Cubic Polynomial $x^3-2$ is Irreducible Over the Field $\Q(i)$

Field theory problems and solution in abstract algebra

Problem 399

Prove that the cubic polynomial $x^3-2$ is irreducible over the field $\Q(i)$.

 
LoadingAdd to solve later

Proof.

Note that the polynomial $x^3-2$ is irreducible over $\Q$ by Eisenstein’s criterion (with prime $p=2$).
This implies that if $\alpha$ is any root of $x^3-2$, then the degree of the field extension $\Q(\alpha)$ over $\Q$ is $3$:
\[[\Q(\alpha) : \Q]=3. \tag{*}\]

Seeking a contradiction, assume that $x^3-2$ is reducible over $\Q(i)$.
Then $x^3-2$ has a root in $\Q(i)$ as it is a reducible degree $3$ polynomial. So let us call the root $\alpha \in \Q(i)$.

Then $\Q(\alpha)$ is a subfield of $\Q(i)$ and thus we have
\[2=[\Q(i) :\Q]=[\Q(i): \Q(\alpha)][\Q(\alpha):\Q]\geq 3\] by (*). Hence we have reached a contradiction.
As a result, $x^3-2$ is irreducible over $\Q(i)$.


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Your email address will not be published.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
Prove that any Algebraic Closed Field is Infinite

Prove that any algebraic closed field is infinite.  

Close