What is $\mathbb Q(p^{1/N})$ where $p$ is a prime?

Question

I'm struggling with this problem: Find an explicit description for the smallest field containing the rationals if you adjoin the $N^{\text{th}}$ root of a prime $p$.

A friend (a retired PhD with wide curiosity) asked me this question, and neither of us are getting anywhere with it. Our main interests were not in field theory, so our expertise doesn't run too deeply there. We suspect it's not too difficult (comes out of a textbook), but we can't find and prove an answer.

The problem came Field Theory and its Classical Problems.

We are looking for a subfield, so once $x = p^{1/N}$ is adjoined, we must also have powers of $x$, and their invoices. We are not including complex roots, only real ones.

The problem is given as stated. I presume that an "explicit description" would mean a way to express all elements of the subfield, such as $\{\text{all} \sum_{i = 0}^{N-1} a_i p^{i/N}, a_i \in \mathbb Q\}$

While that's a subspace of $R$ as a vector space, it's not obvious that it's a subfield. A proof that it's a subfield would solve the problem. Or a proof that some smaller set still containing $p^{1/N}$ is a subfield. That the set contains inverses is the sticking point.

Answer 1

This is a standard and well known construction and any good algebra texts(see Dummit and Foote) contains all of these.The only reason I am providing this answer is because the books may take 10-20 pages to arrive at this result. So I am sort of giving a short quick summary of what the process is and how you "attach roots" to a field.

I'll denote $\Bbb{Q}$ as $F$ as this is a general construction.

Take the polynomial $x^{N}-p$. This is irreducible by Eisenstein.

Then $\dfrac{F[x]}{(x^{N}-p)}$ is a field. Where $(x^{N}-p)$ denotes the maximal ideal generated by this polynomial.

Now consider the projection(or quotient) map $\pi:F[x]\to\dfrac{F[x]}{(x^{N}-p)}$

Then $(q(\pi(x)))=\pi (q(x))\,,\forall q(x)\in F[x]$

So given the polynomial $x^{N}-p=f(x)$.

$(f(\pi(x)))=f(x)\mod f(x)=0$.

So $\pi(x)$ is a root of the polynomial $x^{N}-p$.

Now you can prove that $\{1,\pi(x),(\pi(x))^{2}...(\pi(x))^{N-1}\}$ form a basis for $\dfrac{F[x]}{(x^{N}-p)}$ as an $F-$ vector space.

Now let $K$ be an extension of $F$ containing $\sqrt[N]{p}\,$(Such an extension is guaranteed due to $\mathbb{C}$) and define $F(\sqrt[N]{p})$ to be the smallest subfield containing both $F$ and $\sqrt[N]{p}$. By smallest I mean the intersection of all subfields of $K$ containing $F$ and $\sqrt[N]{p}$.

Then consider the map $\phi:F[x]\to F(\sqrt[N]{p})$ such that $\phi(q(x))=q(\sqrt[N]{p})$.

Then this is a ring homomorphism with kernel $(x^{N}-p)$. And hence the isomorphism theorem gives :-

$\frac{F[x]}{(x^{N}-p)}\cong F(\sqrt[N]{p})$.

Now you can clearly see that $\{1,\sqrt[N]{p},(\sqrt[N]{p})^{2},...,(\sqrt[N]{p})^{N-1}\}$ is a basis for $F(\sqrt[N]{p})$ as an $F-$ vector space. and hence it gives you your explicit description since you can express any element in the field $F(\sqrt[N]{p})$ as a linear combination of the basis elements .

So $F(\sqrt[N]{p})=\{a_{0}+a_{1}\sqrt[N]{p}+a_{2}(\sqrt[N]{p})^{2}+...+a_{N-1}(\sqrt[N]{p})^{N-1}\,|a_{i}\in F\}$