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I took a number theory course this past semester and I found the idea of there being different primes in different fields. The only fields that we did in detail were the set of reals and the set of Gaussian integers.

What other fields are there, and what numbers are prime in those fields?

Mathematical Dummy
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    What do you mean by “field”? – Ted Shifrin Jun 09 '22 at 18:41
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    That's a big question. A book on algebraic number theory would tell you a lot. For starters, the Gaussian integers use $\sqrt{-1}$ to make a larger field. But you adjoin any square root, $\sqrt{3}$, $\sqrt{-3}$, etc. – B. Goddard Jun 09 '22 at 18:42
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    Have you read the introduction here: https://en.m.wikipedia.org/wiki/Field_(mathematics) – Adam Rubinson Jun 09 '22 at 18:43
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    Gaussian integers $\mathbb{Z}[i]$ form the ring of integers in the field $\mathbb{Q}(i)$, but they do not form a field in their own right. – Sammy Black Jun 09 '22 at 18:43
  • Your title question is different. The fields you know are $\Bbb Q,\Bbb R, \Bbb C$. There are indeed many more fields, e.g., finite fields $\Bbb F_q$, or field of fractions, or function fields. Or finite extensions of $\Bbb Q$, like $\Bbb Q(\sqrt{2})$, $\Bbb Q(\sqrt[5]{7})$, $\Bbb Q(i,\sqrt{2},\sqrt[5]{7})$... – Dietrich Burde Jun 09 '22 at 18:54

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Fields don't have any proper ideals, so the idea of looking literally at primes in a field isn't very interesting. The usual construction is to start with a number field $k$ (i.e., a finite extension of $\mathbb{Q}$) and consider the algebraic integers $\cal{O}_k$: those elements $\alpha\in k$ for which the minimal monic polynomial of $\alpha$ over $\mathbb{Q}$ has coefficients in $\mathbb{Z}$. (For $k = \mathbb{Q}$ itself, the minimal polynomial of $\alpha$ is just $X - \alpha$, and so $\cal{O}_\mathbb{Q} = \mathbb{Z}$.) This turns out to be a ring with some nice properties that parallel the situation with $\mathbb{Z}$ and $\mathbb{Q}$:

  • $\operatorname{Frac}{\cal O}_k = k$;
  • Every ideal of ${\cal O}_k$ is generated by (at most) two elements;
  • Every ideal of ${\cal O}_k$ has a unique factorization into prime ideals of ${\cal O}_k$.

Thus ${\cal O}_k$ is almost a principal ideal domain, and it looks like a unique factorization domain if we consider ideals rather than individual elements.

To that end, we can consider the prime ideals inside ${\cal O}_k$. There's a whole area of number theory devoted to figuring out exactly what these prime ideals look like, but one of the basic questions is to figure out what happens to rational primes $p$ (i.e., ordinary primes in $\mathbb{Z}$) inside ${\cal O}_k$: Does the ideal $(p)$ remain prime, or does it split into a product of prime ideals? (The third point in the list above means that there's no ambiguity about this product in the latter case.) The general situation is well understood, especially in the case where $k/\mathbb{Q}$ is Galois. The particular case of $k = \mathbb{Q}(\sqrt{d})$ for squarefree $d$ is particularly easy to state: For an odd rational prime $p$, the ideal $(p)$ remains prime; has $(p) = \mathfrak{p}\mathfrak{q}$ for distinct prime ideals $\mathfrak{p},\mathfrak{q}\subset {\cal O}_k$; or has $(p) = \mathfrak{p}^2$ for some prime ideal $\mathfrak{p}\subset {\cal O}_k$ depending on whether the Jacobi symbol $(d/p)$ is $-1, 1,$ or $0$, respectively. This result is also one of many that leads to or leads from quadratic reciprocity. There's quite a lot more to say about the subject, and there are many other kinds of fields that come up just within number theory, but this is a start.

anomaly
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    All true and does answer the question - I think at a level beyond what the OP's number theory course prepared them for. – Ethan Bolker Jun 09 '22 at 19:23
  • @EthanBolker: Number theory unfortunately has a huge barrier to entry in terms of its complexity (and good luck to anyone interested in arithmetic geometry!), but hopefully there are at least some searchable keywords in the post for anyone who's interested in the subject. I guess a more readable tl;dr form of it would be, "(1) Number fields have 'rings of integers' inside them that parallel the relationship between $\mathbb{Z}$ and $\mathbb{Q}$'; (2) the 'primes' are the prime ideals of that ring; (3) there's a vast but complicated area of number theory that predicts what happens to them." – anomaly Jun 09 '22 at 19:32
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What you are calling a "field" is officially a ring of algebraic integers.

There is lots known about unique factorization and what the primes are in the rings of integers in the fields $\mathbb{Q}(\sqrt{d})$.

See the wikipedia page https://en.wikipedia.org/wiki/Quadratic_field and this stackexchange question: List of quadratic field with the UFD property

Ethan Bolker
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