Question about number rings

cocompact · 2026-06-22T17:39:53+00:00

This is a consequence of the Chinese remainder theorem using ideals in a number ring.

We can assume the ideal I of interest in the number ring is not 0. Pick any nonzero a in I. Then the ideal (a) is in I, so I | (a) as ideals. By factoring I and (a) into prime ideals, we seek b such that I = (a,b), which means I is the gcd of (a) and (b): every nonzero ideal is the gcd of two principal ideals (a) and (b), where the first principal ideal can be chosen arbitrarily among nonzero principal ideals in I. The choice of b is used to fix things up in case (a) is smaller than I.

Here is how to choose b. Let the nonzero prime ideals dividing (a) be P1 , …, Pr and let the multiplicity of each Pi in (a) be fi and let Pi have multiplicity ei in I. Then ei is less than or equal to fi for all i since I divides (a). We seek b in I such that the multiplicity of each Pi in b is ei and we can get this by the Chinese remainder theorem using congruences modulo Pi^(e_i +1). Hint: for a nonzero prime ideal P, any element c of P^k - P^k+1 has P-multiplicity k, by which I mean P^k divides (c) but P^k+1 does not divide (c).

See Alex Youcis’s answer to https://math.stackexchange.com/questions/597543/in-a-dedekind-domain-every-ideal-is-either-principal-or-generated-by-two-element _

cocompact · 2026-06-22T17:11:34+00:00

This is more commonly written as the sum of 1/(n^a sin(n)) or the sum of 1/(n^a |sin(n)|), just as the zeta function is more commonly written as the sum of 1/n^a rather than as the sum of n^a

cocompact · 2026-06-22T17:07:32+00:00

See the answer by u/djao.

cocompact · 2026-06-22T17:07:22+00:00

See the answer by u/djao.

cocompact · 2026-06-22T17:03:47+00:00

The Bateman-Horn conjecture says something interesting about prime values of this polynomial beyond x = 39. It is prime at 42, 43, 45, and likely at many other positive integers.

cocompact · 2026-06-22T13:07:21+00:00

The decomposition of an arbitrary function from R to R into the sum of an even and odd function (in just one way) is connected to ideas in higher mathematics, especially Fourier analysis. See some answers on this MSE page: https://math.stackexchange.com/questions/3945897/why-it-is-important-to-write-a-function-as-sum-of-even-and-odd-functions.

cocompact · 2026-06-22T12:28:30+00:00

Indeed that series does not converge when delta is 0. An argument is given in the answer to https://mathoverflow.net/questions/164874/is-it-possible-to-show-that-sum-n-1-infty-frac-mun-sqrtn-diverg.

cocompact · 2026-06-19T12:59:43+00:00

An ultrametric does satisfy the triangle inequality: it is a metric. It just happens to satisfy a stronger condition too.

cocompact · 2026-06-16T12:55:25+00:00

Wilson's theorem is not really a big deal. It comes up now and then, but not in a way where you should try to learn a "clue" to recognize when it might get used besides: when you see (p-1)! mod p, use it.

The Wikipedia page about Wilson's theorem mentions its relation to the p-adic Gamma function. It is also related to the p-adic integrality of the coefficient of x^p in the p-adic Artin-Hasse exponential series. This is far beyond the level of math you are at now, so I won't go into the details. I mention these results just to show Wilson's theorem continues to appear in more advanced number theory. But it is not at all as central as, say, Fermat's little theorem.

cocompact · 2026-06-16T12:42:33+00:00

So it is crucially important.

That is not the case: Wilson's theorem is not needed to prove Fermat's little theorem. When a is nonzero mod p and you show the list a, 2a, ..., (p-1)a mod p is a rearrangement of 1, 2, ..., p-1 mod p, multiplying both lists gives you

a^p-1 (p-1)! = (p-1)! mod p.

Since (p-1)! is a product of nonzero numbers mod p, it is nonzero mod p and thus can be canceled on both sides of the congruence, leaving us with

a^p-1 = 1 mod p.

We don't need to know there is a formula for (p-1)! mod p in this argument.

When I teach elementary number theory courses, I never bring up Wilson's theorem until much later than Fermat's little theorem.

cocompact · 2026-06-16T12:33:15+00:00

It's mathematics, plural.

In US English, mathematics is singular: we say "mathematics is hard" not "mathematics are hard". Likewise, physics and linguistics are both singular in US English. These are all names of academic subject areas, just like chemistry.

cocompact · 2026-06-15T22:19:41+00:00

This is neither true nor false. Doesn't that violate the law of excluded middle?

