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!


1 Answer 1


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, 2013 at 17:36
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    $\begingroup$ $Gal(\Bbb{Q}(\zeta_{11})/\Bbb{Q})$ 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, 2013 at 19:26
  • $\begingroup$ Thanks! Oh by the way, how to show f remains irreducible in K? $\endgroup$
    – Jing Zhang
    Nov 19, 2013 at 3:45
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    $\begingroup$ $\zeta_{11}$ 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, 2013 at 6:15
  • $\begingroup$ As Giorgio Mossa explained, the field $K$ consists of the constructible numbers. How did I miss that? $\endgroup$ Dec 7, 2013 at 9:09

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