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 $\Q(\theta)$.

 
FavoriteLoadingAdd 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
\[(1+\theta)^{-1}=\frac{1}{4}(\theta^2-\theta+10).\]


FavoriteLoadingAdd 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Your email address will not be published.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 *

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