Give some motivation for the definition of algebraic integer?

[user202729]

While the definition of algebraic number is quite obvious, I feel the definition of algebraic integer is a bit unmotivated.

I think something like the following.

https://math.stackexchange.com/questions/693541/intuition-behind-the-definition-of-algebraic-integers

---

other points:

• we may define that if K = ℚ[α], then 𝒪_K = ℤ[α] for α the "simplest" representation in some sense, but this definition depends on the choice of α and doesn't work out -- for example, if α is √d or a root of unity, it's natural to choose 𝒪_K = ℤ[√d] with squarefree d, but what if α is something like the root of X³-X²-2X-8?

  (remark: this appear in the counterexample of monogenic extension)

  In other words, we want the definition of algebraic integer to be some intrinsic concept of the numbers.

• the algebraic integers should be, in some sense, "sparse" in K, just like how ℤ[i] is sparse in ℚ[i].

  However, what "sparse" means here is not very clear (unlike the case ℚ[i] → ℂ ≅ ℝ² has a natural embedding -- for example, in the case of K = ℚ[√2], K itself is dense in ℝ), so we require the following:

  If we pick any ℚ-vector space basis {β₁, …, βₙ} of K, then there is some ε > 0, such that for all elements c₁ β₁ + ⋯ + c_n β_n ∈ 𝒪_K, then c₁²+c₂²+⋯+cₙ² > ε².

The last point is the important point here, and it turns out to be equivalent to each of the 3 of the following:

• for all α ∈ 𝒪_K, ℤ[α] is a finite ℤ-module.
• minimal polynomial of α has integer coefficients.

---

(remark: another possible definition of "sparse" you can think of is that (𝒪_K ∩ ℚ) is not dense, equivalently, 𝒪_K ∩ ℚ = ℤ. While this is a corollary of the above, the converse need not hold, for example ℤ[(i√7-1)/4] ∩ ℚ = ℤ, but (i√7-1)/4 is not an algebraic integer)

---

Another point, it may be useful to mention that this definition of algebraic integer is useful in the sense that anything smaller than it does not have unique factorization of ideal property. (https://math.stackexchange.com/questions/682912/ideals-in-a-non-dedekind-domain-that-cannot-be-factored-into-product-of-primes/682947#682947) i.e. proving unique factorization of ideal implies integrally closed, if the proof is nice.

[vEnhance]

I think this is overall a good idea, but I don't have the bandwidth right now to write it myself. So anyone is welcome to submit a pull request using any of the above; otherwise I will eventually get to it myself, but not soon.