Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Are all degree $2$ field extensions Galois?

I know that this is true over the rationals. But is it true in general?

share|cite|improve this question
up vote 11 down vote accepted

No - for example, let $K=\mathbb{F}_2(T)$, and let $L=\mathbb{F}_2(\sqrt{T})$. We have that $[L:K]=2$, but the extension $L/K$ is not separable, and therefore not Galois.

However, $[L:K]=2$ implies $L/K$ is Galois when $K$ (and hence $L$) is of any characteristic other than 2. This is because we can take the minimal polynomial $f$ of a primitive element $\alpha\in L$ (i.e., an element such that $L=K(\alpha)$), which will be irreducible of degree 2, and use the quadratic formula (which works because the characteristic isn't 2) to show that the other root of $f$ must be in $L$.

share|cite|improve this answer

It is true over any field of characteristic different from $2$.

If $F$ has characteristic different from $2$, and $K$ is of degree $2$ over $F$, then let $\alpha\in K$, $\alpha\notin F$. Then $K=F(\alpha)$, since $2=[K:F]=[K:F(\alpha)][F(\alpha):F]$, and $[F(\alpha):F]\gt 1$. Let $p(x)$ be the minimal polynomial of $\alpha$ over $K$. Then $p(x) = x^2 + rx+t$ for some $r,t\in F$, and $\alpha = \frac{-r+\sqrt{r^2-4t}}{2}$ or $\alpha=\frac{-r-\sqrt{r^2+4t}}{2}$ (since the characteristic is not $2$). Moreover, the polynomial is irreducible and separable, and $\sqrt{r^2-4t}\notin F$. So $K = F(\sqrt{r^2-4t})$ and $K$ is a splitting field of an irreducible separable polynomial (namely, $x^2 - (r^2-4t)$), hence is Galois over $F$.

If the characteristic is $2$, then the result is true for perfect fields, but not in general, as the examples by Zev Chonoles and Giovanni De Gaetano show.

share|cite|improve this answer
While you were not looking, $F$ has tried to put the blame on characteristic $3$ (line 2). – Georges Elencwajg May 11 '11 at 20:14
@elgeorges: Sigh; thanks for pointing it out! – Arturo Magidin May 11 '11 at 20:16

No, It's not true in general.

Here is a counterexample. Let $F$ a field of characteristic 2, and consider the field $k$ of rational functions over $F$, i.e. $k=F(T)$. Now consider the extension of degree 2 obtained adding the roots of the polynomial $x^2-T=0$. This extension is not separable (remember that $F$ so $k$ has characheristic 2) hence not a Galois extension (this is a result of Artin).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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