Disclaimer: This text is neither a formal
proof of why algebraic equations of degree over
5 are not solvable by radicals, nor a rigorous
text with its content carefully cross validated.
It is just a text the author write for fun.
You may safely skip this text without worrying
about missing some important concepts, and you
must judge by your own if you decide to read.
Some of us might have already heard of the
famous theorem that algebraic equations of degree
over are not solvable by radicals, and by the
text of permutation group
we've known that is simple when it is
on over letters, should these two fact have
connection? The answer is, yes.
Given an algebraic equation of degree :
When it comes to you, what are you thinking
about? Are you meditating on some magical term
to transform and simplify it into form like
,
just as what we were taught in middle school?
According to the fundamental theorem of algebra,
an algebraic equation of degree has
roots. Noted that:
By associating terms of corresponding degree,
we have the Vieta's formula:
While
is itself an linear equation, if we can find
some method to generate more linear
equations in the form of
,
to form a system of linear equations of rank
, the equation is solved equivalently. One
way to do so is to assign values to
that is linearly
independent from all s and previously
assigned value. When an equation is solvable
by radicals, each
is representable by formula of
combined by
finite times of addition, subtraction,
multiplication, division and root extraction.
Substituting these formula into each
will yield us each equation in the system
of linear equations.
Conversely, for any algebraic equation, if you
can prove it is impossible to find such a system
of linear equations, the algebraic equation is
not solvable by radicals.
Symmetric polynomial
Inspired by how we solve the quadratic equation,
we would like to construct such a linearly
independent equation
by expanding the
and see how we can use terms in Vieta's
formula to express it.
In the beginning, things go quiet fluently
for the quadratic equation, since there's:
However, when it comes to the cubic equation,
we encounters the first obstable that:
I won't expand any further since it's enough
for demonstrating the problem. Unlike what
we've done so easily with the quadratic case,
there're terms related with that are
not possible to get rid of.
While you might still disbelieve and argue
that might still be
expressed if more exhaustive reduction is
taken, or another choice of coefficient for
is tried, let's see how the case of
and
are fundamentally different from each other.
For a polynomial of roots
,
define the permutation group action on
polynomial as
,
and obviously there's
.
Polynomial is a symmetric polynomial
if it is unconditionally invariable, that is,
.
All terms from the Vieta's formula are symmetric
polynomials, which can be proved by swapping roots
in and
see how coefficients
are invariable and so do terms in Vieta's formula.
We call the symmetric polynomials from Vieta's
formula elementary symmetric polynomials,
since they are directly connected with
coefficients of the equation we want so solve.
The is also symmetric since
. While
is not symmetric:
randomly choose some cubic equation with root
, there're
and
.
What's more, for two symmetric polynomials
, their combination
,
where defines and depends only on the
result of , is still symmetric since
.
The combination could be addition, subtraction,
multiplication, division, or functions depending
only on value of . So basically you cannot
express a asymmetric polynomial in combination
of symmetric ones.
If we want to construct an equation
linearly independent from
,
there can't be ,
and among them we're always able to select some
. Consider the polynomial
with which we construct a linearly independent
equation, we would like to show when
the polynomial cannot be symmetric.
Let's rewrite the polynomial as
,
where contains all the terms besides
and . After swapping
and , the polynomial turns into
.
The equation has solution
, where
. So if we
want the polynomial to preserve its origin
value after swapping and ,
the following condition must hold:
To simplify, we denote as
,
which means the portion of equals
unconditionally, and denote as
its opposition.
When , the polynomial is
in the form of , by it is not
possible to be symmetric except for :
just choose an equation with solution
, swap and
you will see the result.
When , since merely swapping
and does not touch coefficients
of terms in , it is not possible for
. When , the condition turns
into , in
order for the condition to hold regardless of
the choice of , there must be
, while is
requried as a precondition. So the polynomial
cannot be symmetric when .
When , since
it's always possible to choose another
in . Take
out from and put back to
and we get
.
Repeat the discussion of above
and we will get the same result.
By now, we conclude the proof of polynomial
in the form of
will never be symmetric when .
Lagrange resolvent
The statement above seem to be telling us
equations of degree over are not
solvable by radicals, while in fact the
equations of degree and are actually
solvable, so there must be something wrong
with the reasoning. What is it?
While symmetric polynomial can be constructed
from composition of symmetric polynomials,
there's nothing prohibiting composition of
asymmetric polynomials to be symmetric.
In a cubic equation, consider the polynomial
,
under the permutations of
there're:
And under the permutations of there're:
Expand the polynomial and we get:
Noted that , so
let's combine and we get:
Similarily, let's combine
and we get:
Phew, after verbose mechanical deductions
of and ,
we find their values fit into roots of
quadratic equation
.
Although we are not able to represent
and using terms from Vieta's
formula directly, we are still able to
construct symmetric polynomials from them,
solve their values from a quadratic equation,
and finally form linear equations
and
,
combined with
we are able to solve the values of
.
Consider another asymmetric polynomial
,
under the permutations of there're:
Although we can still construct symmetric polynomials
,
,
and , we
will need to solve another cubic equation
in order to extract their values.
