Product of Primitive Polynomials

Theorem

Let $\Q \left[{X}\right]$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.

Let $f \left({X}\right), g \left({X}\right) \in \Q \left[{X}\right]$ be primitive polynomials.

Then the product of $f$ and $g$ is also a primitive polynomial.

Proof

Let $f$ and $g$ be as follows:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle f\)	\(=\)	\(\displaystyle \sum_{k \in \Z} a_k X^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle g\)	\(=\)	\(\displaystyle \sum_{k \in \Z} b_k X^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

From the definition of primitive polynomial, the coefficients of $f$ and $g$ are all integers.

From the definition of polynomial product:

$\displaystyle f g = \sum_{k \in \Z} c_k \mathbf X^k$

where:

$\displaystyle c_k = \sum_{\substack{p + q = k \\ p, q \in \Z}} a_p b_q$

it is clear that the coefficients of $f g$ are also all integers.

Suppose $f g$ is not primitive.

Then its coefficients must have a GCD greater than $1$.

Therefore there exists some prime $p$ which divides all the coefficients of $fg$.

Now $p$ can not divide all the coefficients of either $f$ or $g$, because they are primitive polynomials.

So:

• Let $i$ be the smallest integer such that $p$ does not divide $a_i$;
• Let $j$ be the smallest integer such that $p$ does not divide $b_j$.

Consider the coefficient $c_{i+j}$ of $f g$:

$\displaystyle c_{i+j} = \sum_{k=0}^{i+j} a_k b_{i+j-k} = a_0 b_{i+j} + a_1 b_{i+j-1} + \cdots + a_i b_j + \cdots + a_{i+j-1}b_1 + a_{i+j} b_0$

From the assumption, it follows that $p \backslash c_{i+j}$, and so:

$\displaystyle p \backslash \sum_{k=0}^{i+j} a_k b_{i+j-k}$

Also, from the choice of $i$ and $j$, we have:

$p \backslash a_m$ whenever $m < i$;

$p \backslash b_n$ whenever $n < j$.

Now all the terms, except for $a_i b_j$, of $\displaystyle \sum_{k=0}^{i+j} a_k b_{i+j-k}$ contain a factor either from $\left\{{a_0, a_1, \ldots, a_{i-1}}\right\}$ or $\left\{{b_0, b_1, \ldots, b_{j-1}}\right\}$.

It follows that we have:

$p \backslash \left({\displaystyle \sum_{k=0}^{i-1} a_k b_{i+j-k} + \sum_{k=i+1}^{i+j} a_k b_{i+j-k}}\right)$

But then, as also $p \backslash c_{i+j}$, it follows that $p \backslash a_i b_j$ as well.

From Euclid's Lemma for Prime Divisors, $p$ has to divide one or the other.

This contradicts the definition of $i$ and $j$. So $i$ and $j$ cannot both exist.

It follows that $p$ divides at least one of $f$ and $g$, one of which is therefore not primitive.

From this contradiction, we conclude that $f g$ must be primitive.

Hence the result.

$\blacksquare$

Sources

• C.R.J. Clapham: Introduction to Abstract Algebra (1969)... (previous)... (next): $\S 6.31$: Theorem $60$