What field is roots of in?
They are obviously not in by
Eisenstein's criterion or rational root
theorem. And since we've already known
are the
solution of it, and it's in ,
so the roots are in . What a
simple question.
But what if I ask, is there any smaller
field than that contains
? Consider
equipped with conventional addition and
multiplication in : clearly
it's closed under addition; since
,
it's closed under multiplication and thus
a commutative ring; since
,
every non-zero element is invertible,
and thus it's a field. It won't be hard
to verify that
is inside, so
is a
field that contains
and smaller than .
What's the point of finding smaller
fields such as
rather than sleeping on ?
You see, in order to reach the field
, we cheat twice in definition.
First, we cheat by allowing "infinite"
times (defined by converging series) of
addition, subtraction, multiplication and
division to be closed in the real numbers
, and then cheat by allowing
even to be solvable by
defining its root , and with
some tricks of complex analysis, we've
proved the fundamental theorem of algebra
that all polynomials over
can find all its roots lying in .
This is just like when you take a taxi
and the driver asks you the destination,
you simply reply "the Earth", something
that is definitely right (since you will
need rocket for space travel) but hollow.
When I take out a random quintic equation,
e.g. , by "not solvable by
radicals" it doesn't mean there's no
solution, it doesn't even mean there's
no explicit formula to express the solution:
clearly it has at least one root in
, and we can approximate the
root by leveraging the Newtonian method
to a infinite series of summation, and the
four roots remaining can be expressed by
this root with radical solution of
quartic equations. By "not solvable by
radicals", I mean you cannot breaking
down into some radical
equations (in the form of ), while
all quadratic, cubic and quartic equations
can. To judge whether an equation is
convertible into radical equations,
especially for quintic equations and higher,
we need Galois theory, in which finding and
studying "smaller fields than "
is the baby step, but a giant step.
Field Extensions
A field extension is a binary
relationship between base field and
extension field such that is
a subring (or we can safely use subfield
in this scenario) of . The slash symbol
can't be intepreted as quotient field
of over .
Conventionally, We also adapt the symbol
to represent the relationship
of two or more fields chained together by
field extensions (aka. Tower of fields),
where the subset symbol does not merely
mean by set inclusion.
We can come up with some instances of field
extensions quite quickly: like
,
and . At the first glance you may
be wondering what good will such a simple
relationship do, but we are about to add
some important details to it.
Smallest subfields of different characteristics
We would like to prove that any field of
characteristic is an extension field of
(which is not necessarily a proper subfield),
such that is isomorphic to .
First, given that is of characteristic
, we have a cyclic subgroup
generated by
, and obviously there's
. Since every
element in can be written as
, by equipping it with ring
multiplication, following the ring's axioms
there must be
,
so is closed under
multiplication and isomorphic to .
Then, obviously is integral domain, since
we've proved that
any field containing integral domain must also include a subring isomorphic to 's fields of fractions ,
we have
,
and we are done.
With analogous rationale, we can prove that
any field of characteristic is an
extension field of such that
: the cyclic
subgroup generated by is
, by
equipping it with ring multiplication we have
,
finally we have
.
In this way, the subfield related to
and created using the methods
we've described above is the smallest.
The term "smallest" means you can't make
smaller subfield, as a field must always contain
the unity , with the addition of
made closed while the elements
inside made invertible. And the structure of
such smallest subfield is completely defined
by the characteristic of field: a field of
characteristic contains the smallest
subfield isomorphic to ; a field
of characteristic contains the smallest
subfield isomorphic to .
On the other hand, since the process of field
extension will never mutate the ring operations
defined in the smallest subfield, a field
extension should never exists if
.
Simple extensions
The process of human's finding solutions to
algebraic equations is the process of finding
"new numbers". Through solving the equation
by the Pytagorean
theorem, Hippasus discovered the root is a
"new number" that is not in .
Today we know it's an irrational number and
in or .
