Rings of Polynomials

Maggie Schildt

27 Rings of Polynomials

Definition. Polynomial Rings

Let $R$ be a ring. A polynomial $f(x)$ with coefficient $x$ in $R$ is an infinite formal sum given as

$\begin{equation*} \sum_{i=0}^{\infty}{a_{i}x^{i}}=a_{0}+a_{1}x+...+a_{n}x^{n}+... \end{equation*}$

where $a_{i}\in R$ and $a_{i}=0$ for all but a finite number of values of $i$ [12].

Here we have $a_{i}$ as a coefficient of $f(x)$ . We call a degree of $f(x)$ the largest value for $i$ such that $a_{i}\neq 0$ . Just like rings have two binary operations such that terms in a ring can be added together of multiplied the same is true for rings of polynomials.

Example:

Find the sum and the product of the given polynomials in the given polynomial ring. $f(x)=2x^{3}+4x^{2}+3x+2$ , $g(x)=3x^{4}+2x+4$ in $\mathbb{Z}_{5}[x]$ [12].

Let us compute the sum first

$\begin{equation*} \begin{split} f(x)+g(x)&=2x^{3}+4x^{2}+3x+2+3x^{4}+2x+4\\ &=3x^{4}+2x^{3}+4x^{2}+5x+6\\ &=3x^{4}+2x^{3}+4x^{2}+0(x)+1\\ &=3x^{4}+2x^{3}+4x^{2}+1 \end{split} \end{equation*}$

Next, let us compute the product of $f(x)g(x)$

$\begin{equation*} \begin{split} f(x)g(x)&=(2x^{3}+4x^{2}+3x+2)(3x^{4}+2x+4)\\ &=6x^{7}+12x^{6}+9x^{5}+6x^{4}+4x^{4}+8x^{3}\\ &+6x^{2}+4x+8x^{3}+16x^{2}+12x+8\\ &=6x^{7}+12x^{6}+9x^{5}+10x^{4}+16x^{3}+22x^{2}+16x+8\\ &=1x^{7}+2x^{6}+4x^{5}+0(x^{4})+1x^{3}+2x^{2}+1x+3\\ &=x^{7}+2x^{6}+4x^{5}+x^{3}+2x^{2}+x+3 \end{split} \end{equation*}$

Thus we can conclude that the sum $f(x)+g(x)=3x^{4}+2x^{3}+4x^{2}+1$ and the product $f(x)g(x)=x^{7}+2x^{6}+4x^{5}+x^{3}+2x^{2}+x+3$

In this example we used a familiar technique for simplification as used in finding the solutions to an integral domain problem. This technique comes into play prominently in Abstract Algebra and we will cover it in depth in our section on Number Theory. For now we look at some basic commonalities between what we know of ring properties and what see are properties of ring polynomials. We can add or multiply polynomials and we will see that there are different families for which these rings may fall, just as we saw in the diagram in the previous section.

Theorem:

The set $R[x]$ of all polynomials in an indeterminate $x$ with the coefficients in a ring $R$ is a ring under polynomial addition and multiplication. If $R$ is commutative, then so is $R[x]$ , and if $R$ has unity $1\neq 0$ , then 1 is also unity for $R[x]$ .

We see here that even for Polynomial rings we can still classify them as commutative if the ring $R$ of the polynomial ring $R[x]$ is commutative. Recall the diagram given in the previous section, we saw that within a portion of commutative rings we found integral domains and within the family of integral domains we found fields. It would be rational to expect that there is a family of commutative rings of polynomials that can be classified as fields as well [12].

Definition. Fields

Let $F$ be a field, then $F[x]$ is an integral domain and as such $f(x)$ is the field of quotients for $F[x]$ . Thus any element in $F(x)$ can be represented as a quotient $\frac{f(x)}{g(x)}$ of two polynomials in $F[x]$ where $g(x)\neq 0$ .

We can relate this definition to polynomial functions by stating that if $F[x_{1},...x_{n}]$ is an integral domain with a field of quotients $F(x_{1},...,x_{n})$ then the field of quotients is a field of rational functions in $n$ indeterminates over $F$ [12].

Now that we have an established understanding of fields as they pertain to polynomials we can look at properties of polynomials [12].

