Dedekind domain

Ideal class group

Let 𝑅 be a Dedekind domain, e.g. the ring of integers of some number field, 𝐼(𝑅) its group of fractional ideals, and 𝑃(𝑅) be the subgroup of principal ideals. The ideal class group is the quotient group ring

Cl⁡(𝑅)=𝐼(𝑅)/𝑃(𝑅)

Equivalently, let 𝐽(𝑅) be the set of nonzero ideals in 𝑅, and define an equivalence relation on 𝑅 so that for 𝔞,𝔟 ∈𝐽(𝑅)

𝔞∼𝔟⟺(∃0≠𝑎,𝑏∈𝑅)[𝑎𝔞=𝑏𝔟]

Then 𝐽(𝑅)/ ∼ ≅Cl⁡(𝑅).¹

Proof

First we verify that ( ∼) defines an equivalence relation on 𝐽(𝑅), where the only condition that isn’t immediately obvious is ^E3. Suppose 0 ≠𝑎,𝑏,𝛽,𝛾 ∈𝑅 so that

𝑎𝔞=𝑏𝔟𝛽𝔟=𝛾𝔠

Then 𝑎𝛽𝔞 =𝑏𝛽𝔟 =𝑏𝛾𝔠, so 𝔞 ∼𝔠, as required.

Next we show that ( ∼) defines a congruence relation on the monoid 𝐽(𝑅). Now suppose 𝔞,𝔞′,𝔟,𝔟′ ∈𝐽(𝑅) such that 𝑎𝔞 =𝛼𝔞′ and 𝑏𝔟 =𝛽𝔟′. Then 𝑎𝑏𝔞𝔟 =𝛼𝛽𝔞′𝔟′ so 𝔞𝔟 ∼𝔞′𝔟′. The quotient monoid 𝐽(𝑅)/( ∼) is thence well-defined.

Now we show that 𝐽(𝑅)/( ∼) is in fact a group. Let 𝔞 ∈𝐽(𝑅) and 0 ≠𝑎 ∈𝔞. Then ⟨𝑎⟩ ⊆𝔞 so 1 ∼⟨𝑎⟩ =𝔞𝔟 for some ideal 𝔟.

Finally we show that these groups are isomorphic, letting 𝐺 and ˜𝐺 denote the constructions with and without fraction ideals respectively. An arbitrary element in 𝐺 is 𝐴𝑃(𝑅) for some fractional ideal 𝐴. But 𝑟𝐴 =𝔞 for some 𝑟 ∈𝑅 and 𝔞 ∈𝐼(𝐽), so 𝔞𝑃(𝑅) =𝐴⟨𝑟⟩𝑃(𝑅). Therefore we can always use an integral ideal as a representative for an element of 𝐺, which itself represents an element of ˜𝐺. Clearly 𝔞𝑃(𝑅) =𝔟𝑃(𝑅) ⟺ 𝔞 ∼𝔟: hence we have an isomorphism.

Results

• Ideal class group of a number field

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tidy | en | SemBr

Footnotes

1. 2022. Algebraic number theory course notes, pp. 21–22 ↩