Gauss's Lemma on Primitive Rational Polynomials

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Theorem

Let $\Q$ be the field of rational numbers.

Let $\Q \sqbrk X$ be the ring of polynomials over $\Q$ in one indeterminate $X$.

Let $\map f X, \map g X \in \Q \sqbrk X$ be primitive polynomials.


Then their product $f g$ is also a primitive polynomial.


Proof 1

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

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

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

From the definition of polynomial product:

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

where:

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

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


Aiming for a contradiction, 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 divides 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 $f g$:

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

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

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

where $\divides$ denotes divisibility.

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

$p \divides a_m$ whenever $m < i$
$p \divides b_n$ whenever $n < j$.

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

It follows that we have:

$p \divides \paren {\displaystyle \sum_{k \mathop = 0}^{i - 1} a_k b_{i + j - k} + \sum_{k \mathop = i + 1}^{i + j} a_k b_{i + j - k} }$

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

$\blacksquare$


Proof 2

Recall Polynomial has Integer Coefficients iff Content is Integer.

By hypothesis $f$ and $g$ have content $1 \in \Z$.

Therefore:

$f, g \in \Z \sqbrk X$

Aiming for a contradiction, suppose that $f g$ is not primitive, say:

$\cont {f g} = d \ne 1$

By the Fundamental Theorem of Arithmetic we can choose a prime $p$ dividing $d$.

Let $\pi : \Z \to \Z / p \Z$ be the canonical epimorphism to the ring of integers modulo $p$.

From Ring of Integers Modulo Prime is Field, $\Z / p \Z$ is a field.


Let $\Pi : \Z \sqbrk X \to \paren {\Z / p \Z} \sqbrk X$ be the induced homomorphism of the polynomial rings.

By construction, $p$ divides each coefficient of $f g$, so:

$\map \Pi {f g} = \map \Pi f \, \map \Pi g = 0$

From Polynomial Forms over Field form Integral Domain, $\paren {\Z / p \Z} \sqbrk X$ is an integral domain.

Thus:

$\map \Pi f = 0$ or $\map \Pi g = 0$

After possibly exchanging $f$ and $g$, we may assume that $\map \Pi f = 0$.


Now by Kernel of Induced Homomorphism of Polynomial Forms, if:

$f = a_0 + a_1 X + \cdots + a_n X^n$

we must have:

$\map \pi {a_i} = 0$

for $i = 0, \ldots, n$.

That is:

$a_i \in p \Z$

for $i = 0, \ldots, n$.

But this says precisely that $p$ divides each $a_i$, $i = 0, \ldots, n$.

Therefore $p$ divides the content of $f$, a contradiction.

Hence our assumption that $f g$ is not primitive was invalid.

The result follows by Proof by Contradiction.

$\blacksquare$


Also see


Source of Name

This entry was named for Carl Friedrich Gauss.