This is one of the exercises during my reading of Ian Stewart's Galois Theory. Whether the following statement is true:

If $K$ is a field of characteristic zero in which every element is a perfect square, then the Galois group of any irreducible $n$-th degree polynomial over $K$ is isomorphic to $A_n$.

I think it might not be true but the examples of such fields in my knowledge are very few. Any thoughts? Thanks!


I proffer the following counterexample.

First we need to construct such a field $K$. Let $K_0=\Bbb{Q}$, and recursively define fields $K_n, n\in\Bbb{N}$ by the recipe that $K_{\ell+1}$ is gotten from $K_\ell$ by adjoining the square roots of all the elements of $K_{\ell}$ to it. Note that we can do this construction inside $\Bbb{C}$. Then we let $$ K=\bigcup_{n\in\Bbb{N}}K_n\subset \Bbb{C}. $$ Then any element of $K$ belongs to some field $K_\ell$, and thus has a square root in $K_{\ell+1}$, hence also in $K$.

If we pick any element $z\in K$, I claim that $[\Bbb{Q}(z):\Bbb{Q}]$ is a power of two. Clearly $z\in K_\ell$ for some $\ell$. W.l.o.g. we can assume that $\ell$ is the smallest natural number with that property. Then there exists a finite set of elements $z_1,z_2,\ldots,z_m\in K_{\ell-1}$ such that $z\in K_{\ell-1}(\sqrt{z_1},\sqrt{z_2},\ldots,\sqrt{z_m})$. This means that we can write $z$ in terms of finitely many element of $K_{\ell-1}$ and their square roots. By induction hypothesis all those elements are algebraic of degree a power of two over $\Bbb{Q}$. Thus so is their compositum. Adjoining the square roots of $z_i, i=1,\ldots,m,$ won't change that fact so we see that $\Bbb{Q}(z)$ is contained in a field of degree a power of two settling the claim.

Let's then consider the eleventh cyclotomic polynomial $$ \phi_{11}(x)=x^{10}+x^9+x^8+\cdots+x^2+x+1. $$ By the known theory of cyclotomic fields we know that $\phi_{11}(x)$ factors into a product of two quintic factors over $F=\Bbb{Q}(\sqrt{-11})$, the only quadratic subfield of the eleventh cyclotomic field. Let $f(x)$ be one of those quintic factors.

By the above observation $f(x)$ remains irreducible over $K$. If $L=K(e^{2\pi i/11})$ is the splitting field of $f(x)$, we easily see that $Gal(L/K)$ is isomorphic to $Gal(\Bbb{Q}(e^{2\pi i/11})/\Bbb{Q}(\sqrt{-11})\cong C_5$.

But $C_5$ is a proper subgroup of $A_5$.

  • $\begingroup$ Thanks! I think this is a great example! The reason why $\phi_{11}(x)$ factors over $F$ into quintics is because the Galois group $Gal(\mathbb{Q}(\zeta_{11})/\mathbb{Q})$ is $C_{10}$ which is generated by some odd permutation thus the discriminant is not a square in $\mathbb{Q}$. I could only vision $F$ as $\mathbb{Q}(\sqrt{disc(f)})$. Is there any quick way of computing the discriminant? Thanks! $\endgroup$
    – Jing Zhang
    Nov 18 '13 at 17:36
  • 1
    $\begingroup$ $Gal(\Bbb{Q}(\zeta_{11})/\Bbb{Q})$ has a unique subgroup of index two, so there is a unique quadratic subfield. The Gauss sum $$S=\sum_{k=0}^{10}\zeta_{11}^{k^2}$$ has the property $S^2=-11$, so we know which quadratic field it is. The real subfield $$L=\Bbb{Q}(\zeta_{11})\cap\Bbb{R}=\Bbb{Q}(\cos(\frac{2\pi}{11}))$$ is the other intermediate field. Its quintic, and we could use the minimal polynomial of $\cos(\frac{2\pi}{11})$ instead of that factor of $\phi_{11}(x)$. $\endgroup$ Nov 18 '13 at 19:26
  • $\begingroup$ Thanks! Oh by the way, how to show f remains irreducible in K? $\endgroup$
    – Jing Zhang
    Nov 19 '13 at 3:45
  • 1
    $\begingroup$ $\zeta_{11}$ is of degree ten over $\Bbb{Q}$ and of degree five over $\Bbb{Q}(\sqrt{-11})\subset K$. Ten is not a power of two, so $\zeta_{11}\notin K$. But $[K(\zeta_{11}):K]$ has to be a factor of $5$, five is a prime. $\endgroup$ Nov 19 '13 at 6:15
  • $\begingroup$ As Giorgio Mossa explained, the field $K$ consists of the constructible numbers. How did I miss that? $\endgroup$ Dec 7 '13 at 9:09

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.