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Polynomial Rings

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Prove that if f(x) and g(x) are polynomials with rational co-efficients whose product f(x)g(x) has integer co-efficients, then the product of any co-efficient of g(x) with any coefficient of f(x) is an integer.

My initial thoughts on this are that the exercise seems to be set up for an application of Gauss Lemma since we have that Z is a UFD with field of fractions Q and further we have \(\displaystyle p(x) \in Z[x] \) where p(x) = f(x)g(x) and \(\displaystyle f(x), g(x) \in Q[x] \).

Thus we apply Gauss Lemma (see attached) so

p(x) = (rf(x))(sg(x))

where \(\displaystyle rf(x), sg(x) \in Z[x] \)

But .... where to from here ... can someone please help me advance from here ...

Peter




[This problem has also been posted on MHF]
 
Last edited:

TheBigBadBen

Active member
May 12, 2013
84
Prove that if f(x) and g(x) are polynomials with rational co-efficients whose product f(x)g(x) has integer co-efficients, then the product of any co-efficient of g(x) with any coefficient of f(x) is an integer.

My initial thoughts on this are that the exercise seems to be set up for an application of Gauss Lemma since we have that Z is a UFD with field of fractions Q and further we have \(\displaystyle p(x) \in Z[x] \) where p(x) = f(x)g(x) and \(\displaystyle f(x), g(x) \in Q[x] \).

Thus we apply Gauss Lemma (see attached) so

p(x) = (rf(x))(sg(x))

where \(\displaystyle rf(x), sg(x) \in Z[x] \)

But .... where to from here ... can someone please help me advance from here ...
The one thing I would think of is that given

\(\displaystyle p(x) \in \mathbb{Z}[x] \) where \(\displaystyle p(x)=f(x)g(x)\) and \(\displaystyle f(x), g(x) \in \mathbb{Q}[x]-\mathbb{Z}[x] \), there exist some \(\displaystyle r,s \in \mathbb{Z}\) so that \(\displaystyle rf(x), sg(x) \in \mathbb{Z}[x] \) are both primitive polynomials. Gauss's lemma tells us that \(\displaystyle r s \cdot p(x) = rf(x) \cdot sg(x)\) must be a primitive polynomial itself. It follows that \(\displaystyle r,s=\pm 1\), which means that we have deduced that f and g must have integer coefficients after all.

Clearly, if what I said holds, the statement would follow. Either there is some flaw in my logic, or Gauss's Lemma is too powerful a tool for this problem...
 

TheBigBadBen

Active member
May 12, 2013
84
The one thing I would think of is that given

\(\displaystyle p(x) \in \mathbb{Z}[x] \) where \(\displaystyle p(x)=f(x)g(x)\) and \(\displaystyle f(x), g(x) \in \mathbb{Q}[x]-\mathbb{Z}[x] \), there exist some \(\displaystyle r,s \in \mathbb{Z}\) so that \(\displaystyle rf(x), sg(x) \in \mathbb{Z}[x]\) are both primitive polynomials. Gauss's lemma tells us that \(\displaystyle r s \cdot p(x) = rf(x) \cdot sg(x)\) must be a primitive polynomial itself. It follows that \(\displaystyle r,s=\pm 1\), which means that we have deduced that f and g must have integer coefficients after all.

Clearly, if what I said holds, the statement would follow. Either there is some flaw in my logic, or Gauss's Lemma is too powerful a tool for this problem...
I made a mistake here: r and s are not necessarily integers. Since $r\,f(x)$ and $s\,g(x)$ are primitive, we can only guarantee that $r,s \in \mathbb{Q}$. This is still, however, sufficient; following the proof, we still find that $r\,s=\pm1$, which is enough to tell us that the product of a coefficient from one and a coefficient from the other is an integer.