Through radical solution to cubic equations,
Cardano adopted the existence of square root
of , which is also a "new number" to
. Although Cardano said it was
"an agony" dealing with the "new number" and
took only root in real numbers, people were
unaware of the usefulness of the "new number"
until the proof of fundamental theorem of
algebra by Gauss, naming the "new number"
to imaginary unit and enlarging
the number field to .
The phenomenon behind finding "new numbers" is
astonishing: whenever there's polynomial with
no root in the base field, an extension
containing the root field always come up in time,
e.g. has root
,
has root
.
While it is not just a coincidence, we will
present the theorem that for a field and an
irreducible polynomial such that
,
there's always a field extension such that
.
Given that there're already real instances of
this theorem, e.g.
and ), why don't we
observe them first? In the field
, we can always use
as a term that remains intact under
the field operations of ,
except for replacing with .
Analogous thing happens for , in
which we will need to replace
with . Actually, it won't be hard to find that
we can operate and in
the way just like the indeterminate of the
polynomial ring . When it comes to
replacement, we will just need to the evaluate
the canonical homomorphisms of
and ,
where there're
and
.
Finally, it won't be hard to verify that
defined by
and
is field isomorphism, and so does
defined by
and
.
In this way,
and
describe
the structure of and
completely.
Such observation can be elevated to our case:
since is constrained to be irreducible
polynomial, by the theory of polynomial rings
we know
is maximal, therefore
is field. By our assumption, if the extension
field truly exists, it won't be hard to
verify that there's ring homomorphism
defined by
and
. Since
is already a field,
the kernel of is only zero or the whole
field, while it's clearly not the latter case,
so must contain a subfield isomorphic to
. In the meantime,
is the root of , which can be
easily verified by:
In this way, the field
is the
extension field of such that
,
so the existence of the extension field is
proved. However, we would like to take one
more step to claim and prove that
,
which means it's an -dimensional vector
space over . From the theory of polynomial
rings we know
,
where all polynomials ,
are mapped to distinct coset ,
and all polynomials
are mapped to
with being obtained by dividing by ,
using the division algorithm of . And by
applying the ring isomorphism of we have
.
The field
is the extension field of we want, and
is called a simple extension
of , obtained by adjoining element
to . Clearly the polynomials
associate with shares the same root of
and generate the same principal
ideal as , for convenience
we pickup the monic one among them and
call it the minimal polynomial of
over , while is called
the degree of over ,
describing the dimension of when
it's viewed as finite dimensional vector
space over .
One can easily verify that
.
So for covenience, we allow to
appear in the slot of element to adjoin such
that , where the minimal polynomial
of over is , and the
degree of over is .
Consider a simple extension
where the minimal polynomial of
over is , we would like to
prove any polynomial that has
root in must be divisible
by . Let's do Euclidean algorithm in
so that we have
.
Given that is irreducible,
might only be or . Assume
it's the former case, then we have
,
which is a contradiction. So it's only the
latter case and we are done.
In this way, we can prove the minimal
polynomial of over is unique:
assume there were distinct minimal
polynomials of over
, we have ; since both
and are irreducible, they can
only be associate; since they are both
monic, there can only be , which is
a contradiction to 's distinction.
Conversely, the element defined by
minimal polynomial may not be unique:
both are the root of
; both
are the root of
; and both
are the root of . More
interestingly, in the field extension
that contains distinct roots
of the same minimal
polynomial , given that there's
,
the extension fields obtained by
adjoining to must be
isomorphic, and with little effort one will
see the isomorphism is
defined by
and
. Such pair of
are said to be
conjugate over . This means one may
not tell one extension field from another
without prior knowledge of and operating
merely on "alphabetic" level (that is,
presuming the root of to be
in ).
Algebraic extensions and transcendental extensions
In a field extension , an element
is said to be algebraic over
if it's the solution to any algebraic
equation in ; otherwise it's said to be
transcendental. The field extension
is said to be algebraic extension
if all elements in is algebraic over , and
transcendental extension otherwise.
We can quickly come up with some instances:
the extension
is algebraic, by every equation
can be converted into
, and
; the extension
is transcendental, since
is transcendental and
not the root of any algebraic equation
in . An transcendental extension
may contain algebraic elements (e.g.