Theorem: Evaluation Homomorphism

Let $E$ and $F$ be fields, with $F\leq E$ . Suppose that $f(x)\in F[x]$ factors in $F[x]$ , so that $f(x)=g(x)h(x)$ for $g(x), h(x)\in F[x]$ and let $\alpha \in E$ . For the evaluation homomorphism $\phi_{\alpha}$ , there is $f(\alpha)=\phi_{\alpha}(f(x))=\phi_{\alpha}(g(x)h(x))=\phi_{\alpha}(g(x)) \phi_{\alpha}(h(x))=g(\alpha)h(\alpha)$ . Thus if $\alpha \in E$ , then $f(\alpha)=0$ if and only if either $g(\alpha)=0$ or $h(\alpha)=0$ .

From this we can draw out that the purpose of homomorphisms is much the same as it always has been but we evaluate now by using the value of $\alpha$ and although we may be solving for a zero of $f(x)$ we may run into polynomials that have a remainder and this simply means that whatever the value of the remainder is, we claim as our solution. We can see this in the following example.

Example:

Let $F=E=\mathbb{Z}_{7}$ . Compute for $\phi_{3}[(x^{4}+2x)(x^{3}-3x^{2}+3)]$ [12].

Given our understanding of the evaluation homomorphisms for field theory consider that $\alpha = 3$ then we have the following.

$\begin{equation*} \begin{split} \phi_{3}[(x^{4}+2x)(x^{3}-3x^{2}+3)]&=(3^{4}+2(3))(3^{3}-3(3)^{2}+3)\\ &=(81+6)(27-27+3)\\ &=87\cdot 3\\ &=261 \end{split} \end{equation*}$

Since $p=7$ for $\mathbb{Z}_{7}$ , we can simplify 261 if we divide 261 by 7 we get a remainder of 2 thus we can conclude that $\phi_{3}[(x^{4}+2x)(x^{3}-3x^{2}+3)]=2$ .

Just as we discussed in the definition for the Evaluation Homomorphism for field theory, we are given a polynomial with a distinguished $\alpha$ which we used to solve the function to a quantifiable value 261. We then used the field $\mathbb{Z}_{7}$ which we knew meant that we have to simplify 261 by factors of 7 which would either result in 0 or would result in the remainder between 261 and $n7$ where $n$ is the number of times 7 is multiplied to produce the value closest but smaller than 261.

We can continue our understanding of polynomials over fields by discussing the factorization of polynomials. It might seem backwards at first but the best way to find the factor of $f(x)$ is by division, thus it is only natural for there to be a theorem for the division of $F[x]$ [12].

Theorem: Division Algorithm

Let $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{0}$ and $g(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+...+b_{0}$ be two elements of $F[x]$ , with $a_{n}$ and $b_{n}$ both nonzero elements of $F$ and $m>0$ . Then there are unique polynomials $q(x)$ and $r(x)$ in $F[x]$ such that $f(x)=g(x)q(x)+r(x)$ , where either $r(x)=0$ of the degree of $r(x)$ is less than the degree $m$ of $g(x)$ .

Example:

Find $q(x)$ and $r(x)$ as described by the division algorithm $f(x)=x^{4}+5x^{3}-3x^{2}$ and $g(x)=5x^{2}-x+2$ in $\mathbb{Z}_{11}[x]$ so that $f(x)=g(x)q(x)+r(x)$ with $r(x)=0$ or of degree less than the degree of $g(x)$ .[12]

Consider the long division

$\begin{equation*} \begin{split} &9x^{2}+5x+10\\ 5x^{2}-x+2& \overline{)x^{4}+5x^{3}-3x^{2}}\\ &\ \underline{x^{4}+2x^{3}+7x^{2}}\\ &\ \ \ \ \ \ \ \ 3x^{3}+x^{2}\\ &\ \ \ \ \ \ \ \ \underline{3x^{3}-5x^{2}+10x}\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6x^{2}+x\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underline{6x^{2}-10x+9}\\ &\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \end{split} \end{equation*}$

Thus we have found $q(x)=9x^{2}+5x+10$ and $r(x)=2$ .

This example is a wonderfully simple way of understanding the relationship between $f(x), g(x), q(x)$ and $r(x)$ . At first this long division might look wrong but we need to keep in mind that we have $n=11$ so we can alter values in terms of multiples of 11 with remainder, again this will be explained more in depth in our section on number theory.

