Polynomial Rings - Gauss's Lemma - Point re Proof

Jan 2010
594
5
Hobart, Tasmania, Australia
Gauss' Lemma is stated and proved on pages 303-304 of Dummit and Foote (see attachment)

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Gauss Lemma

Let R be a UFD with field of fractions F and let \(\displaystyle p(x) \in R[x] \). If p(x) is reducible in F[x] then p(x) is reducible in R[x].

More precisely, if p(x) = A(x)B(x) for some nonconstant polynomials \(\displaystyle A(x), B(x) \in F[x] \), then there are nonzero elements \(\displaystyle r, s \in F \) such that rA(x) = a(x) and sB(x) = b(x) both lie in R[x] and p(x) = a(x)b(x) is a factorization in R[x].

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In the proof of Gauss' Lemma on page 304 (see attachment) we find:

"Assume d is not a unit and write d as a product of irreducibles in R, say \(\displaystyle d = p_1p_2 ...p_n \). Since \(\displaystyle p_1 \) is irreducible in R, the ideal \(\displaystyle (p_1) \) is prime (cf Proposition 12, section 8.3 - see attachment), so by Proposition 2 above (see attachment) the ideal \(\displaystyle p_1R[x] \) is prime in R[x] and \(\displaystyle (R/p_1R)[x] \) is an integral domain.

Reducing the equation dp(x) = a'(x)b'(x) modulo \(\displaystyle p_1 \) we obtain the equation \(\displaystyle 0 = \overline{a'(x)} \overline{b'(x)} \), hence one of the two factors, say \(\displaystyle \overline{a'(x)} \) must be zero. ... ... (see attachment) .. "

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Problem!

In the proof we read:

"Since \(\displaystyle p_1 \) is irreducible in R, the ideal \(\displaystyle (p_1) \) is prime (cf Proposition 12, section 8.3 - see attachment), so by Proposition 2 above (see attachment) the ideal \(\displaystyle p_1R[x] \) is prime in R[x] and \(\displaystyle (R/p_1R)[x] \) is an integral domain. ..."

I can see that this is the case, BUT why is this needed and how does the rest of the proof connect to this???

The next part of the proof is

"Reducing the equation dp(x) = a'(x)b'(x) modulo \(\displaystyle p_1 \) we obtain the equation \(\displaystyle 0 = \overline{a'(x)} \overline{b'(x)} \), hence one of the two factors, say \(\displaystyle \overline{a'(x)} \) must be zero. ... ... (see attachment) .. "

But to do this we only need to divide by \(\displaystyle p_1 \) and take the remainder as defining the coset. Is the section above confirming in some way that we can divide successfully???

Can someone please clarify?

Peter
 

Attachments

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Jan 2010
594
5
Hobart, Tasmania, Australia
I have been thinking about my own question above.

A possible answer is as follows:

When we reduce the equation dp(x) = a'(x)b'(x) modulo \(\displaystyle p_1 \) we are then dealing with elements in the ring \(\displaystyle (R/p_1R)[x] \). That is, when we are dealing with the equation \(\displaystyle 0 = \overline{a'(x)} \ \overline{b'(x)} \) we are working with elements in \(\displaystyle (R/p_1R)[x] \).

When we argue that one of the two factors in the equation \(\displaystyle 0 = \overline{a'(x)} \ \overline{b'(x)} \), say \(\displaystyle \overline{a'(x)} \) must be 0, we need to be sure that a'(x) and b(x) are not zero divisors and that means we need the ring \(\displaystyle (R/p_1R)[x] \) to be an integral domain.

Can someone please confirm for me that my reasoning with respect to the proof of Gauss' Lemma is correct.

Peter
 
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