Gauss's Lemma on Irreducible Polynomials

Theorem
Let $\Z$ be the ring of integers.

Let $\Z \sqbrk X$ be the ring of polynomials over $\Z$.

Let $h \in \Z \sqbrk X$ be a polynomial.


 * $(1): \quad h$ is irreducible in $\Q \sqbrk X$ and primitive
 * $(2): \quad h$ is irreducible in $\Z \sqbrk X$.

1 implies 2
Suppose first that $h$ is not irreducible in $\Z \sqbrk X$.

Let $\ds h = a_0 + a_1 X + \cdots + a_n X^n$.

If $\deg h = 0$, then the content of $h$ is:
 * $\cont h = \gcd \set {a_0} = \size {a_0}$

Since $h$ is primitive, we have $h = \pm 1$.

Now by Units of Ring of Polynomial Forms over Field, the units of $\Q \sqbrk X$ are the units of $\Q$.

Thus $h$ is a unit of $\Q \sqbrk X$.

Therefore $h$ is irreducible.

If $\deg h \ge 1$, then by Units of Ring of Polynomial Forms over Integral Domain, the units of $\Z \sqbrk X$ are the units of $\Z$.

Therefore $h$ is not a unit of $\Z \sqbrk X$.

Thus since $h$ is reducible, there is a non-trivial factorization $h = f g$ in $\Z \sqbrk X$, with $f$ and $g$ both not units.

If $\deg f = 0$, that is, $f \in \Z$, then $f$ divides each coefficient of $h$.

Since $h$ is primitive, this means that $f$ divides $\cont h = 1$.

But the divisors of $1$ are $\pm 1$, so $f = \pm 1$.

But then $f$ is a unit in $\Z \sqbrk X$, a contradiction.

Therefore $\deg f \ge 1$, so $f$ is a non-unit in $\Q \sqbrk X$.

Similarly, $g$ is a non-unit in $\Q \sqbrk X$.

Therefore $h = fg$ is a non-trivial factorization in $\Q \sqbrk X$.

2 implies 1
Suppose now that $h$ is not irreducible in $\Q \sqbrk X$.

That is, $h$ has a non-trivial factorization in $\Q \sqbrk X$.

Since the units of $\Q \sqbrk X$ are the units of $\Q$, this means that $h = f g$, with $f$ and $g$ both of positive degree.

Let $c_f$ and $c_g$ be the contents of $f$ and $g$ respectively.

Define $\tilde f = c_f^{-1} f$ and $\tilde g = c_g^{-1} g$.

By Content of Scalar Multiple, it follows that $\cont {\tilde f} = \cont {\tilde g} = 1$.

Moreover by Polynomial has Integer Coefficients iff Content is Integer we have $\tilde f, \tilde g \in \Z \sqbrk X$.

Now we have:
 * $\tilde f \tilde g = \dfrac {f g} {c_f c_g} = \dfrac h {c_f c_g}$

Taking the content, and using Content of Scalar Multiple we have:
 * $\cont {\tilde f \tilde g} = \dfrac 1 {c_f c_g} \cont h$

By Gauss's Lemma on Primitive Rational Polynomials we know that $\cont {\tilde f \tilde g} = 1$.

Moreover, by Irreducible Integer Polynomial is Primitive, $\cont h = 1$.

Therefore we must have $c_f c_g = 1$.

Thus we have a factorization in $\Z \sqbrk X$:
 * $\tilde f \tilde g = h$

This is a non-trivial factorization of $h$, as both $f$ and $g$ have positive degree.

Thus $h$ is not irreducible in $\Z \sqbrk X$.