One of the main motivations of this study is to determine the roots of a polynomial over a field. It is obvious that the roots of a product of polynomials is just their union (and in fact the multiplicities sum). Thus a good first step is to determine whether or not the given polynomial is a product of lower degree polynomials.
Recall we say a non-constant polynomial
is reducible if there exist non-constant polynomials
and
such that
. Otherwise, the polynomial is said to be irreducible. It is immediate that linear, i.e. degree 1, polynomials are irreducible. For low degree polynomials, it is easy to determine whether or not they are irreducible.
- Lemma 4.2.1
If
is a degree 2 or degree 3 polynomial, it is reducible if and only if it has a root.
Proof. This simply amounts to noting that if
is of degree at most 3 then, any decomposition of the form
must have at least one of
or
be linear.
Note that the statement does not hold for higher degree polynomials. For example
has no roots in the rationals but
so it is reducible.
A case we are particularly interested in is polynomials over the rationals. A very useful theorem for this is the Rational Root Theorem which is a consequence of Gauss' Lemma.
Proof. First we show that if
is reducible over
then it must be reducible over
. Suppose we have
where
and
are non-constant polynomials in
. Suppose
is the lowest common multiple of the denominators coefficients of the right hand side. Then
![{\displaystyle df(x)=a(x)b(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c16d000ba4ba75e71d3427acb715fadc3df7ccf)
with
![{\displaystyle a(x),b(x)\in \mathbb {Z} [x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8eff11d6583aaf75df09cc55a16b35c251cc571)
. If
![{\displaystyle d=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/958b426c5ace91d4fb7f5a3becd7b21dba288d50)
, we are done. So suppose
![{\displaystyle d>1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f3dc6d8c61675b7e83b9bba7a236569af521f6a)
. We can write
![{\displaystyle d}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e85ff03cbe0c7341af6b982e47e9f90d235c66ab)
as a product of primes
![{\displaystyle d=p_{1}p_{2}\cdots p_{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e4bf634b5a4928a9138fd69771e5c5ff5d0fee3)
. Modding out by
![{\displaystyle p_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9b58f22283ca46dd5da309cc34303b06a797783)
we get
![{\displaystyle 0={\overline {a}}(x){\overline {b}}(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/846e138a3dc02ab30b2d6b35ace1c445dde95992)
where
![{\displaystyle {\overline {a}}(x),{\overline {b}}(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf2b2c9c841e1d5909b318ac2143a87ea5a95ceb)
are the corresponding polynomials in
![{\displaystyle \mathbb {Z} /p_{1}\mathbb {Z} [x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/204b372bb62fd184edd3ae41bc9186ae3f1a113a)
(in other words we mod each of the coefficients by
![{\displaystyle p_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9b58f22283ca46dd5da309cc34303b06a797783)
). Since
![{\displaystyle \mathbb {Z} /p_{1}\mathbb {Z} [x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/204b372bb62fd184edd3ae41bc9186ae3f1a113a)
is an integral domain, this means that at least one of the factors is 0. Without loss of generality we can assume that
![{\displaystyle {\overline {a}}(x)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/945944244fbafbd7a9165ba5a878df4e01defaf8)
. But this means that all of its coefficients are a multiple of
![{\displaystyle p_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9b58f22283ca46dd5da309cc34303b06a797783)
. Therefore we can cancel out
![{\displaystyle p_{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9b58f22283ca46dd5da309cc34303b06a797783)
from both sides of the equation
![{\displaystyle df(x)=a(x)b(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c16d000ba4ba75e71d3427acb715fadc3df7ccf)
. This leaves
![{\displaystyle n-1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fbd0b0f32b28f51962943ee9ede4fb34198a2521)
primes on the left hand side. We can apply the same argument and conclude via induction that
![{\displaystyle f(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/202945cce41ecebb6f643f31d119c514bec7a074)
is reducible over
![{\displaystyle \mathbb {Z} [x]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d4da3ac703cc7721ebba91a53f6752de7157124)
.
