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. ◻