I am studying basic algebraic number theory these days and I am curious if the concept of “integral extension” is important in purely number theoretic sense.
Of course, integral extension is a central object in commutative algebra so we can deduce lots of properties from this concept.
However, as many number theory texts emphasize, commutative algebra originates from algebraic number theory and algebraic geometry. I am curious about the motivation of integral extensions.
In number theory, one extends $\mathbb{Z}$ to some larger ring $O$ to tackle problems on $\mathbb{Z}$. Indeed, this point of view greatly motivates the development of algebraic number theory as Kummer used $\mathbb{Z}[\zeta_n]$ to prove Fermat’s last theorem for regular primes. ($\zeta_n$ is an $n$-th primitive root of unity)
Let $K$ be a number field. Then, is $O_K$ important in purely number-theoretic sense? I think knowing that “$\mathbb{Z}[S]$ is a Dedekind domain and ramification theory works for the extension $\mathbb{Z}[S]/\mathbb{Z}$” should be a main focus in algebraic number theory, so if not many such rings were $O_K$ then the theory of integral extension would be a bit useless in number theory. For example, if $\mathbb{Z}[\zeta_n]$ were not $O_K$ where $K=\mathbb{Q}(\zeta_n)$ ($\zeta_n: n$-th primitive root of unity) and $\mathbb{Z}[w]$ were not $O_K$ where $K=\mathbb{Q}(w)$ ($w$: quadratic integer) then would the concept integral extension still be important in number theory? I don’t get any number-theoretic importance of the definition of integral closure itself. Let $K$ be a number field. Then, what does the collection of solutions of monic integral polynomials in $K$ have to do with number theory?
Thank you in advance!
Your reason for caring about integral extensions seems backwards. Kummer did not care about rings of integers because $ \mathbf Z[\zeta_n] $ was one, he cared about $ \mathbf Z[\zeta_n] $ because it was the ring of integers of $ \mathbf Q(\zeta_n) $.
To illustrate how the ring of integers of $ K $ becomes a natural object of study, you may think of the prime ideals of $ \mathcal O_K $ as the finite places of $ K $, i.e as different nonarchimedean valuations on the number field $ K $. In that case, if you are given a nonarchimedean valuation, it is fairly easy to see that it has to restrict to a $ p $-adic valuation on $ \mathbf Q $. In addition, we may check directly that elements integral over $ \mathbf Z $ always lie in the unit ball of any such valuation. If we say that $ \mathcal O_v $ is the unit ball corresponding to the valuation $ v $, then clearly $ \mathcal O_v $ is a ring, and we may define
$$ \mathcal O = \bigcap_{v} \mathcal O_v $$
where the intersection is over all nonarchimedean valuations $ v $ of $ K $. (This definition recovers $ \mathcal O = \mathbf Z $ if we choose $ K = \mathbf Q $.) It turns out that $ \mathcal O $ is in fact the ring of integers of $ K $ in the usual sense. It is easy to see that $ \mathcal O_K \subset \mathcal O $, whereas checking the equality is a bit more involved. Nevertheless, this gives an intrinsic reason for why the ring of integers is a natural object to attach to a number field. The reason that $ \mathcal O $ turns out to be a fairly regular domain in our case is that the valuations in the number field case turn out to be discrete; one can see this using Newton polygons, for instance. Then, the components $ \mathcal O_v $ are discrete valuation rings and thus integrally closed, and so is their intersection $ \mathcal O $. In fact, by showing that the number field $ K $ has "enough" valuations through embedding it into $ \bar {\mathbf Q_p} $ for different $ p $, we can directly show that $ \mathcal O = \mathcal O_K $ without knowing that $ \mathcal O_K $ has unique factorization of ideals.
Why do we care about $ \mathcal O $? It is because, in some sense, $ \mathcal O $ has "all of the relevant information" about the arithmetic of the field $ K $. You may think of each $ \mathcal O_v $ as capturing a piece of the behavior, and taking intersections allows you to capture all of it in a single subring of $ K $. If $ \mathcal O $ was any smaller than it was here, then we would once again lose sight of the behavior of primes dividing the conductor of this order. For instance, the subring $ \mathbf Z[\sqrt{5}] $ of $ \mathbf Q(\sqrt{5}) $ gives us incorrect information about the behavior of $ 2 $ in this field: it tells us that $ 2 $ should be ramified, while it is not.
Another way to justify the integral closure property is that if a domain has unique factorization of ideals, it is necessarily integrally closed (in fact, it is necessarily a Dedekind domain), as it must locally be a DVR everywhere, and integral closure is a local property. In this sense, to recover a weaker version of unique factorization requires us to impose the condition of integral closure on our ring, which we may again think of as a regularity condition. As the two other conditions for Dedekindness are always satisfied for finitely generated (over $\mathbf Z$) subrings of $ K $, we may take any integrally closed, finitely generated subring and work with it. If we try to take the largest such subring, this must be the integral closure of $ \mathbf Z $ in $ K $.
The short answer to your question is that we don't really care about integral closure in itself, but integral closure (like smoothness) turns out to be a regularity condition that's necessary for your objects to be sufficiently well behaved (in this case, subrings of $ K $ which have unique factorization of ideals). Integral extension, on the other hand, is often the "right" amount to extend the ground ring (precisely, this is the case if every valuation of the ground field extends to a valuation on the larger field, as it is with $ \mathbf Q $ and $ K $) if we care about understanding local behavior.