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Is $\mathbb Q(\pi,i\pi):\mathbb Q$ a simple extension?

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Any finite extension of $\mathbb{Q}$ is a simple extension. See the primitive element theorem. – Daenerys Naharis Mar 16 '11 at 5:22
@Joseph: I believe this theorem speaks about algebraic extensions, which is clearly not the case here. – Asaf Karagila Mar 16 '11 at 5:29
Oops, didn't read the question closely. I'll leave the comment so others can see my shame. – Daenerys Naharis Mar 16 '11 at 5:44

No. Let $x \in \mathbb Q(\pi,i\pi) = \mathbb Q(i,\pi).$ If $x$ is algebraic, then $\mathbb Q(x)$ is finite over $\mathbb Q$, hence an algebraic extension, and so does not contain the transcendental element $\pi.$ On the other hand, if $x$ is transcendental over $\mathbb Q$, then $\mathbb Q(x)$ is isomorphic to the field of rational functions in one variable over $\mathbb Q$. The algebraic closure of $\mathbb Q$ in this extension is equal to $\mathbb Q$ itself, and so this field does not contain $i$.

Thus $\mathbb Q(\pi,i\pi)$ is not of the form $\mathbb Q(x)$ for any of its elements $x$, and so is not a simple extension of $\mathbb Q$.

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Suppose there exists a $x\in\mathbb Q(\pi,i)$ such that $\mathbb Q(\pi,i)=\mathbb Q(x)$. Since $\mathbb Q(\pi,i)$ is infinite over $\mathbb Q$, $x$ must be trascendental. Lüroth's theorem, then, tells us that every subfield of $\mathbb Q(x)$ properly containing $\mathbb Q$ is itself a simple trascendental extension of $\mathbb Q$. But this then applies to $\mathbb Q(i)\subseteq\mathbb Q(x)$. Of course, this is absurd.

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in my opinion this invocation of Luroth may be overkill. Isn't it rather clear that the polynomial $t^2+1$ does have a root in $\mathbb{Q}(i)$ and does not have a root in $\mathbb{Q}(x)$? This seems accessible to a wider audience. – Pete L. Clark Mar 16 '11 at 6:57
@Pete: I don't have a big problem with overkillness :P But you are of course right. – Mariano Suárez-Alvarez Mar 16 '11 at 7:02

Let's see. Assume otherwise, thus $\mathbb Q(\pi,i\pi) = \mathbb Q(\pi,i) = \mathbb Q(x)$ for some x. This means that they are equivalent as modules over $\mathbb Q$, and thus $i = a_nx^n + \cdots + a_0$ for some polynomial with $n \geq 1$, and so we have that $(a_nx^n + \cdots + a_0)^2 + 1 = 0$ is a rational polynomial of nonzero degree with $x$ as a root, thus $x$ is algebraic. But $\pi$ is transcendental, and since the algebraic numbers are a field we cannot write $\pi$ as a polynomial in $x$, contradicting the assumption that $\mathbb Q(\pi,i) = \mathbb Q(x)$. So the answer is no.

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Dear Alex, You have to be a little careful, because the putative expression for $i$ could actually be a ratio of two polynomials in $x$. One then has to write this ratio as ratio of coprime polynomials, leading to an equation $f(x)^2 + g(x)^2 = 0$ for coprime polynomials $f$ and $g$. This is impossible unless $x$ is algebraic, and then your argument continues as before. (In my answer I skip over this step somewhat glibly, by simply asserting that the algebraic closure of $\mathbb Q$ in the field of rational functions is just $\mathbb Q$.) Regards, – Matt E Mar 16 '11 at 5:44
@Matt: Thanks. I was caught up in thinking of the extensions as modules and forgot about the inversion operation. But the proof is essentially the same, multiplying both sides of the equation by the square of the denominator in the expression for $i$. Edit: You explained this while I was writing my comment. – Alex Becker Mar 16 '11 at 5:49

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