Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
add comment

3 Answers

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.

share|improve this answer
    
Oh right! thank you very much –  Belgi Mar 28 '12 at 18:39
add comment

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.

share|improve this answer
add comment

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 $p$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.