), but must contain at
least one transcendental element.
For a field extension that with an element
, then let be the
smallest field including every elements in and
the element , we would like to show that
is algebraic iff there's
where
is the minimal polynomial of over ,
and is transcendental iff there's
.
To make the smallest field, we first take the
"molds" of possible addition, subtraction and
multiplication operations for the element
to adjoin with existing elements in
, and the "molds" form the set ; then
we put the element into the molds to
evaluate the smallest ring, this is done by
evaluation homomorphism
with its domain
trimmed to , resulting in the ring
; finally any field
that contains must also
contain its field of fractions, so the smallest
field containing and must be
isomorphic to
.
When is algebraic, we can find its
minimal polynomial over ; and since
we know
when 's domain is trimmed to
, we can draw the conclusion that
;
finally since
is already a field, we have
,
and therefore the simple extension of
adjoining coincides with the
smallest field containing and .
When is transcendental, there's no
non-zero polynomial such that
; so we know there's
,
and thus ;
finally by
we can draw the conclusion that the smallest
field containing and is
isomorphic to , which is the field
of rational functions over .
Finally, since 's being algebraic or
transcendental are mutually exclusive and
complementary, there's no other case of
structure of and we can safely
identify whether is algebraic or
transcendental by the sturcture of .
Finitely generated field extensions
In the field extension , for elements
,
we define
as the smallest field containing and
,
and call it the finitely generated
field extension. Obviously such a smallest
field can be formed by substituting in
indeterminates
of with
.
But we would like to show such a smallest
field can also be formed by extending with
consequently in arbitrary order, by adjoining
elements one at a time.
First we would like to show the equivalence
of indeterminates for substition, that there's
,
assume they are not equal by some rational function
such that
,
but given that there's rational function
(tips: define a rational function that
takes in the parameter in the order of
mitigating the permutation by
) such that
,
the assumption is false.
Then we would like to show the field
can be formed by adjoining to
,
that there's
.
First we have by simply aggregating coefficients of
, thus we have
under the meaning of set inclusion.
Conversely, we have by the operational rules of
,
thus we have
under the meaning of set inclusion.
Finally, by they contain each other under
set inclusions, we have
.
In this way, we've shown that
can be obtained by adjoining
to in arbitrary order, forming a tower
of fields that
.
Finite extensions
Let's consider an instance of finitely
generated field extension, that on the top
of , we may adjoin
to it, forming the field
which is
also also a finite dimensional vector space
over the base field with basis
, which is
obtained by multiplying basis
of over
and the basis of
over
. The same thing
will happen if we adjoin to
, as we've proved.
In order to generalize such kind of field
extensions, we introduce the concept of
finite extension: field extension
is said to be a finite extension when
is finite dimensional vector space over
, with its dimension onventionally
denoted as .
We would like to prove that all finite
extensions are algebraic. Let there
be , assume there's
transcendental element , then
for any , there's
by 's transcendence, which means
forms a basis of -dimensional vector
space. This is not true when ,
since is -dimensional vector space
over , and the elements
taken from are always linearly dependent,
impossible to build up the basis.
And we would like to prove the Tower law,
that if are both finite extensions,
is also finite extension with
.
Let there be
,
the basis of be
,
so that the elements of are in the form of
,
and the basis of be
,
so that the elements of are in the form of
.
So eventually the elements of can be
represented as
.
To prove that
are linearly independent, assume they are
linear dependent instead by
for a set of that're not all
zeroes. Combine the coefficients of
s by
,
since
are linearly independent, not all of the
are zeroes.
But this will imply
for a set of that are not all zeroes,
and that
are linearly depdent, which is a contradiction.
In this way, we've proved that is a
finite extension of with basis
,
and of dimension .
The instance we've came up with is a also a good
instance of the Tower Law, that there's
while
and
.
