# Proof of Gauss' Lemma - Issue re Proof

#### Peter

##### Well-known member
MHB Site Helper
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 [TEX] p(x) \in R[x] [/TEX]. 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 [TEX] A(x), B(x) \in F[x] [/TEX], then there are nonzero elements [TEX] r, s \in F [/TEX] 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 [TEX] d = p_1p_2 ...p_n [/TEX]. Since $$\displaystyle p_1$$ is irreducible in R, the ideal [TEX] (p_1) [/TEX] is prime (cf Proposition 12, section 8.3 - see attachment), so by Proposition 2 above (see attachment) the ideal [TEX] p_1R[x] [/TEX] is prime in R[x] and [TEX] (R/p_1R)[x] [/TEX] is an integral domain.

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

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Problem!

"Since $$\displaystyle p_1$$ is irreducible in R, the ideal [TEX] (p_1) [/TEX] is prime (cf Proposition 12, section 8.3 - see attachment), so by Proposition 2 above (see attachment) the ideal [TEX] p_1R[x] [/TEX] is prime in R[x] and [TEX] (R/p_1R)[x] [/TEX] 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 [TEX] p_1 [/TEX] we obtain the equation [TEX] 0 = \overline{a'(x)} \overline{b'(x)} [/TEX], hence one of the two factors, say [TEX] \overline{a'(x)} [/TEX] must be zero. ... ... (see attachment) .. "

But to do this we only need to divide by [TEX] p_1 [/TEX] and take the remainder as defining the coset. Is the section above confirming in some way that we can divide successfully???

Peter

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#### Peter

##### Well-known member
MHB Site Helper
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 [TEX] p_1 [/TEX] we are then dealing with elements in the ring [TEX] (R/p_1R)[x] [/TEX]. That is, when we are dealing with the equation [TEX] 0 = \overline{a'(x)} \ \overline{b'(x)} [/TEX] we are working with elements in [TEX] (R/p_1R)[x] [/TEX].

When we argue that one of the two factors in the equation [TEX] 0 = \overline{a'(x)} \ \overline{b'(x)} [/TEX], say [TEX] \overline{a'(x)} [/TEX] 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 [TEX] (R/p_1R)[x] [/TEX] 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

#### Opalg

##### MHB Oldtimer
Staff member
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 [TEX] p_1 [/TEX] we are then dealing with elements in the ring [TEX] (R/p_1R)[x] [/TEX]. That is, when we are dealing with the equation [TEX] 0 = \overline{a'(x)} \ \overline{b'(x)} [/TEX] we are working with elements in [TEX] (R/p_1R)[x] [/TEX].

When we argue that one of the two factors in the equation [TEX] 0 = \overline{a'(x)} \ \overline{b'(x)} [/TEX], say [TEX] \overline{a'(x)} [/TEX] 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 [TEX] (R/p_1R)[x] [/TEX] 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
That is correct. Glad you managed to work it out for yourself.