# Factorization of Polynomials - Gauss's Lemma

• Nov 5th 2012, 12:37 AM
Bernhard
Factorization of Polynomials - Gauss's Lemma
I am reading Anderson and Feil - A First Course in Abstract Algebra - Section 5.3 Polynomials with Integer Co-effficients

On page 60 A&F state Gauss's Lemma in the following way (see attachment)

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Theorem 5.5 Gauss's Lemma

If $\displaystyle f \in \mathbb{Z} [x]$ and f can be factored into a product of non-scalar polynomials in $\displaystyle \mathbb{Q} [x]$, then f can be factored into a product of non-scalar polynomials in $\displaystyle \mathbb{Z} [x]$ ; each factor in $\displaystyle \mathbb{Z} [x]$ is an associate of the corresponding factor in $\displaystyle \mathbb{Q} [x]$

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Then at the bottom of page 60 A&F state the following:

"We can rephrase Gauss's Lemma in the following way: To see whether a polynomial in $\displaystyle \mathbb{Z} [x]$ can be factored non-trivially in $\displaystyle \mathbb{Q} [x]$, we need only check to see if it can be factored non-trivially in $\displaystyle \mathbb{Z} [x]$. The latter is presumably an easier task."

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My problem is that at first glance this does not seem like a restatement of Gauss's Lemma - if we check and find that f can be factored non-trivially in $\displaystyle \mathbb{Z} [x]$ then we can conclude nothing since the implication of Gauss's Lemma goes the other way i.e. it says if f can be factored in $\displaystyle \mathbb{Q} [x]$ then it can be factored in $\displaystyle \mathbb{Z} [x]$! Problem!

UNLESS what A&F are referring to is the contrapositive of Gauss's Lemma - if you find that f cannot be factored non-trivially in $\displaystyle \mathbb{Z} [x]$ then it cannot be factored non-trivially in $\displaystyle \mathbb{Q} [x]$ ???

Can anyone clarify this matter for me? Would appreciate the help.

Peter
• Nov 5th 2012, 01:03 AM
emakarov
Re: Factorization of Polynomials - Gauss's Lemma
Quote:

Originally Posted by Bernhard
if we check and find that f can be factored non-trivially in $\displaystyle \mathbb{Z} [x]$ then we can conclude nothing since the implication of Gauss's Lemma goes the other way i.e. it says if f can be factored in $\displaystyle \mathbb{Q} [x]$ then it can be factored in $\displaystyle \mathbb{Z} [x]$!

A factorization in ℤ[x] is obviously a factorization in ℚ[x] since ℤ ⊂ ℚ.
• Nov 5th 2012, 01:16 AM
Bernhard
Re: Factorization of Polynomials - Gauss's Lemma
Hmm ... yes OK ...

Makes me wonder about the point of A&Fs statement that they say is a rephrasing of Gauss's Lemma - that is

"We can rephrase Gauss's Lemma in the following way: To see whether a polynomial in $\displaystyle \mathbb{Z} [x]$ can be factored non-trivially in $\displaystyle \mathbb{Q} [x]$ , we need only check to see if it can be factored non-trivially in $\displaystyle \mathbb{Z} [x]$. The latter is presumably an easier task."
• Nov 5th 2012, 01:30 AM
emakarov
Re: Factorization of Polynomials - Gauss's Lemma
Let's write Q(p) to mean that a polynomial p can be factored non-trivially in ℚ[x] and Z(p) to mean that p can be factored in ℤ[x]. Then Gauss's Lemma says ∀p, Q(p) → Z(p). Also, we have trivially that ∀p, Z(p) → Q(p). So, an easy corollary of Gauss's Lemma is that ∀p, Q(p) ↔ Z(p).

The rephrasing, as I understand it, says ∀p, (Z(p) → Q(p)) ∧ (¬Z(p) → ¬Q(p)), which is also equivalent to ∀p, Q(p) ↔ Z(p). The second conjunct, in particular, is the contrapositive of Gauss's Lemma and is therefore equivalent to it.

Above, ∀ means "for all," ∧ means "and," and ¬ means "not."
• Nov 5th 2012, 01:34 AM
Bernhard
Re: Factorization of Polynomials - Gauss's Lemma

I am left puzzling why Gauss Lemma is not simply stated as ∀p, Q(p) ↔ Z(p).

But anyway, thanks to you I am hally with the logic showing that ∀p, Q(p) ↔ Z(p).