Finally, we would like to show that every
finitely generated algebraic extension,
the finitely generated field extension with
all elements to adjoin being algebraic, is
a finite extension, and every finite
extension is a finitely generated
algebraic extension. Consider the finitely
generated algebraic extension
,
the intermediate extensions
are simple extension and thus finite
extensions, and by the Tower Law we know
must also be a finite extension. Conversely,
let there be finite extension with basis
,
and all elements of are in the form of
,
which must be in
.
But if we allow to be proper subset of
,
given that
are all elements of and their field
operations are made closed in , it's a
contradiction to
's
being smallest field containing
and ,
so there must only be
.
And since
are
all algebraic, must be a finitely
generated algebraic extension. In this way
we've shown the interconnection between
finitely generated algebraic extensions and
finite extensions.
Splitting fields
Inspired by the fundamental theorem of field
extensions, for any algebraic equation
, which can
be asserted to be monic, without losing
generality, we try to extract the roots of
by factorizing into
over some "sufficiently large" field, using
the following algorithm:
- Initialize .
- If , halt; otherwise
increment by .
- Take out any irreducible factor ,
of over .
- If is of degree , then there's
,
so we just need to update
directly and go back to step 2.
- Otherwise has no root in ,
but we can make simple extension
such that
is the minimal polynomial of
over . So we will update
and go back to step 2.
In the algorithm, we factorize by
plucking factor
off ,
one at a time, resulting in a factorization
of with exactly factors of
over that's "sufficiently
large". Thus the algorithm will halt exactly
at with
.
Apparently the algorithm sheds light on the
existence of the splitting field
of , which
is the smallest field (not just "sufficiently
large", but perfect at size) containing all
roots of . We
can also say that
splits .
Problem of uniqueness of splitting field
Let's evaluate the splitting field of
following the steps
of our algorithm. It's already irreducible by
Eisenstein's criterion, and
is one of the roots and on the number axis.
But given that there're other conjugate roots
off the number axis on complex plane, the
extension field
does not split . But now we have
over .
The roots of
are
the other two conjugate roots
of , where
is the primitive cubic root. And by
merely adjoining
to we get
,
in which there's
.
So
is the splitting field of with
.
On the other hand, you can first adjoin
to ,
and then there's
,
where the
contains the other two conjugate roots
and is irreducible for being a
cyclotomic polynomial of prime order.
And when we adjoin to
,
clearly the
contains all of the roots and thus
splits . In fact, given that we've
known the three roots
,
simply adjoin all of them to
and then
can be expected.
In general, given a field extension
where contains
all roots of any polynomial
(the is referred
as the algebraic closure of ,
which we will not formalize in this text
immediately though), e.g.
, then the
splitting field of must be uniquely
.
Otherwise for any other splitting field
of
whose roots
differ
from for at
least one element. So has
roots, or is not UFD by
,
choose your favorite contradiction.
In such manner, given that splitting
field of must be unique as
with
's roots in being
, what's
problem of "uniqueness of splitting field",
as is mentioned in the title?
The problem is that we presume there's a
unique algebraic closure of ,
identified by set elements. Such
hallucination is rooted in our stereotype
that polynomials over should
be eventually splitted over .
To cure the hallucination, we need our old
friend .
Consider the field
and algebraic equation
over .
Clearly the root is not in and we
need to extend. For the extension field,
apparently a candidate would be
,
which splits for
containing both
.
Meantime another candidate is
,
where every non-zero element is
invertible by
.
It also splits
for containing both roots
. There will
never be a unique algebraic closure
containing both
and
at the same time despite their being
"conjugate", otherwise
is observed. Therefore neither the
uniqueness of splitting field nor the
uniqueness of algebraic closure could
be expected. The uniqueness is identified
by set elements.
By now, should we conclude that the
splitting field is not unique? Well, it
is still in the midway of the story. The
two splitting fields
are actually isomorphic by
,
as their structure are totally defined by
the minimal polynomial .
In fact, we are able to prove the diverging
splitting fields are isomorphic, and their
uniqueness can be identified by
field structure.
Splitting field is unique up to isomorphism
It won't be hard to verify that there's
,
while the "roots" are defined by
,
and the minimal polynomial
and of them are interchangeable
under the isomorphism of .
