Calculate $[K(\alpha^2):K]$

Question

Let $K\subseteq F$ be a field extension. $K(\alpha)/\ K$ an extension of degree $4$ where $\alpha^2$ is not a root of $m_{(\alpha,K)}(x)$. I am asked to calculate $[K(\alpha^2):K]$

Using the towers formula $$[K(\alpha^2):K]=[K(\alpha^2):K(\alpha)][K(\alpha):K]$$

And we already know that $[K(\alpha):K]=4$

How do I find $[K(\alpha^2):K(\alpha)]$?

Answer 1

The correct tower formula is $$4=[K(\alpha):K]=[K(\alpha):K(\alpha^2)][K(\alpha^2):K]$$

Since $[K(\alpha):K(\alpha^2)]\le2$, we have $[K(\alpha^2):K]\ge 2$ and so there are two possibilities:

• $[K(\alpha^2):K]=2$: this happens iff $K(\alpha^2)\ne K(\alpha)$. For instance, for $K=\mathbb Q$ and $\alpha$ a root of $x^4-2$. Then $\alpha^2$ is a root of $x^2-2$ and is not a root of the minimal polynomial of $\alpha$.

• $[K(\alpha^2):K]=4$: this happens iff $K(\alpha^2)=K(\alpha)$. For instance, for $K=\mathbb Q$ and $\alpha$ a root of $x^4-2x-2$. Then $\alpha^2$ is a root of $x^4 - 4 x^2 - 4 x + 4$, which is irreducible, and so $\alpha^2$ is not a root of the minimal polynomial of $\alpha$.

$x^4-2x-2$ was chosen because then clearly $\alpha$ is a polynomial in $\alpha^2$ and so $K(\alpha^2)=K(\alpha)$. It is irreducible by Eisenstein's criterion with $p=2$.

$x^4 - 4 x^2 - 4 x + 4$ is the characteristic polynomial of the map $x \mapsto \alpha^2 x$ and so must be irreducible since $\alpha^2$ has degree $4$.

Answer 2

Let $f(X)\in K[X]$ (resp. $g(X$)) be the minimal polynomial of $\alpha$ (resp. $\alpha^2$) over $K$. Because of the degree formula in a tower, $[K(\alpha^2):K] =1,2$ or $4$. The value $1$ is excluded because it would mean that $\alpha^2\in K$, hence $[K(\alpha):K] =2$, contrary to the hypothesis.

Suppose $[K(\alpha^2):K]=4$, so that $K(\alpha)=K(\alpha^2)$. This would imply that the $K$-homomorphism $K(\alpha)\to K(\alpha^2)$ determined by $\alpha \to \alpha^2$ is an automorphism, in other words, the homomorphism $K[X]\to K[X]$ determined by $X\to X^2$ transforms the principal ideal $(f(X))$ onto $(g(X))$: contradiction with the hypothesis that $\alpha^2$ is not a root of $f(X)$. In conclusion, $[K(\alpha^2):K]=2$.