Product of Rational Polynomials
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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 $\map f X, \map g X \in \Q \sqbrk X$.
Using Rational Polynomial is Content Times Primitive Polynomial, let these be expressed as:
- $\map f X = \cont f \cdot \map {f^*} X$
- $\map g X = \cont g \cdot \map {g^*} X$
where:
Let $\map h X = \map f X \map g X$ be the product of $f$ and $g$.
Then:
- $\map {h^*} X = \map {f^*} X \map {g^*} X$
Proof
From Rational Polynomial is Content Times Primitive Polynomial:
- $\cont h \cdot \map {h^*} X = \cont f \cont g \cdot \map {f^*} X \map {g^*} X$
and this expression is unique.
By Gauss's Lemma on Primitive Rational Polynomials we have that $\map {f^*} X \map {g^*} X$ is primitive.
From Content of Rational Polynomial is Multiplicative:
- $\cont f \cont g = \cont f \cont g > 0$.
The result follows.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Theorem $62$