In fact, such observation is the starting
point of the whole proof.
Consider there're two isomorphic fields
connected by field isomorphism
, it won't be hard to see
there's an induced field isomorphism
defined by
.
Obviously for an irreducible polynomial
, must also
be irreducible, and vice versa. So both
and
must be fields. Given that there's
,
by observing the cosets we can simply
conclude that
.
Finally by simply substituting in the
roots defined by
and
, we
can conclude that
.
For polynomial , the existence
of some field that's large enough to
contain all roots of is obvious. Without
losing generality, assume there're distinct
fields that contain all roots of ,
we can pluck off
, which are both defined by the same
minimal polynomial over ,
there must be ,
defining field isomorphism
such that
.
And for any factorization
over
, there must also be
over
where
is
the field isomorphism defined by
.
If already splits , then
we are done. Otherwise we can pluck
off , which is defined
by minimal polynomial ,
meantime we pluck off
defined by minimal polynomial
. In this way,
there must be
defining field isomorphism , and
defining field isomorphism .
Continue on such process of factorization
until splits
in , there must be an isomorphic
field
splitting in , while internally
is unique in
and
is unique in . In this way, we've
proved that the splitting field of
must be unique up to isomorphism.
And clearly the proof does not rely on
identical base fields, for isomorphic base
fields connected by field
isomorphism , the statement
still holds that for the splitting field
of and splitting field
of , there's
, where
is induced ring isomorphism between polynomial
rings by . This establish the statement
that splitting field
or of in
our example and the
splitting field of
are the same
up to isomorphism.
Finite fields
One of the most important object in field
theory is finite field. A finite field
is, namely a field of finite order. We may
also call it the Galois field in
memory of Evariste Galois. It can be shown
that a finite field can only be of order
, where is prime and is
positive integer, and all finite field of
the same order are isomorphic. Such
non-triviality makes finite field prominent.
Finite field can only be of prime power order
For a finite field , since it's finite,
it cannot be of characteristic , so
its characteristic must be some prime ,
and we've already shown it contains a
subfield .
Let and
be elements outside . Clearly there's
no transcendental element in over ,
so we can make finite extension
, which is
of order when
.
On the other hand, all elements of
has been adjoined, so it can only
be .
In this way, when is of characteristic
, and , the
order of can only be .
Separable polynomials
A polynomial is said to be
separable when it has no duplicate
root in its splitting field. Such kind of
polynomials plays an important role in
proving the isomorphism of finite field
of the same order.
The formal derivative on polynomial
ring is a map
defined by
,
We would like to claim and prove that
polynomial is separable iff
.
Let's derive some basic properties of the
formal derivative that we will use first:
So the formal derivative and the derivative
in calculus have operation rules in common.
However formal derivative does not rely on
the feasibility of defining limits, which is
generally not possible for fields, especially
for finite fields.
Let separable polynomial be
factorized into
with distinct roots
in its splitting field. It won't be hard
to find there's:
And for every irreducible factor
of , it divides every
factor
except for . So
whenever is separable, there's
.
Conversely, without losing generality,
let polynomial
be of duplicate root , and it
won't be hard to find there's:
Apparently there's
,
and we are done.
Finite fields of the same order are isomorphic
More precisely, we would like to show all
fields of order are isomorphic to
the splitting field of
.
First, it won't be hard to find there's
,
and thus
,
which means the polynomial
must be separable in its splitting field,
with distinct roots.
Then we would like to show that the roots
of forms a field. One should
remember the Freshman's dream
that in commutative
ring of characteristic . And given
that , we can repeat such
process so that there's
.
In this way, we can easily conclude there's
in of characteristic . Assume
are the roots of
, we have:
- and .
- .
- .
- When , .
- When , .
- When , .
In this way, the roots of
forms a subfield of its splitting field,
with elements and it's the extension
field of . However, recall
how will we make a splitting field: for roots
outside , we adjoin it to
until it splits ,
and we've already known the elements in the
splitting field are obtained by field operations
of with roots outside
. We have just proved they
will result in another root, thus the splitting
field is subfield of , and is the
splitting field of itself. In
this way, we've proved the existence of the
finite field of order , whose structure
is the splitting field of polynomial
.
