unsure that http://latex.codecogs.com/png.latex?%20p_1R[x] is an ideal!) Can anyone show explicitly and rigorously why this is true?

Would appreciate help.

Peter
• Apr 13th 2013, 10:46 PM
rushton
Re: Polynomial Rings - Gauss's Lemma - Point 2 of a previous post
Hey mate I havent had a chance to go over the post you had this morning, but i did cover this question in my answers. $\displaystyle p_{1}R[x]$ is the ideal $\displaystyle (p_{1})$ because if you take any $\displaystyle r \in R[x]$ and $i \in I$ for you ideal I then $\displaystyle r \cdotp i \in I$.
So multiplying a whole ring by a prime element is obviously going to give you a prime ideal (think of what (2) in Z actually is). According to this (http://web.science.mq.edu.au/~chris/...olynomials.pdf) a prime polynomial is a irreducible polynomial, I am going to ask my lecturer on Tuesday also so if I get a different answer I will get back to you. I then gave you the proof for R/I is an integral domain iff I is prime in the other post.

Edit, now that I think ok it a little more it a irreducible poly must be a prime element just by the definition of what prime means. If p is prime then p=qr we must have q or r is a unit which is the definition of a irreducible polynomial.