Transforming from a cubic equation to
another cubic equation is not reducing
the problem and is useless.
So the asymmetric polynomials
and
are
deliberately constructed, they have special
symmetries inside, reducing the problem
of solving a cubic equation to solving
a quartic equation.
Lagrange, who carefully studied the
solution for cubic equation published by
Cardano, and the solution for quartic
equation by Ferrari, was the first to
reveal the affinity between solution
by radicals and symmetries of roots, and
came up with the method of finding a
solution by radicals as above. Such
method is also called the Lagrange
resolvent. In Lagrange's theory,
he also came up with the Lagrange
resolvents for cubic and quartic
equations.
Criteria of symmetries
So what kind of symmetries should the
asymmetric polynomials in a Lagrange
resolvent has? For an algebraic
equation of degree , first of all,
we need a polynomial
of linear combination of roots, whose
th root contributes an equation to
the system of linear equations we used
to extract the roots. The polynomial
must be deliberatedly selected so that
it is helpful for reducing the problem.
For the symmetry group acting on
, it either preserve its original
form, or transform into another form.
The permutations preserving 's
original form forms a group
:
- (Closed) .
- (Identity) .
- (Invertible) .
- (Associative) By definition of and 's associativity.
For each left coset , it's
obvious that
.
and .
For convienience, denote the polynomials
result from different left cosets as
,
and each one is distinct.
Assume 's acting on
results in different polynomials
. Then over
these polynomials we build a symmetric
polynomial depending only on asymmetric
polynomials' values and masking out
their asymmetries. That is:
Which means 's acting on symmetric
polynomial is the same as permutating
asymmetric polynomials
in , and
due to 's permutation invariant nature
the result of keeps unchanged. But
none of these s may be dropped or
duplicated, otherwise by combining the
duplicate 's asymmetry, and 's
nature of allowing only permutation of
inputs, we can easily construct
contradiction to 's being symmetric.
Have you noticed this is an invitation
for constructing a group homomorphism?
It's the joint force of these polynomials
to define a map
such that
permutation of the roots are mapped to
permutation of these polynomials.
Let's focus on what happens to paramters of
in . For any ,
there's some ,
when acts on , it
permutates parameters of that
.
Then for any , there's
some , when
acts on ,
it permutates the parameters again that
.
Since
and ,
there's ,
and is a group homomorphism.
So the symmetrizing requires
each polynomial corresponds
to an element in quotient set
. Noted that
since
and is amomg these , there's
.
Finally by
we have . So in order
to create appropriate polynomials
for ,
must be held.
This is what we have exactly done with
cubic equation. Consider the subnormal
series of :
The polynomial
is fixed under , and under
there're
and
corresponding to each one of .
With this setup we derive the solution
by radicals for cubic equation.
While for the quartic equation, things
go complex. Consider the subnormal series
of :
If we are able to (though in fact not
able to) directly construct polynomials
in the form of
which is symmetric under and
linearly independent from
, there're
only linearly
independent equations generated and
is obviously insufficient.
So instead of creating polynomial
symmetric under directly, we
first create polynomial
which is symmetric under
and generate asymetric polynomials
under :
Under normal circumstance, we will need
to solve a sextic equation to extract
their values. But we are lucky this time
since we have in between. Let's
bundle them three by three so that:
Now we are able to build
which is symmetric under . The value
of can be extracted with a
quadratic equation. Then to extract
from and
from , we set
to Vieta's formula of cubic equation,
and extract the value of s with
cubic equations.
Since the work of solution to quartic
equation is too complex however well
studied, we will not talk too much about
it in this text. You can get more
detailed information from here .
Finally, for algebraic equation of degree
over , consider its subnormal series:
Similarily, if we are able to create
linearly independent polynomials which is
symmetric under , we only get
more linearly independent
equations. This is far from the requirement
of collecting more linearly
independet equations. We also can't follow
the route of solving quartic equations by
symmetrizing into intermediate polynomials
since they also require normal subgroups
in the middle to work but is simple.
So finally algebraic equation of degree
over is not solvable by radicals.
Conclusion
In this text, we characterize the step of
solving an algebraic equation as follow:
- Come up with some linearly independent
polynomial of roots
.
- Find what other polynomials can
transform into under permutations of
, so we have a set of polynomials
.
- Symmetrize these polynomials and see
how the coefficients of the polynomial
can be matched to them. Since the values
of each are
distinct, in order to extract their values,
we will need to symmetrize them into
Vieta's formula of equation of order
. Sometimes intermediate polynomials
are needed and we would have solved
multiple algebraic equations before we
can extract the value for
.
- With the extracted values of these
, we generate
more linearly independent equations.
- Solve the system of linear equations to
extract the values of
.
For an algebraic equation of degree over
, to extract the value of any linearly
independent polynomial we create for it,
we will need to solve an algebraic equation
of order or . The former is not
reducing the problem, the latter is not
providing sufficient equations to the
system of linear equations. So there's no
way we can solve an algebraic equation
of degree over by radicals.
Although this text is far from a formal
proof, actually when compared with the
formal Galois theory you will find there're
many points missing, but I hope you can
see in what way group theory objects can
be connected to other mathematical objects
from this text.