The converse is easy to see since a decomposition over
is in particular a decomposition over
. Since the coefficients don't share a factor this means that the decomposition is into a product of non-constant polynomials (in particular we avoid cases like
which is a non-trivial decomposition in
but a trivial one in
since
is only a unit in the latter ring).
In particular this makes it easy to determine when a polynomial over the rationals has a rational root. Given
, suppose it is monic with integer coefficients. Then if
has a root we can write
where both factors have integer coefficients by Gauss' Lemma. Therefore in particular
is an integer and must be a factor of the constant polynomial. If
is not monic and its has a rational root
then we would be able to write
![{\displaystyle f(x)=(px-q)(a_{n-1}x^{n-1}+\dots +a_{0})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6115c8fee8edbf3ff9c5cc0ea959ac31526b77bf)
In particular
is a factor of the leading coefficient of
and
is a factor of the constant term. This is known as the Rational Root Theorem. By trying all these possibilities, one can immediately determine whether or not there exist rational roots of a given polynomial over the rationals. (even if the coefficients are rational, one can multiply the polynomial by an integer to obtain a polynomial with integer coefficients and then work with this scaled polynomial, which has the same roots as the original).
Now that we know trying to reduce polynomials over
is (essentially) equivalent to reducing them over
, it's useful to have some irreducibility criteria from the latter case. A very useful result for this is Eisenstein's criterion.
Proof. Suppose
were reducible. Then there exist non-constant, monic polynomials
such that
. Consider the polynomial in the quotient
. We find that
![{\displaystyle x^{n}={\overline {g}}(x){\overline {h}}(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64048cd53f9c9f19c68b12d2535fb0137b30c658)
where
and
are the respective polynomials modulo
(in other words
where
is the canonical projection map). Since
and
were taken to be monic, we know their reductions
and
are also monic and hence non-constant. In particular then
is a non-trivial decomposition of
.
By comparing coefficients above, we see that the product above has no constant term. Therefore at least one of
and
has no constant term (this is where we use the fact that
is prime so in particular
is an integral domain). Suppose one of them does have a non-zero constant term. Then their product would contain lower degree terms but we know their product is exactly
. Therefore both
and
have no constant term. But this means the constant terms of
and
are both multiples of
so in particular their product
is a multiple of
leading to a contradiction.
Example: From Eisenstein's criterion, it is immediate that
is irreducible over
and hence over
(by Gauss' Lemma). This is one way to show that
are all irrational for
.
Example: Here is a more sophisticated example. Consider the polynomial
![{\displaystyle f(x)={\frac {x^{p}-1}{x-1}}=x^{p-1}+x^{p-2}+\cdots +x+1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c679a8d60381c1a2709ee646c2d2872c56f9f05b)
where
![{\displaystyle p}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81eac1e205430d1f40810df36a0edffdc367af36)
is a prime. We observe that
![{\displaystyle f(x+1)={\frac {(x+1)^{p}-1}{x}}={\frac {1}{x}}\sum _{k=1}^{p}{\binom {p}{k}}x^{k}=\sum _{k=0}^{p-1}{\binom {p}{k+1}}x^{k}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58f7689e79149fe536eb5f8f03f5920c6946eb16)
In particular then
![{\displaystyle f(x+1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0b442c73d179ec2a0a715c1dfd848a7794c8e5b)
is a monic polynomial where every non-leading coefficient is a multiple of
![{\displaystyle p}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81eac1e205430d1f40810df36a0edffdc367af36)
and the constant term is exactly
![{\displaystyle p}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81eac1e205430d1f40810df36a0edffdc367af36)
. Therefore by Eisenstein's criterion
![{\displaystyle f(x+1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e0b442c73d179ec2a0a715c1dfd848a7794c8e5b)
is irreducible which in turn means that
![{\displaystyle f(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/202945cce41ecebb6f643f31d119c514bec7a074)
is irreducible.
Once we have an irreducible polynomial, we know we can't decompose it any further so we need to start working with field extensions and splitting fields.
- Show that
is irreducible over
.
- Find all irreducible polynomials of degree at most 3 over
and
.
- Show that
is irreducible over
.
- Show that if
is irreducible then
is prime. Hint: Prove the contrapositive.