Rings of Polynomials

Abigail Huettig

26 Rings of Polynomials

A ring of polynomials $R[x]$ is a ring consisting of all polynomials with coefficients in the ring $R$ [7]. Similarly, the set of polynomials over a field $F$ , denoted $F[x]$ , consists of all polynomials with coefficients in the field $F$ [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 $f(x)\in F[x]$ is reducible over $F$ if we can factor it as $f(x)=g(x)h(x)$ for some $g(x),h(x)\in F[x]$ 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 $\mathbb{R}$ 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 $x^4+4$ can be factored into linear factors in $\mathbb{Z}_5[x]$ . Find this factorization.

Solution
Since we are in $\mathbb{Z}_5[x]$ , addition is defined as $a+b=a+b$ mod(5) and multiplication as $ab=ab$ mod(5). So, we have

$\begin{align*} x^4+4&=x^4+5x^2+4 &&\text{since 5 mod(5)=0}\\ &=(x^2+4)(x^2+1)\\ &=(x^2+5x+4)(x^2+5x+6) &&\text{since 5 mod(5)=0 and 6 mod(5)=1}\\ &=(x+4)(x+1)(x+3)(x+2)\\ \end{align*}$

Therefore, $x^4+4=(x+4)(x+1)(x+3)(x+2)$ in $\mathbb{Z}_5[x]$ .

In the example above, we see that each factor is of a degree less than 4 and has coefficients that are elements of $\mathbb{Z}_5$ , 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 $f(x)$ is not reducible over a field $F$ . In this case, we say that $f(x)$ is irreducible over $F$ [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

If $f(x)\in \mathbb{Z}[x]$ , then $f(x)$ factors into a product of two polynomials of lower degees $r$ and $s$ in $\mathbb{Q}[x]$ if and only if it has such a factorization with polynomials of the same degrees $r$ and $s$ in $\mathbb{Z}[x]$ .

Corollary V.15.1

If $f(x)=x^n+a_{n-1}x^{n-1}+...+a_0$ is in $\mathbb{Z}[x]$ with $a_0\neq 0$ and if $f(x)$ has a zero in $\mathbb{Q}$ , then it has a zero $m$ in $\mathbb{Z}$ , and $m$ must divide $a_0.$

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 $f(x)=x^4-22x^2+1$ is irreducible over $\mathbb{Q}$ .

Proof.
If we could reduce $f(x)$ over $\mathbb{Q}$ , then $f(x)$ would either have a cubic factor and a linear factor or two quadratic factors. This is because $4=3+1$ and $4=2+2$ . So, to multiply two factors together to get $x^4-22x^2+1$ , we need either

$(x^3+ax^2+bx+c)(x+d)$

or

$(x^2+ax+b)(x^2+cx+d).$

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

Suppose $f(x)$ has a linear factor in $\mathbb{Q}[x]$ . That is,

$x^4-22x^2+1=(x^3+ax^2+bx+c)(x+d),$

where $a$ , $b$ , $c$ , and $d$ are rationals. If $x^4-22x^2+1=0$ , we have

$\begin{align*} 0&=x^4-22x^2+1\\ &=(x^3+ax^2+bx+c)(x+d)\\ &=(x+d) &&\text{if $(x^3+ax^2+bx+c)\neq 0$}\\ -d&=x \end{align*}$

So, $x=-d$ is a zero, and $-d=-1\cdot d$ is rational because $d$ and $-1$ are rational and $\mathbb{Q}$ is closed under multiplication. This means $f(x)$ has a zero in $\mathbb{Q}[x]$ and thus, a zero in $\mathbb{Z}$ that must divide 1 by corollary V.15.1. So, $x=1$ or $x=-1$ must be a zero. But,

$\begin{align*} f(\pm 1)&=(\pm 1)^4-22(\pm 1)^2+1\\ &=1-22+1\\ &=-20\\ &\neq 0 \end{align*}$

Because $f(x)$ 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 $f(x)$ has two quadratic factors in $\mathbb{Q}[x]$ . By Theorem V.15, $f(x)$ has a factorization of the form $(x^2+ax+b)(x^2+cx+d)$ in $\mathbb{Z}[x]$ . Expanding this, we have

$\begin{align*} (x^2+ax+b)(x^2+cx+d)&=x^4+cx^3+dx^2+ax^3+acx^2\;+\\ &\;\quad adx+bx^2+bcx+bd\\ &=x^4+(c+a)x^3+(ac+b+d)x^2\;+\\ &\,\quad(ad+bc)x+bd \end{align*}$

Because our original polynomial is $x^4-22x^2+1$ , we have

$\begin{align*} c+a&=0\\ ac+b+d&=-22\\ ad+bc&=0\\ bd&=1 \end{align*}$

Because $bd=1$ and $b$ and $d$ are integers, it follows that $b=d=\pm 1$ . So, we have

$ac+2b=-22.$

If $b=1$ , $ac=-24.$ If $b=-1$ , $ac=-20.$ This implies $a$ is negative and $c$ is positive or vice versa. Because $c+a=0$ ,

$|a|=|c|.$

However, there do not exist $a,c\in\mathbb{Z}$ such that $|a|=|c|$ and $ac=-24$ or $ac=-20$ . Thus, this factorization is impossible as well. Therefore, $f(x)=x^4-22x^2+1$ is irreducible over $\mathbb{Q}$ .

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)

A polynomial $f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0\in \mathbb{Z}[x]$ is irreducible if for some prime $p$ , the following all hold:

$a_n$ is not divisible by $p$ .
$a_k$ , for $k=0,...,n-1$ , is divisible by $p$ .
$a_0$ is not divisible by $p^2$ .

Keep in mind that the Eisenstein Criterion only applies to polynomials with integer coefficients. If we can find a prime number $p$ 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 $f(x)=x^3+10x+5$ in $\mathbb{Z}[x]$ satisfies an Eisenstein criterion for irreducibility over $\mathbb{Q}$ .

Solution
We have that $a_0=5$ , $a_1=10$ , $a_2=0$ , and $a_3=1$ . Let $p$ be the prime number 5. Because

$\begin{align*} 5&=5(1)\\ 10&=5(2)\\ 0&=5(0)\\ 1&=5(0)+1 \end{align*}$

we have that $a_0$ , $a_1$ , and $a_2$ are divisible by 5, but $a_3$ is not divisible by 5. Thus, the first and second conditions of the criterion are satisfied. Furthermore,

$5=25(0)+5$

so $a_0$ is not divisible by $5^2=25$ . Therefore, $f(x)=x^3+10x+5$ in $\mathbb{Z}[x]$ satisfies an Eisenstein criterion for irreducibility over $\mathbb{Q}$ .

26 Rings of Polynomials

Irreducibility

Eisenstein Criterion

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