Irreducible Polynomial $x^3+9x+6$ and Inverse Element in Field Extension

Problems and Solutions in Field Theory in Abstract Algebra

Problem 334

Prove that the polynomial
\[f(x)=x^3+9x+6\] is irreducible over the field of rational numbers $\Q$.
Let $\theta$ be a root of $f(x)$.
Then find the inverse of $1+\theta$ in the field $\Q(\theta)$.

 
LoadingAdd to solve later

Sponsored Links

Proof.

Note that $f(x)$ is a monic polynomial and the prime number $3$ divides all non-leading coefficients of $f(x)$. Also the constant term $6$ of $f(x)$ is not divisible by $3^2$. Hence by Eisenstein’s criterion, the polynomial $f(x)$ is irreducible over $\Q$.

We divide the polynomial $f(x)$ by $x+1$ and obtain
\[x^3+9x+6=(x+1)(x^2-x+10)-4\] by long division.

Then it follows that in the field $\Q(\theta) \cong \Q[x]/(f(x))$ (note that $f(x)$ is the minimal polynomial of $\theta$), we have
\[0=(\theta+1)(\theta^2-\theta+10)-4,\] and hence this yields that we have the inverse
\[(1+\theta)^{-1}=\frac{1}{4}(\theta^2-\theta+10).\]


LoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Field Theory
Problems and Solutions in Field Theory in Abstract Algebra
Explicit Field Isomorphism of Finite Fields

(a) Let $f_1(x)$ and $f_2(x)$ be irreducible polynomials over a finite field $\F_p$, where $p$ is a prime number. Suppose...

Close