# Abstract Algebra/Factorization

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.

- Theorem 4.2.2 (Gauss' Lemma)

Let be a primitive polynomial. Then is irreducible over if and only if it is irreducible over .

**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. Thenwith . If , we are done. So suppose . We can write as a product of primes . Modding out by we getwhere are the corresponding polynomials in (in other words we mod each of the coefficients by ). Since is an integral domain, this means that at least one of the factors is 0. Without loss of generality we can assume that . But this means that all of its coefficients are a multiple of . Therefore we can cancel out from both sides of the equation . This leaves primes on the left hand side. We can apply the same argument and conclude via induction that is reducible over .

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

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.

- Lemma 4.2.3 (Eisenstein's Criterion)

Let be a prime in and is a polynomial in . Suppose divides all the and does not divide the constant term . Then is irreducible over and over .

**Proof.** Suppose were reducible. Then there exist non-constant, monic polynomials such that . Consider the polynomial in the quotient . We find that

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 where is a prime. We observe that In particular then is a monic polynomial where every non-leading coefficient is a multiple of and the constant term is exactly . Therefore by Eisenstein's criterion is irreducible which in turn means that 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.

## Exercises

edit- 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.*