# Polynomial which is Irreducible over Integers is Irreducible over Rationals

## Theorem

Let $\Z \sqbrk X$ be the ring of polynomial forms over the integers in the indeterminate $X$.

Let $\Q \sqbrk X$ be the ring of polynomial forms over the field of rational numbers in the indeterminate $X$.

Let $\map f X \in \Z \sqbrk X$ be irreducible in $\Z \sqbrk X$.

Then $\map f X$ is also irreducible in $\Q \sqbrk X$.

## Proof

Aiming for a contradiction, suppose $\map f X = \map g X \, \map h X$ for some $\map g X, \map h X \in \Q \sqbrk X$.

$\map f X \in \Z \sqbrk X$

and so by definition has coefficients all of which are integers.

But from Factors of Polynomial with Integer Coefficients have Integer Coefficients it follows that $\map f X$ can be expressed as:

$\map f X = \map {g'} X \, \map {h'} X$

where both $\map {g'} X$ and $\map {h'} X$ are elements of $\Q \sqbrk X$ which have coefficients all of which are integers.

That is:

$\map {g'} X \in \Z \sqbrk X$

and:

$\map {h'} X \in \Z \sqbrk X$

This contradicts the statement that $\map f X$ is irreducible in $\Z \sqbrk X$.

Hence the result by Proof by Contradiction.

$\blacksquare$