Is \(\pi\) Algebraic of Degree 3 Over Any Extension of \(\mathbb{Q}\)?

[Kiwi1]

Q7. Name an extension of \(\mathbb{Q}\) over which \(\pi\) is algebraic of degree 3.

I have a very simple answer to this (below) but is it correct? Is it likely to be what the text is after?

\(\mathbb{Q}(\pi^3)\) is an extension of \(\mathbb{Q}\), it is not an algebraic extension but I am not asked in the first 5 words for it to be algebtaic.

Now \(\mathbb{Q}(\pi)\) is an algebraic extension of \(\mathbb{Q}(\pi^3)\) with degree 3 because \(\pi\) is a root of: \(x^3-\pi^3\)

Is this what they are after? I can't see any way that pi can be algebraic over Q directly.

 

[Deveno]

Pi is *not* algebraic over $\Bbb Q$, that is to say, it is transcendental over $\Bbb Q$ ($\Bbb Q[\pi] \cong \Bbb Q[x]$), so your answer looks fine.

The question is probably just asking you to realize not all extensions of a field are algebraic extensions, and that being an algebraic extension depends *crucially* on the underlying field. Deciding "which" field extensions ARE algebraic can often be difficult, as there are many real numbers, for example, for which it is unknown whether or not they are algebraic over $\Bbb Q$.

The only thing missing (and I do not know if your instructor expects this) is showing that $x^3 - \pi^3$ is, in fact, irreducible over $\Bbb Q(\pi^3)$ (hint: $\Bbb Q(\pi^3) \subset \Bbb R$).