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Let $f(x)\in \mathbb{Z}[x]$. The set content of $f(x)=a_{n}x^{n}+ \cdots +a_{0}$ is defined as the greatest common divisor of $a_{0}, ..., a_{n}$ and is denoted by cont$(f(x))$.

I want to show that cont$(f(x)g(x))=$cont$(f(x))$cont$(g(x))$. I saw a proof of it, but I have some doubts. This is:

We know that $$\text{mcd}\left( \frac{a}{\text{mcd}(a, b)}, \frac{b}{\text{mcd}(a, b)}\right)=1$$ without loss of generality let us cont$(f(x))=1$ and cont$(g(x))=1$ where:

\begin{align*} f(x) = & a_{n}x^{n}+ \cdots +a_{0} \\ g(x) = & b_{n}x^{n}+ \cdots +b_{0} \end{align*}

Supose that $f(x)g(x)\neq 1$ then there is no unity $p \in \mathbb{Z}$ that divides all the coefficients of $f(x)g(x)$. Well, it is not very clear to me why such a unit satisfies that.

Let $i$ the smallest integer such that $p \nmid a_{i}$ and $j$ the largest integer such that $p \nmid b_{j}$ and take the coeficient $x^{i+j}$ in $f(x)g(x)$, from where

\begin{equation*} f(x)g(x)=a_{0}b_{i+j}+a_{1}b_{i+j-1}+\cdots +a_{i-1}b_{j-1}+a_{i}b_{j-1}+\cdots +a_{i+j-1}b_{1}+a_{i+j}b_{0} \end{equation*}

In the summands to the left of $a_{i}b_{j}$ the factor $a_{k}$, with $k<i$, is divisible by $p$, and to the right of $a_{i}b_{j}$ the factor $b_{k}$ with $k<j$ is divisible by $p$. So $p \mid a_{i}b{j}$ i.e. $p \mid a_{i}$ or $p \mid b{j}$ but this contradicts our previous assumption. Therefore, it cannot be the case that $f(x)g(x)\neq 1$, then $f(x)g(x)= 1$ now cont$(f(x)g(x))=$cont$(f(x))$cont$(g(x))$.

Firstly, the proof is correct? and secondly, I am left with the doubt why the factors $a_{k}$ and $b_{k}$ are divisible by $p$, in addition to the doubt I mentioned before.

Any help is welcome!

Wrloord
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  • This form of Gauss's Lemma is already proved here many times, e.g. see the linked dupe (and its links). Compare your argument to these common proofs, and if any doubts remain then post questions on the exissting proofs (so they can be improved if need be), and if you get no answer then post a new question on specifically what you doubt, and why so. – Bill Dubuque Mar 30 '23 at 23:09
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Mar 30 '23 at 23:09

1 Answers1

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If there is some prime $p$ that divides each of the coefficients of $f(x)g(x)$, then we have $f(x)g(x) \equiv 0 \pmod p$. But $(\Bbb Z / p \Bbb Z)[x]$ is an integral domain (this can easily be seen by looking at the product of the leading coefficients of $f$ and $g$), so that means either $f(x) \equiv 0 \pmod p$ or $g(x) \equiv 0 \pmod p$, contradicting your hypothesis that $\operatorname{cont}(f) = \operatorname{cont}(g)=1$.

Robert Shore
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