Polynomial over Field is Reducible iff Scalar Multiple is Reducible
Theorem
Let $K$ be a field.
Let $K \sqbrk X$ be the ring of polynomial forms over $K$.
Let $P \in K \sqbrk X$.
Let $\lambda \in K \setminus \set 0$.
Then $P$ is irreducible in $K \sqbrk X$ if and only if $\lambda P$ is also irreducible in $K \sqbrk X$.
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Proof
Necessary Condition
Let $P$ be irreducible.
Suppose further that $ \lambda P$ has a non-trivial factorization:
- $\lambda P = Q_1 Q_2$
that is, such that $Q_1$ and $Q_2$ are not units of $K \sqbrk X$.
By Units of Ring of Polynomial Forms over Field it follows that $\deg Q_1 \ge 1$ and $\deg Q_2 \ge 1$.
Let $Q_1' = \lambda^{-1} Q_1$.
This implies that:
- $P = Q_1' Q_2$
with $\deg Q_1' = \deg Q_1 \ge 1$.
But this is a non-trivial factorization of $P$ in $K \sqbrk X$.
This contradicts our supposition that $P$ is irreducible.
Therefore $\lambda P$ has no non-trivial factorization, that is, $\lambda P$ is irreducible.
$\Box$
Sufficient Condition
Let $\lambda P$ be irreducible.
Let $Q = \lambda P$.
From the necessary condition, we know that any scalar multiple of $Q$ is irreducible.
In particular:
- $\lambda^{-1} Q = \lambda^{-1} \lambda P = P$
is irreducible, the required result.
$\blacksquare$