- #1

swampwiz

- 476

- 41

https://cims.nyu.edu/~kiryl/Algebra/Section_3.10--Polynomials_Over_The_Rational_Field.pdf

Unfortunately, I am stuck on Lemma 3.10.1 that concludes that the product of a pair of primitive polynomials is itself primitive.

I understand about how there is some index in each polynomial at which the polynomial becomes imprimitive, and that the term c

_{j + k}is of the form:

c

_{j + k}= a

_{j}b

_{k}+ T

_{a}g

_{a}+ T

_{b}g

_{b}

where g

_{a}& g

_{b}are the Greatest Common Divisor of the terms below a

_{j}& b

_{k}, respectively (here, I use T

_{a}& T

_{b}to denote general numbers that are multiplied by g

_{a}& g

_{b}to get the terms for the summation of the products of a & b terms whose index is less than i & j, respectively)

So the proof follows that it presumes that c

_{j + k}has a prime divisor. Now, if I presume that this divisor is the same as g

_{a}& g

_{b}, then I get the contradiction that c

_{j + k}has no divisor. Fine. However if I only presume that the divisor is the same as g

_{a}, but not g

_{b}, I get

(here U is a general number like T)

c

_{j + k}= g

_{a}U = a

_{j}b

_{k}+ T

_{a}g

_{a}+ T

_{b}g

_{b}

a

_{j}b

_{k}= g

_{a}( U - T

_{a}) - g

_{b}T

_{b}

Now I don't know what prime factors a

_{j}or b

_{k}have, and it seems that I have no way of proving that the RHS of this doesn't coincidentally sum up to a number that happens to be a multiple of whatever prime factors a

_{j}or b

_{k}

What am I missing here?