We continue our discussion on polynomials by introducing another way to look at our long division problem [12].

Theorem: Factor Theorem

An element $a\in F$ is a zero of $f(x)\in F[x]$ if and only if $x-a$ is a factor of $f(x)$ in $F[x]$ .

We understand this theorem easily because we know that $a$ can only be a zero if it is a factor of $f(x)$ otherwise there is a remainder. This can be seen when we look at the formula for $f(x)=(x-a)q(x)+r(x)$ if $a$ is a zero then we can simplify this formula to $f(a)=0=0q(a)+r(a)\Rightarrow f(a)=0+r(a)=0 \Rightarrow f(a)=0$ . There is a formal proof for this theorem but we will use the brief explanation, allowing us to focus our attention on the next vital definition in Rings of Polynomials.

Definition.

A non-constant polynomial $f(x)\in F[x]$ is irreducible over $F$ or is irreducible polynomial in $F[x]$ if $f(x)$ cannot be expressed as a product $g(x)h(x)$ of two polynomials $g(x)$ and $h(x)$ in $F[x]$ both of lower degree than the degree of $f(x)$ . If $f(x)\in F[x]$ is a non-constant polynomial that is not irreducible over $F$ , then $f(x)$ is reducible over $F$ .

Example:

Demonstrate that $x^{4}-22x^{2}+1$ is irreducible over $\mathbb{Q}$ [12].

Firstly, we see that $f(x)$ is of degree 4. The only possibility for $f(x)$ to be irreducible in $\mathbb{Q}$ is if, either there are no roots in $\mathbb{Q}$ or $f(x)$ can not be factored in a polynomial of degree 2. Since $a_{0}=1$ then we can say if $f(x)$ has a zero in $\mathbb{Q}$ then $f(x)$ also has a zero in $\mathbb{Z}$ that must divide 1. This leaves us with only two options $-1$ or $1$ .

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

Since there are no zeros $f(x)$ has no linear factors over $\mathbb{Q}$ . Now let us address $\mathbb{Z}$ where $a,b,c,d\in \mathbb{Z}$ . Let

$\begin{equation*} x^{4}-22x^{2}+1=(x^{2}+ax+b)(x^{2}+cx+d) \end{equation*}$

we can expand this equivalency to result in

$\begin{equation*} \begin{split} x^{4}-22x^{2}+1&=x^{4}+ax^{3}+bx^{2}+cx^{3}+acx^{2}+bcx+dx^{2}+adx+db\\ &=x^{4}+(a+c)x^{3}+(b+ac+d)^{2}+(bc+ad)x+db \end{split} \end{equation*}$

If we evaluate the newly simplified form we see that $a+c=0$ , $b+ac+d=-22$ , $bd+ad=0$ , and $db=1$ . Firstly, if $bd=1$ then $b=d=1$ or $b=d=-1$ . Let $b=d=1$ , then $-22=1+ac+1$ and if we subtract $2$ from both sides we get $-24=ac$ . Since $a+c=0$ then $a=-c$ and as such $-c^{2}=-24$ which is impossible for $c\in \mathbb{Z}$ . Now let $b=d=-1$ then $-22=-1+ac-1$ . If we subtract $2$ from both sides we get $-20=ac$ and since $a=-c$ then $-c^{2}=20$ this is also not possible for $c\in \mathbb{Z}$ . Therefore we can conclude that $f(x)$ has no quadratic factors over $\mathbb{Z}$ nor $\mathbb{Q}$ and thus we say $f(x)$ is irreducible over both.

What we can derive from this example is that if the polynomial $f(x)$ is irreducible then there can be no two polynomials, which must be of a lower degree, whose product is $f(x)$ . When we say of lower degree we say in the previous example that $f(x)$ was given a degree of 4, this is because the leading $x$ variable had an exponent of 4. So the two polynomials $g(x)$ and $h(x)$ that might produce $f(x)$ must have a leading variable with an exponent $n<4$ . In the example we saw that both polynomials had an exponent equal to 2.

This last theorem is another way we can classify a polynomials under a ring. All of these classifications help us better understand functions that arise and what properties they have based on their classification. For example, if you were to be given an $f(x)$ and told nothing other than the fact it was reducible over $\mathbb{Z}$ you would instantly know that $f(x)$ was a polynomials in a field that could be factored into two separate polynomials.

27 Rings of Polynomials

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