Is $\mathbb Q(\pi,i\pi):\mathbb Q$ a simple extension?
$\mathbb{Q}(\pi, i\pi)$ over $\mathbb{Q}$
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2Oops, didn't read the question closely. I'll leave the comment so others can see my shame. – 2011-03-16
3 Answers
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 $\mathbb Q$.
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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0@Pete: I don't have a big problem with overkillness :P But you are of course right. – 2011-03-16
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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0@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. – 2011-03-16