Suppose {𝛼𝑖}𝑛𝑖=1 ⊂O𝐾 is an Integral basis for 𝐾.
It suffices to show that {𝜄(𝛼𝑖)}𝑛𝑖=1 form a basis for ℝ𝑛.
To this end, let
𝐴1=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣𝜎1(𝛼1)⋯𝜎1(𝛼𝑛)⋮⋱⋮𝜎𝑟1(𝛼1)⋯𝜎𝑟1(𝛼𝑛)ℜ𝜏1(𝛼1)⋯ℜ𝜏1(𝛼𝑛)ℑ𝜏1(𝛼1)…ℑ𝜏1(𝛼𝑛)⋮⋱⋮ℜ𝜏𝑟2(𝛼1)⋯ℜ𝜏𝑟2(𝛼𝑛)ℑ𝜏𝑟2(𝛼1)⋯ℑ𝜏𝑟2(𝛼𝑛)⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦∈M𝑛,𝑛(ℂ)be the matrix containing all these embeddings of the 𝛼𝑖.
We now apply the following elementary row operations:
- Add 𝑖ℑ𝜏𝑗(𝛼𝑘) to ℜ𝜏𝑗(𝛼𝑘) giving
𝐴2=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣𝜎1(𝛼1)⋯𝜎1(𝛼𝑛)⋮⋱⋮𝜎𝑟1(𝛼1)⋯𝜎𝑟1(𝛼𝑛)𝜏1(𝛼1)⋯𝜏1(𝛼𝑛)ℑ𝜏1(𝛼1)…ℑ𝜏1(𝛼𝑛)⋮⋱⋮𝜏𝑟2(𝛼1)⋯𝜏𝑟2(𝛼𝑛)ℑ𝜏𝑟2(𝛼1)⋯ℑ𝜏𝑟2(𝛼𝑛)⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦.
- Multiply ℑ𝜏𝑗(𝛼𝑘) by −2𝑖 giving
𝐴3=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣𝜎1(𝛼1)⋯𝜎1(𝛼𝑛)⋮⋱⋮𝜎𝑟1(𝛼1)⋯𝜎𝑟1(𝛼𝑛)𝜏1(𝛼1)⋯𝜏1(𝛼𝑛)−2𝑖ℑ𝜏1(𝛼1)…−2𝑖ℑ𝜏1(𝛼𝑛)⋮⋱⋮𝜏𝑟2(𝛼1)⋯𝜏𝑟2(𝛼𝑛)−2𝑖ℑ𝜏𝑟2(𝛼1)⋯−2𝑖ℑ𝜏𝑟2(𝛼𝑛)⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦.
- Add 𝜏𝑗(𝛼𝑘) to −2𝑖ℑ𝜏𝑗(𝛼𝑘) giving
𝐴4=⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣𝜎1(𝛼1)⋯𝜎1(𝛼𝑛)⋮⋱⋮𝜎𝑟1(𝛼1)⋯𝜎𝑟1(𝛼𝑛)𝜏1(𝛼1)⋯𝜏1(𝛼𝑛)――𝜏1(𝛼1)⋯――𝜏1(𝛼𝑛)⋮⋱⋮𝜏𝑟2(𝛼1)⋯𝜏𝑟2(𝛼𝑛)―――𝜏𝑟2(𝛼1)⋯―――𝜏𝑟2(𝛼𝑛)⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦.We see now that 𝐴4 =𝑇(𝛼1,…,𝛼𝑛) as defined in Discriminant of a separable extension,
and thus
covol𝜄(OK)=|det𝐴1|=2−𝑟2|det𝑇(𝛼1,…,𝛼𝑛)|=2−𝑟2√|Δ𝐾:ℚ|≠0as required.