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Question: Let $f(x)$ be a non zero polynomial of degree at most $3$ in $F_p[x]$. Prove that if $f(x) = g(x)h(x)$, where neither $g(x)$ nor $h(x)$ is a unit, then one of $g(x)$ or $h(x)$ must have degree $1$.

Case 1: $deg(f(x)) = 1$

$deg(g(x)) = 1$ and $deg(h(x)) = 0$ or vice versa. Wouldn't that mean that the $deg 0$ polynomial would have to be a unit since we are working in $F_p[x]$?

Case 2: $deg(f(x)) = 2$

$$deg(g(x)) = deg(h(x)) = 1.$$

Or... $g = 2$, $h = 0$. I have the same question as Case $1$.

Case 3: $deg(f(x)) = 3$

etc.

vallev
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sbxsgxo
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    The is silly. The units are precisely the polynomials of degree $0$, and the degree of the product equals the sum of the degrees. If both $g$ and $h$ have positive degree and the degrees add up to no more than three... The hypotheses cannot be satisfied if $\deg(f)\leq 1$, so the implication is true by vacuity. – Arturo Magidin Sep 30 '24 at 01:20
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    Duplicate. Beware that it fails when the coefficient ring has zero divisors, e.g. here. $\ \ $ – Bill Dubuque Sep 30 '24 at 03:09

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