Proof. We start by defining the following
map:
Note that this map is clearly a
ring homomorphism. For all
![$ p(x),q(x) \in L[x]$ $ p(x),q(x) \in L[x]$](http://images.planetmath.org:8080/cache/objects/4723/l2h/img13.png)
:
Thus, the
kernel of

is an
ideal of
![$ L[x]$ $ L[x]$](http://images.planetmath.org:8080/cache/objects/4723/l2h/img17.png)
:
Note that the kernel is a
non-zero ideal. This fact relies on the fact that

is a finite extension of fields, and therefore it is an
algebraic extension, so every element of

is a
root of a non-zero polynomial

with
coefficients in

, this is,

.
Moreover, the ring of polynomials
is a principal ideal domain (see example of PID). Therefore, the kernel of
is a principal ideal, generated by some polynomial
:
Note that the only
units in
![$ L[x]$ $ L[x]$](http://images.planetmath.org:8080/cache/objects/4723/l2h/img28.png)
are the constant polynomials, hence if

is another
generator of

then
Let

be the
leading coefficient of

. We define

, so that the leading coefficient of

is

. Also note that by the previous remark,

is the unique generator of

which is monic.
By construction,
, since
belongs to the kernel of
, so it satisfies
.
Finally, if
is any polynomial such that
, then
. Since
generates this ideal, we know that
must divide
(this is property
).
For the uniqueness, note that any polynomial satisfying
and
must be a generator of
, and, as we pointed out, there is a unique monic generator, namely
.