For the uniqueness of finite field of order
, first we know its multiplicative group is
cyclic, so every elementsin fulfils
, and
every element in fulfils
including .
Then we also know contains a subfield
, and can be
viewed as a polynomial over . When we "grow"
the splitting field of towards ,
by adjoining root of current irreducible
polynomials merely from until it splits
, we will find the splitting field
is a subfield of with the same amount of
elements as , so coincides with the
splitting field of . Finally, since
and both and are
splitting fields of the same polynomial up to
isomorphism, we have .
In this way, we've proved the finite field of
the same order must be unique up to isomorphism,
thanks to the uniqueness of splitting fields.
Conventionally, we denote the unique finite
field of order as
or . Obviously we can derive
from the process of deduction above.
Subfields of finite fields
We would like to show
contains a unique subfield isomorphic to
iff .
First, if would like to
contain any subfield, clearly it can only
be , and
both
and are
finite extensions. By the Tower law, we have
,
so such a subfield
exists only if there's .
Then, for the existence of such subfield,
consider the modulo- reduction
of polynomial to
,
obviously the image of factor polynomial
of over is the
factor polynomial of over
. And we know there's
over , then we have
over . Obviously
is also separable so there're
roots contained in the factor
of . Obviously
those elements in are roots
of , we can "grow" the splitting
field in the direction of splitting
first, and then clearly the
subfield that splits is exactly
the same as . In this way,
has a subfield isomorphic
to if there's .
Finally, for the uniqueness of such subfield,
obviously the is
the subgroup of .
Since the multiplicative group is cyclic,
the subgroup of order is unique as
, so the subfield
must also be unique.
Finite fields can be obtained from simple extensions
We would like to show that for every finite
field , let
be the primitive root and
be the minimal
polynomial of over , then
.
Which means every finite field is extension
field obtained by a simple extension adjoining
primitive root over .
First, obviously is algebraic over
, by the theory of algebraic
extension we know there's
,
all we need to do is to find the image of
.
Then, for the evaluation homomorphism
, from the domain of
coefficient and the element taken into
evaluation we know the image is subset of
. And by
and
,
we know the is surjective on
its image , thus we have
and
.
Finally, when these two pieces are putting
together, we can conclude that there's
and we are done.
The importance of this theorem is, it provides
a very simple way to visualize finite fields
of and do field
operations on it, which is done by polynomial
ring operations of modulo
minimal polynomial of the primitive root.
Conclusion
In this text, we discuss some basic properties
of field extensions. First enlightened by
human's solving equations we discovered a kind
of field extension called the simple extension,
claiming that for each irreducible polynomial
that has not root over the base
field , there's inevitably a extension field
contains a
root of , by the PID property of polynomial
ring over . Then we introduced the concept
of algebraic and transcendental elements and
extensions, showing the structure of field
with single algebraic or transcendental adjoined
to it. Finally we introduced the concept of
finitely generated extensions and finite
extensions, proving that finitely generated
extension adjoining the same set of elements
to base field is unique and all finite
extension must be finitely generated,
and vice versa.
We then dig into human's solving equation and
introduced the concept of splitting field,
which is the smallest field containing the
base field and all roots of the equation,
and splits the polynomial of the equation.
We also show that the splitting field is
always not unique when identified by set
elements, but always unique up to isomorphism.
Finally, equipped with tools of field
extensions we explore the properties of
finite fields. We've shown that all finite
fields has the order of prime power ,
and isomorphic to the splitting field of
. We've shown
finite field has a unique
subfield iff
there's . And we've shown every
finite field can actually
be obtained by simple extension of
adjoining primitive root
, providing an alternative way of
doing field operation as doing polynomial
ring operations modulo minimal polynomial
of the primitive root.
The concepts and properties of field extensions
and finite fields are the foundation of fields
and Galois theory.