Let $F =\mathbb F_{2}(u)$ be the field of rational functions over the prime field $\mathbb F_{2}$. Prove that $x^2-u$ is irreducible over $F$ and that it has a double root in a splitting field.

Attempted answer:

Since every polynomial has its splitting field, so does $x^2-u$. Denote its splitting field by $K =\mathbb F_{2}(\sqrt{u})$. Suppose that $x^2-u$ is reducible over F for proof by contradiction.

Then polynomial factors of $x^2-u$ has degree at most 1. Since degree of $x^2-u$ over F is equal to 1, [K:F] = 1, but $\sqrt{u} \notin F_{2}(u)$ = K, so K cannot equal F. Hence, contradiction. Therefore, $x^2-u$ is irreducible over F. Also consider the derivative of $x^2-u$, which is 0. Therefore by definition, $x^2-u$ has multiple roots.

I'm not really sure if this answer is right, can anyone give me some tips on how to do this problem? Any help is appreciated, thanks.


Suppose it is reducible, write $x^2-u=(x-a)(x-b)=x^2-(a+b)x+ab$.

This implies that $a=b$ since the characteristic is 2. Thus $a^2=u$. This is impossible since the degree of $a^2$ is even. Here the degree of a rational function in $u$ is the difference between the degree of the numerator and the denominator.

The root in the splitting field is multiple since the derivative of $x^2-u$ is zero.

  • $\begingroup$ Thank you for your answer, I understand everything besides why $a^2 = u$ is impossible because the degree is even. Could you elaborate on that? $\endgroup$ – James Blake May 19 '16 at 23:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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