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What is the minimal polynomial over $\mathbb{Q}$ of $\bigotimes_{j=0}^{\infty}\mathbb{Q}(\zeta_{j})$, where $\zeta_j$ is a $j$-th primitive root of unity for each $j$?

I want to say it should be $\phi_{\infty}$ where $\phi_{n}$ is the $n$-th cyclotomic polynomial but this would be a power series instead of polynomial.

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closed as off-topic by user26857, Eric Wofsey, Claude Leibovici, SchrodingersCat, tired Nov 24 '15 at 9:37

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    $\begingroup$ Are you talking about the compositum of all the cyclotomic extensions of $\mathbb{Q}$? It only makes sense to talk about minimal polynomials of specific elements of $\overline{\mathbb{Q}}$. $\endgroup$ – D_S Oct 16 '15 at 14:23
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If you’re thinking of the roots of unity in the complex domain, then the set of all these is dense in the unit circle, so any analytic function that vanishes at them all must be identically zero.

However, if you’re just asking about the $p^n$-th roots of unity (all $n$, but for a fixed $p$), in the $p$-adic domain, then there is a perfectly good power series that has for its roots all the $\zeta-1$, as $\zeta$ runs through the $p$-power roots of unity. And it’s the wonderful logarithmic series that you know from Calculus: $$ L(x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots=\sum_{j=1}^\infty(-1)^{j+1}\frac{x^j}j\,. $$ It involves moderately advanced $p$-adic knowldege to prove that this series does what I claim, but the reason it all works is that in the $p$-adic domain, the roots of unity are not dense in any way like the roots of unity in the complex domain; indeed, they’re a discrete set.

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I think that by the minimum polynomial $p$ of a field extension $L$ of $\mathbb{Q}$ you mean a polynomial such that $L$, when viewed as vector space over $\mathbb{Q}$ is spanned by all products of roots of $p$. Given $L$, the polynomial $p$ need not be unique but we should not worry about that.

What we should worry about is that for $p$ to exist, $L$ must be finite dimensional. Of course there is no bound on the number of products of roots you can write down, but whenever you encounter a root $\alpha$ raised to the power the degree of $p$ you can replace it by a linear combination of smaller powers of $\alpha$. Explicitly:

Let $p(x) = x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0$ then $\alpha^n = -a_{n-1}\alpha^{n-1} - ... - a_0$.

We conclude that $L$ is at most $n^n$-dimensional over $\mathbb{Q}$. (Of course you can be smarter and get a sharper bound if you want, but at least here we see that any field extension of this form has finite $\mathbb{Q}$-dimension.)

Your field on the other hand is, viewed as vector space, a tensor product of infinitely many spaces, infinitely many of which have dimension at least 2. So clearly this product is infinite dimensional. It follows that no polynomial can be found whose roots span the whole thing.

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