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I am new in field theory and i began studying it recently.

I have a question about a finite field extension

The basis of the extension $\mathbb{Q}$$(\sqrt{1+\sqrt{3}})$/$\mathbb{Q}$ is the set $\{1,\sqrt{1+\sqrt{3}},\sqrt{1+\sqrt{3}}^2,\sqrt{1+\sqrt{3}}^3\}$ becase of the monic polynomial $(x^2-1)^2-3=0$ ?

Or am i missing something?

Thank you in advance.

Jyrki Lahtonen
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Marios Gretsas
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  • Nothing wrong with your question (+1), you only need to convince yourself and your teacher that the monic quartic is irreducible. FYI: the tag [tag:finite-fields] is meant to be about questions of those fields that have only finitely many elements. $\Bbb{Q}$ and its extension fields have infinitely many elements, and won't qualify :-) – Jyrki Lahtonen Sep 15 '16 at 11:05

2 Answers2

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Yes, the extension is biquadratic, i.e., has degree $4$, because of the arguments given in the answer to this duplicate, for $a=1$ and $b=3$. Indeed, the polynomial $(x^2-1)^2-3$ is irreducible, because neither $b=3$ is a square nor $a^2-b=-2$ is a square.

Community
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Dietrich Burde
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Since $\bigl(\sqrt{1+\sqrt{3}}\,\bigr)^2=1+\sqrt{3}$, we see that $\mathbb{Q}(\sqrt{3})\subseteq\mathbb{Q}(\sqrt{1+\sqrt{3}}\,)$.

Thus if the given extension is not of degree $4$, it has degree $2$. On the other hand, the decomposition of $(x^2-1)^2-3=x^4-2x^2-2$ over the reals is $$ (x-\sqrt{1+\sqrt{3}}\,)(x+\sqrt{1+\sqrt{3}}\,)(x^2-1+\sqrt{3}\,) $$ and the last factor has nonreal roots. Since the product of the two linear factors is $x^2-1-\sqrt{3}$, we conclude that the polynomial has no factor of degree $2$ with rational coefficients.

Hence $x^4-2x^2-2$, being irreducible, is indeed the minimal polynomial of $\sqrt{1+\sqrt{3}}$.

egreg
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  • This is indeed the same answer as by anonymous in the duplicate. – Dietrich Burde Sep 15 '16 at 14:25
  • @DietrichBurde Really? I felt there's no need to apply a general result (not that useful, by the way). – egreg Sep 15 '16 at 14:26
  • I like your answer, so that is no problem. But why is the other answer not useful? With $a=1$ and $b=3$ we get the same formulas - and the same result. – Dietrich Burde Sep 15 '16 at 14:29
  • @DietrichBurde We could write long tables of similar formulas, who's the one that can remember them? On the other hand, such irrationals were studied in depth by Euclid and perhaps some general methods can be found in the Elements. – egreg Sep 15 '16 at 14:30