Gauss's Lemma on Primitive Rational Polynomials
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:
| \(\ds f\) | \(=\) | \(\ds \sum_{k \mathop \in \Z} a_k X^k\) | ||||||||||||
| \(\ds g\) | \(=\) | \(\ds \sum_{k \mathop \in \Z} b_k X^k\) |
From the definition of primitive polynomial, the coefficients of $f$ and $g$ are all integers.
From the definition of polynomial product:
- $\ds f g = \sum_{k \mathop \in \Z} c_k \mathbf X^k$
where:
- $\ds c_k = \sum_{\substack {p \mathop + q \mathop = k \\ p, q \mathop \in \Z} } a_p b_q$
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 $f g$:
- $\ds c_{i + j} = \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k} = a_0 b_{i + j} + a_1 b_{i + j - 1} + \cdots + a_i b_j + \cdots + a_{i + j - 1} b_1 + a_{i + j} b_0$
From the assumption, it follows that $p \divides c_{i + j}$, and so:
- $\ds p \divides \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k}$
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 $\ds \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k}$, 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:
- $p \divides \paren {\ds \sum_{k \mathop = 0}^{i - 1} a_k b_{i + j - k} + \sum_{k \mathop = i + 1}^{i + j} a_k b_{i + j - k} }$
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 $d$.
From Quotient Epimorphism from Integers by Principal Ideal, let $\pi : \Z \to \Z / p \Z$ be the quotient 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.