# 26 Rings of Polynomials

A ring of polynomials is a ring consisting of all polynomials with coefficients in the ring [7]. Similarly, the set of polynomials over a field , denoted , consists of all polynomials with coefficients in the field [17].

One idea we are often interested in is whether a particular polynomial can be factored over a given field. Consider the following definition by Macauley [17].

**Definition V.14**

A polynomial is

**reducible**over if we can factor it as for some of strictly lower degree.

We are already familiar with this idea. Whenever we factored a polynomial in a high-school math class, that polynomial was reducible over the field because every coefficient was a real number. Now, we are generalizing this same process to apply to any field.

Let’s look at an example to demonstrate this. This is a problem I completed for this paper that comes from Fraleigh and Katz [7].

**Example 58**

The polynomial can be factored into linear factors in . Find this factorization.

**Solution**

Since we are in , addition is defined as mod(5) and multiplication as mod(5). So, we have

Therefore, in .

In the example above, we see that each factor is of a degree less than 4 and has coefficients that are elements of , which is consistent with definition V.14.

As was also seen in high-school math classes, not every polynomial can be factored. Such polynomials are said to be irreducible.

#### Irreducibility

Sometimes, a polynomial is not reducible over a field . In this case, we say that is irreducible over [17].

One theorem, along with its corollary, is very helpful in determining if a polynomial is irreducible. Both the theorem and corollary are given by Fraleigh and Katz as follows [7].

**Theorem V.15**

**Corollary V.15.1**

These will make more sense after looking at an example that utilizes them. The following problem is one I completed for this paper and comes from Fraleigh and Katz [7].

**Example 59**

Show that is irreducible over .

*Proof.*

If we could reduce over , then would either have a cubic factor and a linear factor or two quadratic factors. This is because and . So, to multiply two factors together to get , we need either

or

We will now show why neither of these options is possible.

Suppose has a linear factor in . That is,

where , , , and are rationals. If , we have

So, is a zero, and is rational because and are rational and is closed under multiplication. This means has a zero in and thus, a zero in that must divide 1 by corollary V.15.1. So, or must be a zero. But,

Because does not have a zero that divides 1, we have a contradiction of corollary V.15.1, and this factorization is impossible.

Now suppose that has two quadratic factors in . By Theorem V.15, has a factorization of the form in . Expanding this, we have

Because our original polynomial is , we have

Because and and are integers, it follows that . So, we have

If , If , This implies is negative and is positive or vice versa. Because ,

However, there do not exist such that and or . Thus, this factorization is impossible as well. Therefore, is irreducible over .

The process we used in this example cannot always be used, and even if it could, it is a long, inefficient method. There has got to be a better way, and in fact, there is, known as the Eisenstein Criterion.

#### Eisenstein Criterion

The Eisenstein Criterion is given by Macauley as follows [17].

**Theorem V.16** (Eisenstein Criterion)

- is not divisible by .
- , for , is divisible by .
- is not divisible by .

Keep in mind that the Eisenstein Criterion only applies to polynomials with integer coefficients. If we can find a prime number that divides every coefficient except the first one that, when squared, does not divide the last one, then the polynomial is irreducible. Let’s look at the following problem that I completed for this paper and comes from Fraleigh and Katz [7].

**Example 60**

Determine whether the polynomial in satisfies an Eisenstein criterion for irreducibility over .

**Solution**

We have that , , , and . Let be the prime number 5. Because

we have that , , and are divisible by 5, but is not divisible by 5. Thus, the first and second conditions of the criterion are satisfied. Furthermore,

so is not divisible by . Therefore, in satisfies an Eisenstein criterion for irreducibility over .