Confused about Gauss's Lemma of Irreducibility

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There seems to be different forms of this lemma which conclude different things, yet they all seem to go through the same steps, so I figure that they must be related somehow.

I'll start with the form that I understand:

Considering a pair of primitive polynomials A( x ), B( x ), such that C( x ) = A( x ) B( x ), and presume that C( x ) is imprimitive, which means that there must be some prime number that divides all the coefficients. With respect to this unknown prime number, there must be a coefficient in both A( x ) & B( x ) which is the first as per the index to not be divisible by this prime number - since if there is no such first index, all the coefficients would be divisible by this number (call it j & k), and thus the polynomial would not be primitive. So with the first-non-prime-divisible indices, looking at the coefficient of C( x ) whose index is the sum of those first-not-divisible indices, there are 3 sets of terms that sum up to it.

Any coefficient of C( x ) is going to be the sum of 3 sets of terms which are products of coefficients of A( x ) & B( x ): the first set of terms is a single term which is the product of the j-th & k-th terms of A( x ) & B( x ), respectively, and since both of these terms are the first non-prime-divisible coefficients, this term is not divisible by the prime. The only sets of terms are terms which have the indices for one of the polynomials going up and the other going down from this first non-prime-divisible coefficient, but since all of those terms contain a coefficient that is divisible by the prime (i.e., because of the definition), the products are divisible, and thus the fact that there is only term that is non-divisible while all the others are divisible results in the whole sum being non-divisible, thus there exists a coefficient of C( x ) that is not divisible by the prime that it was supposed to have been divisible by, as a consequence of being imprimitive, therefore C( x ) must be imprimitive. Whew!

So moving on, in the Stewart book "Galois Theory", p. 39, there is a LEMMA 3.17

"Let f be a polynomial over Z that is irreducible over Z. Then f, considered as a polynomial over Q, is also irreducible over Q."

The book goes on, and seems to do just about the same busywork as what I have presented here, so it seems as though I should be able to set up a simply corollary based on my work here that concludes with this lemma, but I don't see it.
 

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  • #2
andrewkirk
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First note that your conclusion:
therefore C( x ) must be imprimitive.
is the wrong way around and should say 'primitive'. IOTW, the product of two primitive polynomials must be primitive.

Now let C[x] be a polynomial over Z and assume that it factorises to A[x]B[x] over Q. Any polynomial P[x] over Q can be converted to a polynomial over Z by multiplying by ##m_P##, the LCM of the denominators of coefficients in P[x]. So ##m_AA[x]## and ##m_BB[x]## are polynomials over Z. Further, they must be primitive since, if one of them, say ##m_AA[x]##, had all coeffs divisible by p>1, then ##m_A## could not be the lowest common multiple of the denominators of coeffs of A[x], as we could divide by p and still have a common multiple.

We multiply the equation:
$$A[x]B[x]=C[x]$$
by ##m_Am_B## to get:
$$(m_AA[x])(m_BB[x])=(m_Am_B)C[x]$$
Both factors on the LHS are primitive polynomials in Z[x]. So the RHS must also be primitive (using the theorem you wrote above). Since the coefficients of C[x] are integers, that means that ##m_Am_B## must equal 1, from which it follows that both A[x] and B[x] must be in Z[x]. Hence it is reducible over Z.

We have shown that a polynomial over Z that is reducible over Q is also reducible over Z.

The contrapositive of that is that a polynomial over Z that is irreducible over Z is also irreducible over Q.
 
  • #3
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Yes, I meant to say that C( x ) is primitive.

It seems the whole problem is that the text in Stewart is a bit nebulous in what was meant in his book. I found this:

https://web.ma.utexas.edu/users/lpbowen/m373k/notes20.pdf
that describes the lemma as referring to "a primitive polynomial that is reducible over Q". This extra attribute makes all the difference in the world, so I have completed this proof.
 
  • #5
Erland
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Any polynomial P[x] over Q can be converted to a polynomial over Z by multiplying by ##m_P##, the LCM of the denominators of coefficients in P[x]. So ##m_AA[x]## and ##m_BB[x]## are polynomials over Z. Further, they must be primitive since, if one of them, say ##m_AA[x]##, had all coeffs divisible by p>1, then ##m_A## could not be the lowest common multiple of the denominators of coeffs of A[x], as we could divide by p and still have a common multiple.
This part of the argument is wrong. It is not a priori excluded that the numerators of the coefficients of ##P[x]## (being ##A[x]## or ##B[x]## in this case) have a common prime factor. Then, ##m_PP[x]## needs not be primitive. For example, let ##P[x]=\frac23+\frac25x##. Then ##m_P=15## and ##m_PP[x]=10+6x##, which is not primitive.
 

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