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Learning for an upcoming exam on basic algebra I came across the following problem:

Test for $\alpha = \sqrt{2+\sqrt{2}}$ and $\alpha = \sqrt{1+\sqrt{3}}$ whether the field extension $\mathbb{Q}(\alpha)/\mathbb{Q} $ is normal.

I feel, that I should be able to solve it, but I am running against a wall here. Can you explain to me, how one would generally approach this kind of problem?

Best Regards

  • 1
    List the roots of the minimal polynomial of $\alpha$, and see if they are in $\mathbb{Q}(\alpha)$ ? – 2017-01-25

4 Answers 4

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Take the first one, $\alpha =\sqrt {2+\sqrt 2}$. Let's find it's minimal polynomial. We remark that $$\alpha^2=2+\sqrt 2\implies \left(\alpha^2-2\right)^2=2$$ so we have $$\alpha^4-4\alpha^2+2$$ That is effectively quadratic and we can solve to get the roots $$\pm \sqrt {2\pm \sqrt 2}$$.

Letting $\beta=\sqrt {2-\sqrt 2}$ we easily see that $$\alpha \beta = \sqrt 2\implies \beta =\frac {\alpha^2-2}{\alpha}\implies \beta\in \mathbb Q(\alpha)$$

The other instance should follow along similar lines.

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The minimal polynomial of $\alpha=\sqrt{2+\sqrt2}$ is $f(x)=x^4-4x^2+2=0$ and roots of $f(x)$ is $\sqrt{2+\sqrt2},-\sqrt{2+\sqrt2},\sqrt{2-\sqrt2},-\sqrt{2-\sqrt2} $ and all roots belong to $\mathbb Q(\alpha)$,then $\mathbb Q(\alpha)$ is normal.

While The minimal polynomial of $\alpha =\sqrt{1+\sqrt3}$ is $f(x)=x^4-2x^2-2=0$ and roots is $\sqrt{1+\sqrt3},-\sqrt{1+\sqrt3},i\sqrt{\sqrt3 -1},-i\sqrt{\sqrt3 -1}$, we note that $i\sqrt{\sqrt3 -1}\notin \mathbb Q(\alpha)$ then $\mathbb Q(\alpha)$ isnot normal

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Put

$$r=\sqrt{2+\sqrt2}\implies r^2-2=\sqrt2\implies r^4-4r^2+2=0$$

It's easy to see $\;p(x)=x^4-4x^2+2\in\Bbb Q[x]\;$ is irreducible, so it is the minimal polynomial of $\;r\;$ over the rationals.

Clearly also $\;-r\;$ is a root of $\;p(x)\;$ , and we can then divide with residue

$$p(x)=(x-r)(x+r)g(x)=\left(x^2-(2+\sqrt2)\right)(x^2-(2-\sqrt2))$$

so the extension is normal iff $\;\pm\sqrt{2-\sqrt2}\in\Bbb Q(r)\;$ . But

$$\sqrt{2-\sqrt2}=\frac{\sqrt2}r\;,\;\;\text{and}\;\;\sqrt2=r^2-2\in\Bbb Q(r)$$

so we indeed have $\;\Bbb Q(r)/\Bbb Q\;$ is normal.

You now try the other one.

Added for $\;r=\sqrt{1+\sqrt3}\;$ : we proceed as before

$$r^2-1=\sqrt3\implies r^4-2r^2-2=0\implies f(x)=x^4-2x^2-2\in\Bbb Q[x]$$

is the minimal polynomial, but this time

$$f(x)=(x-r)(x+r)g(x)=\left(x^2-(1+\sqrt3)\right)\left(x^2-(1-\sqrt3)\right)$$

Can you see what goes wrong now and why we have no normality here?

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Since $\mathbb{Q}$ has characteristic zero, you only have to check that $\alpha$'s minimal polynomial has no multiple roots.

  • 1
    No. Normal $\implies$ Galois because any finite extension of $\mathbb{Q}$ is separable – 2017-01-25
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    Wrong. The minimal polynomial of $\root3\of2$ is $p(x)=x^3-2$. Its roots are simple. Yet the extension $\Bbb{Q}(\root3\of2)$ is not normal, because it does not contain the complex roots of $p(x)$. – 2017-01-25