Let $F=\Rats(\sqrt{2})$ . Then the extension $F/\Rats$ is normal because $F$ is clearly the splitting field of the polynomial $f(x)=x^2-2$ . Furthermore $F/\Rats$ is a Galois extension with $\Gal(F/\Rats)\cong \Ints/2\Ints$ .

Now, let $2^{1/4}$ denote the positive real fourth root of $2$ and define $K=F(2^{1/4})$ . Then the extension $K/F$ is normal because $K$ is the splitting field of $k(x)=x^2-\sqrt{2}$ , and as before $K/F$ is a Galois extension with $\Gal(K/F)\cong \Ints/2\Ints$ .

However, the extension $K/\Rats$ is neither normal nor Galois. Indeed, the polynomial $g(x)=x^4-2$ has one root in $K$ (actually two), namely $2^{1/4}$ , and yet $g(x)$ does not split in $K$ into linear factors.

$$g(x)=x^4-2=(x^2-\sqrt{2})\cdot(x^2+\sqrt{2})=(x-2^{1/4})\cdot(x+2^{1/4})\cdot(x^2+\sqrt{2})$$

The Galois closure of $K$ over $\Rats$ is $L=\Rats(2^{1/4},i)$ .