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I am reading the book abstract Algebra and the book claims that if $p(x)$ is irreducible polynomial over a field $F$ and $g(x)$ is polynomial of smaller degree then $\gcd(p,g)$ is invertible.

for example: $F=\mathbb{Q}$, $p(x)=x^3-2$ and $g(x)=x+1$ $\implies$ $\gcd(p,g)=1$

Why is this true ?

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up vote 2 down vote accepted

The gcd of $p$ and $g$ must divide both $p$ and $g$. Because it must divide $g$, its degree is at most the degree of $g$, hence it is of degree strictly less than $p$.

Since $p$ is irreducible, its only divisors are units and associates of $p$; the associates of $p$ have the same degree as $p$. Since $\gcd(p,g)$ is a divisor of $p$ and is of degree strictly smaller than $p$, $\gcd(p,g)$ must be a unit.

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Oh right! thank you very much – Belgi Mar 28 '12 at 18:39

Hint $\:$ The only divisors of an irreducible $p(x)\in F[x]$ of smaller degree are "constants" $0\ne c\in F$ (else $p$ would be reducible). But elements $c\ne 0$ in a field $F$ are invertible. Thus since $\gcd(p,g)$ is a divisor of $p$ of smaller degree $(\le \deg\ g < \deg\ p),\:$ it is a constant $0\ne c\in F,\:$ hence invertible.

This is a polynomial analog of the well-known fact that, for integers, if $d$ is a divisor of a prime $p$ that is strictly smaller $|d|<p$, then $|d| = 1$, so $d$ is invertible, hence $\gcd(d,p) = 1$ is invertible.

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By definition, $\gcd(p,g)$ divides $p$. If $p$ is irreducible, that means the gcd can only be (up to a constant) $p$ or $1$. But since $g$ has lower degree, it's not $p$.

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