Factors of Polynomial with Integer Coefficients have Integer Coefficients

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Theorem

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

Let $\map h X \in \Q \sqbrk X$ have coefficients all of which are integers.


Let it be possible to express $\map h X$ as:

$\map h X = \map f X \, \map g X$

where $\map f X, \map g X \in \Q \sqbrk X$.


Then it is also possible to express $\map h X$ as:

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

where:

$\map {f'} X, \map {g'} X \in \Q \sqbrk X$
the coefficients of $\map {f'} X$ and $\map {g'} X$ are all integers
$\map {f'} X = a \map f X$ and $\map {g'} X = b \map f X$, for $a, b \in \Q$.


Proof

Let $\cont h$ denote the content of $\map h X$.

From Polynomial has Integer Coefficients iff Content is Integer:

$\cont h \in \Z$

Let $\map h X = \map f X \, \map g X$ as suggested.

Then from Rational Polynomial is Content Times Primitive Polynomial:

\(\ds \map h X\) \(=\) \(\ds \cont f \cont g \cdot \map {f'} X \, \map {g'} X\) Rational Polynomial is Content Times Primitive Polynomial
\(\ds \) \(=\) \(\ds \cont h \cdot \map {f'} X \, \map {g'} X\) Content of Rational Polynomial is Multiplicative


From the above, $\map {f'} X$ and $\map {g'} X$ are primitive.

Hence by definition:

$\cont {f'} = \cont {g'} = 1$

From Polynomial has Integer Coefficients iff Content is Integer, both $\map {f'} X$ and $\map {g'} X$ have coefficients which are all integers.

We also have by definition of content that $\cont f$ and $\cont g$ are rational numbers.

The result follows.

$\blacksquare$


Sources

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