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I'm studying Galois theory with Lang's textbook. I found that there are many topics in Galois theory related to algebraic number theory. But, I couldn't understand the connection precisely. So my question is: Why is Galois theory related to algebraic number theory? i.e. How does algebraic number theory use Galois theory?

Robin
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hxrd72
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    This is too broad. I can't imagine doing algebraic number theory without Galois theory... – lulu Aug 05 '23 at 12:28
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    Well, algebraic number theory is in solid part concerned with number fields, which are finite field extensions of $\mathbb{Q}$, and Galois theory is the basic tool to study finite field extensions. So the connexion seems rather logical. – Captain Lama Aug 05 '23 at 12:28
  • Because G.T. is intimately related to the zeros of a polynomial like $a_nx^n+a_{n-1}x^{n-1}+\cdots + a_1x+a_0$. – Piquito Aug 05 '23 at 15:03

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One of the important themes in Galois theory is to study a finite extension by enlarging it to a finite Galois extension when possible, i.e., when the extension is separable, and this always happens in characteristic zero. In particular, every finite extension of $\mathbf Q$ is contained in a finite Galois extension of $\mathbf Q$, so it's reasonable to expect to learn things about all finite extensions of $\mathbf Q$ by understanding finite Galois extensions of $\mathbf Q$.

What's special about a finite Galois extension $K/\mathbf Q$ as compared to other finite extensions of $\mathbf Q$ is that ${\rm Gal}(K/\mathbf Q)$ naturally acts on lots of things related to algebraic number theory. Inside $K$ alone, ${\rm Gal}(K/\mathbf Q)$ acts on the ring of integers $\mathcal O_K$, its ideals (especially its prime ideals), its unit group $\mathcal O_K^\times$, and so on. In algebraic geometry, varieties defined over number fields lead to many more objects that Galois groups act on.

The normal basis theorem in Galois theory says $K$ has a $\mathbf Q$-basis that is the Galois orbit of some element of $K$, but a more subtle question is whether $\mathcal O_K$ has a $\mathbf Z$-basis that is a Galois orbit of some element of $\mathcal O_K$ (a counterexample is $K = \mathbf Q(i)$ since in $\mathbf Z[i]$ there's no $\mathbf Z$-basis of the form $\{a+bi,a-bi\}$). The failure of the "integral" normal basis theorem in some Galois extensions of $\mathbf Q$ is related to wild ramification.

Inside ${\rm Gal}(K/\mathbf Q)$ we have the decomposition and inertia groups (and the higher ramification groups) associated to each nonzero prime ideal in $\mathcal O_K$. When ${\rm Gal}(K/\mathbf Q)$ acts on things, so do those subgroups, and this leads to lots of interesting math. The simplest thing it tells us is that the residue field degrees at all the prime ideals in $\mathcal O_K$ over a common prime number are equal. (We also have equality of all the ramification indices at the prime ideals over a given prime number, but that's not as striking since those ramification indices are $1$ for all but finitely many prime ideals in every finite extension of $\mathbf Q$, even when the extension is not Galois.) In particular, the action of decomposition groups on prime ideals leads to arguably the most important concept in basic algebraic number theory: the Frobenius element in ${\rm Gal}(K/\mathbf Q)$ that is associated to each prime ideal in $\mathcal O_K$ that is unramified over $\mathbf Q$. Frobenius elements are a link between Galois theory of number fields and Galois theory of finite fields.

When you consider Galois groups of infinite-degree extensions of number fields, not just finite extensions, Galois representations enter the story in a profound way. See, for instance, this MO question and this one too.

KCd
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