Peter
• Nov 5th 2012, 07:24 AM
Deveno
Re: Factorization of Polynomials - Gauss's Lemma
sometimes it is. although the usual paraphrase i come across is: if a polynomial with integer coefficients factors over Q, it factors over Z.

that said, the way Gauss' lemma is actually USED, in practice, is to show a polynomial in Z[x] is irreducible over Q by showing it does not factor over Z.

this gives a way of proving, for example, that x2 - 2 is irreducible over Q (and thus that the square root of 2 is irrational):

if x2 - 2 = (x - a)(x - b), we may assume a and b are INTEGERS, thus:

a + b = 0
ab = -2

thus we have a2 = 2 for an integer a, which has no solution (0,-1, and 1 do not work, and all other integers are "too big in size": a2 ≥ 4).

the factors must be monic, of course, because if we had instead an integer factorization of the form:

(cx - a)(dx - b), we have cd = 1, which means that c is a unit, and so c = ±1. if c = -1, d = -1, and we can replace the factorization:

(-x - a)(-x - b) by: (x - (-a))(x - (-b)), and -a,-b are integers whenever a,b are.

this is a rather slick proof of the irrationality of √2 which seems to avoid the use of prime factorization. of course, the "dirty details" have been shifted to the proof of Gauss' lemma, where the UFD property of Z is invoked (and in fact, a version of Gauss' lemma holds in any UFD: if R is a UFD, and F its field of fractions, then a monic polynomial in F[x] is irreducible if and only if it is irreducible in R[x]-the "monic" condition can be extended to the notion of primitive polynomial: that is, the gcd of all the coefficients is a unit of R).

the proof in the general case is substantially the same as in the case for Z and Q.
• Nov 5th 2012, 10:02 PM
Bernhard
Re: Factorization of Polynomials - Gauss's Lemma
Thanks Deveno ... now getting an understanding of the point of Gauss's Lemma ...

Thanks also for the very neat proof of the irrationality of $\displaystyle \surd 2$!!

Regarding UFDs I note that on page 303 of Dummit and Foote Gauss's Lemma is stated as folows:

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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 A(x), B(x) $\displaystyle \in$ F[x], then there are nonzero elements r, s $\displaystyle \in$ 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 factorisation in R[x]

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I must now go and try to understand the field of fractions.

I note that Gallian states Gauss's Lemma as

"The product of two primitive polynomials is primative"

... so I also need to go and understand how this is equivalent to A&F above and toe D&F

Peter
• Nov 6th 2012, 07:23 AM
Deveno
Re: Factorization of Polynomials - Gauss's Lemma
the field of fractions of an integral domain D (often written Q(D), where Q is supposed to remind you of "quotient") is constructed EXACTLY like the rationals are constructed from the integers:

we start with DxD*, ordered pairs of elements of D, where the "second coordinate" is non-zero. this, as it turns out, is "too big", so we need to form an equivalence relation to get it to the right size. so we form DxD*/~ where ~ is the equivalence relation:

(we are going to think of the equivalence class [(a,b)] as the "fraction" a/b. just as with the rationals, a and b do not uniquely determine a rational number: for example in the rational numbers we have 1/2 = 2/4 = 3/6 = .....etc. in in our field of fractions of D, it is easy to check that:

[(a,b)] = [(ac,bc)] (we can "cancel a common factor")).

it is straight-forward to check this IS an equivalence relation, and i urge you to do so.

right now, all we have is a set, and the whole point of this is to get a field that somehow "extends" D (i will explain what this means more precisely in a bit). so we need some binary operations to give it some structure. we define:

[(a,b)] + [(c,d)] = [(ad+bc,bd)] (note that: [(a,1)] + [(c,1)] = [(a+c,1)] this will be useful, later).

[(a,b)]*[(c,d)] = [(ac,bd)].

it is tedious to check that these are well-defined (they depend on only the equivalence class, not the particular representative (a,b)), and satisfy the ring axioms, but it is worth doing at least ONCE in your life. once is probably enough.

claim 1: Q(D) so defined is a field.
claim 2: Q(D) contains a sub-ring isomorphic to D (this is what i meant by Q(D) extends Q).

to prove claim 1, it suffices to show that if [(a,b)] is in Q(D)*, [(a,b)] is a unit.

if [(a,b)] ≠ 0, then a ≠ 0 (hopefully, if you prove Q(D) is a ring, you will discover its additive identity is [(0,b)] (which is equal to [(0,c)] for any non-zero c, because....?)).

so, [(b,a)] is in Q(D)*, and: [(a,b)]*[(b,a)] = [(ab,ba)] = [(ab,ab)] (because D is an integral domain, so it is commutative)

= [(1,1)] because 1(ab) = (ab)1. you'll need to show in your proof that Q(D) is a ring that the equivalence class [(b,b)] = [(1,1)] is the unity (multiplicative identity) of Q(D).

thus U(Q(D)) = Q(D)*, so Q(D) is a field. (all we are "really saying" is that the inverse of a/b is b/a).

claim 2 proof: left to you- show d --> [(d,1)] is an injective ring homomorphism.

having established this, it is customary to write a/b instead of [(a,b)], and certainly "looks neater".