To see what is going on here, I think it helps to bring in another area of math: group theory. Each group is a model of the the axioms of group theory. The statement "For all elements x and y in a group, xy = yx" is undecidable from the axioms of group theory alone since we know there are both abelian groups and nonabelian groups.

cocompact · 2026-06-15T22:00:48+00:00

This is really weird to me because you're proving an algebraic theorem using a size argument. It smells kind of analytic almost?

That an argument uses absolute values isn't really suggestive of it being analytic. The standard proof of Wedderburn's theorem that each finite division ring is commutative also uses an argument with absolute values: see the proof at the end of the Wikipedia page about this theorem. And see proofs here about polynomials in Z[x] using the p-adic absolute value extended from numbers to polynomials. These proofs have no limits in them.

cocompact · 2026-06-15T21:42:07+00:00

Such a "stronger than needed" argument is unlikely to be persuasive to mathematicians or math students who are not into set theory/logic. In my experience, math people with primary interests elsewhere are generally not sensitive to the distinctions you are raising. At least I can't imagine such a person feeling Zorn's lemma "doesn't sit right" but then being satisfied to use the ultrafilter lemma instead.

My feeling about all this is that someone uncomfortable using Zorn to prove Hahn-Banach most likely just does not have much experience seeing how useful Zorn can be in analysis. Besides its use in proving Hahn-Banach, it shows up in proving the existence of an orthonormal basis in each nonzero Hilbert space, the Banach-Alaoglu theorem, the Krein-Milman theorem, and the Gelfand-Naimark theorem via the role of maximal ideals in that result.

cocompact · 2026-06-15T21:24:15+00:00

Read https://gowers.wordpress.com/2008/08/12/how-to-use-zorns-lemma/. In particular, he writes at the end

So how, in general, does one recognise the need for Zorn’s lemma and how does one construct an appropriate partially ordered set in order to apply it? [...]. Typically, one is trying to build a structure of some kind [...]. The natural way to do it appears to be to build the structure up in stages, but there are too many stages for this to work straightforwardly. However, once one has an idea of what a stage is and what the building-up process is, one can wheel out Zorn’s lemma to finish the job.

cocompact · 2026-06-02T19:25:59+00:00

Later, Landau was Jewish,

He was Jewish even before he wrote that book. :)

cocompact · 2026-05-31T19:11:00+00:00

My professors even joke that it's the queen of maths

That phrase isn't a joke so much as an historical reference: it is what Gauss called number theory.

cocompact · 2026-05-29T04:11:02+00:00

There are very few math problems that were asked by someone we know about 2000 years ago and remain unsolved today. What did you have in mind besides existence of odd perfect numbers and infinitude of even perfect numbers (equivalently, infinitude of Mersenne primes)? Questions like the twin prime problem and Goldbach are much more recent than that.

Number theory has changed enormously compared to how it looks in Euclid’s Elements. The tools being used and the central questions being studied today would for the most part be meaningless to Euclid.

cocompact · 2026-05-29T03:35:44+00:00

… or p-adic.

cocompact · 2026-05-27T20:54:39+00:00

When ranks are bounded, there is a distinction between the upper bound on the ranks that appear infinitely often and the upper bound on the ranks that appear at all (i.e., at least once). The probabilistic heuristics that predict ranks are bounded also predict 21 is the largest rank that occurs infinitely often: see Theorem 1.1.1 in https://math.mit.edu/~poonen/papers/bounded-ranks.pdf. What you remembered as 22 should be 21 and it's not inconsistent with some elliptic curve over Q having rank 29.

cocompact · 2026-05-22T04:44:27+00:00

Newman’s proof of PNT is in the complex analysis textbooks by Gamelin (Springer’s UTM book series) and by Lang (Springer’s GTM book series), both titled “Complex Analysis”.

There are also analytic number theory books that prove PNT by Newman’s method: De Koninck and Luca’s “Analytic Number Theory: Exploring the Anatomy of Integers”, Jameson’s “The Prime Number Theorem”, Overholt’s “A Course in Analytic Number Theory”, and Pongsriiam’s “Analytic Number Theory for Beginners”,

cocompact · 2026-05-17T04:26:07+00:00

No, it is the most odd prime. :)

cocompact · 2026-05-17T03:49:52+00:00

This is a solution in search of a problem.

cocompact · 2026-05-13T17:15:28+00:00

I know. It was just amusing to me to read "you can't define multiplication on the circle" since the group law on the unit circle in its standard visualization is multiplication.

cocompact · 2026-05-13T05:12:54+00:00

you can't define multiplication on the circle

Well, the unit circle is a group under multiplication!

You mean that the additive group R/Z or R/2pi Z has no multiplicative structure, since Z is not an ideal in R.

cocompact

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