This post offers a summary note on the norm, trace, characteristic polynomial and discriminant of an element , a number field of degree over .
-linear transformation: For any , we can define a -linear transformation , such that for some . Such an can be represented as a matrix (with respect to a basis for over ), denoted by . We define the and of as follows: The characteristic polynomial is defined as, where is the indeterminate and is an identity matrix.
Example-1: When we exactly know the representation of the algebraic number with respect to a basis in the number field , where , . acts on the basis elements as, which gives,
Example-2: When we do not know the algebraic number in terms of its basis elements, but know its minimal polynomial over , denoted as . Let be the root of the irreducible polynomial . What are the trace and norm of ? Since is irreducible, let’s choose the basis . acts on these basis elements as, which gives, So far so good! What about the characteristic polynomial ? As per Eq. ,
This is not a coincidence that is identical to the minimal polynomial of over . In fact,
Proposition-1: for some .
Proof sketch. Prove using Caley-Hamilton theorem for , then apply induction. ◻
Counting embeddings: There exist alternative equivalent ways to define trace and norm via embeddings of a number field in an algebraic closure, say . For our purpose here, an embedding is an injective homomorphism from to that fix pointwise.
Let be embeddings of in , where , we can also define trace and norm of as,
Well-definedness of trace and norm: How to see that these two definitions (via -linear transformation and through embeddings) are equivalent?
Proof sketch. As per Eq. , is the sum of eigenvalues of , which is identical to sum of the roots of . As per Eq. , is the sum of roots of . The relationship between characteristic and minimal polynomials of via proposition-1, therefore, completes the proof. Similar argument works for the norm. ◻
Discriminant: Two (equivalent) definitions of the discriminant of , where for all ,
, if we think embeddings only.
Well-definedness of discriminant: .
Proof sketch. , and then apply properties of determinant. ◻