Discriminant of a number field

Discriminant of an algebraic integer

Let 𝛼 be an algebraic integer of degree 𝑛 with minimal polynomial 𝑓𝛼(𝑥) ∈ℤ[𝑥] and 𝐾 =ℚ(𝛼). The discriminant of 𝛼 is then #m/thm/num/alg

Δ𝐾:ℚ(𝛼)=(−1)(𝑛2)N𝐾:ℚ⁡(𝑓′(𝛼))

where N𝐾:ℚ is the field norm and 𝑓′(𝑥) ∈ℤ[𝑥] is the formal derivative.

Proof

Let 𝐹(𝛼) :=( −1)(𝑛2)N𝐾:ℚ⁡(𝑓′(𝛼)), 𝜎𝑖 :𝐾 ↪ℂ be the 𝑛 distinct embeddings of 𝐾 in ℂ, and thus 𝛼𝑖 =𝜎𝑖(𝛼) be the 𝑛 conjugates of 𝛼. Expanding the definition of the field norm,

𝐹(𝛼)=(−1)(𝑛2)𝑛∏𝑖=1𝜎𝑖(𝑓′(𝛼))=(−1)(𝑛2)𝑛∏𝑖=1𝑓′(𝛼𝑖),

where since

𝑓(𝑥)=𝑛∏𝑖=1(𝑥−𝛼𝑖),

it follows

𝑓′(𝛼𝑖)=(∏1≤𝑗<𝑖(𝛼𝑖−𝛼𝑗))(∏𝑖<𝑗≤𝑛(𝛼𝑖−𝛼𝑗)),

and thus

𝑛∏𝑖=1𝑓′(𝛼𝑖)=(∏1≤𝑗<𝑖≤𝑛(𝛼𝑖−𝛼𝑗))(∏1≤𝑖<𝑗≤𝑛(𝛼𝑖−𝛼𝑗))=(∏1≤𝑗<𝑖≤𝑛(𝛼𝑖−𝛼𝑗))2(−1)(𝑛2).

Now the term being squared is precisely the determinant of the Vandermonde matrix

𝑇(𝛼)=⎡⎢ ⎢ ⎢⎣1𝛼1⋯𝛼𝑛−11⋮⋮⋱⋮1𝛼𝑛⋯𝛼𝑛−1𝑛⎤⎥ ⎥ ⎥⎦

therefore 𝐹(𝛼) =det𝑇(𝛼)2 =Δ(𝛼).

In particular, if the minimal polynomial is of the form

𝑚𝛼(𝑥)=𝑥𝑛+𝑎𝑥+𝑏∈ℤ[𝑥]

then we have

Δ(𝛼)=(−1)𝑛(𝑛−1)2((−1)𝑛−1(𝑛−1)𝑛−1𝑎𝑛+𝑛𝑛𝑏𝑛−1).
Proof

Let 𝛽 =𝑓′(𝛼) =𝑛𝛼𝑛−1 +𝑎 ∈O𝐾. Since 𝑓 annihilates 𝛼, we have

𝛽=−(𝑛−1)𝑎−𝑛𝑏𝛼−1,𝛼=−𝑛𝑏𝛽+(𝑛−1)𝑎;

whence 𝐾 =ℚ(𝛼) =ℚ(𝛼−1) =ℚ(𝛽) and degℚ⁡𝛽 =𝑛. Now since

0=𝑓(𝛼)=𝑓(−𝑛𝑏𝛽+(𝑛−1)𝑎)=(−𝑛𝑏)𝑛(𝛽+(𝑛−1)𝑎)−𝑛−𝑎𝑛𝑏(𝛽+(𝑛−1)𝑎)−1+𝑏,

we can multiply by (𝛽 +(𝑛 −1)𝑎)𝑛 to get

0=(−𝑛𝑏)𝑛−𝑎𝑛𝑏(𝛽+(𝑛−1)𝑎)𝑛−1+𝑏(𝛽+(𝑛−1)𝑎)𝑛,

whence

𝑔(𝑥)=(𝑥+(𝑛−1)𝑎)𝑛−𝑎𝑛(𝑥+(𝑛−1)𝑎)𝑛−1+(−𝑛)𝑛𝑏𝑛−1∈ℤ[𝑥]

is a monic annihilating polynomial for 𝛽, which must be minimal since deg𝑥⁡𝑔(𝑥) =𝑛 =degℚ⁡𝛽. Again invoking the fact 𝐾 =ℚ(𝛽), we have

(−1)𝑛N𝐾:ℚ⁡(𝛽)=𝑔(0)=(𝑛−1)𝑛𝑎𝑛−(𝑛−1)𝑛−1𝑛𝑎𝑛+(−𝑛)𝑛𝑏𝑛−1=−(𝑛−1)𝑛−1𝑎𝑛+(−𝑛)𝑛𝑏𝑛−1,

whence Δ(𝛼) =( −1)𝑛(𝑛−1)2((−1)𝑛−1(𝑛−1)𝑛−1𝑎𝑛+𝑛𝑛𝑏𝑛−1).

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