8.2 Polynomial Rings

Another set of examples of integral domains in the secondary curriculum are polynomial rings. We let \[\mathbb{R}[x]= \left\{ a(x)=a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n =\sum_{j=0}^n a_j x^j \middle \vert a_j \in \mathbb{R} \right\}\]

The concept of a polynomial can be generalized to not just having real number coefficients. For instance, we will sometimes require that the polynomials have integer coefficients or possibly coefficients that are complex numbers. So we will generalize the definition of a polynomial to be able to be used in these cases.

Definition 8.4 Let $R$ be a ring. We define a polynomial $f(x)$ with coefficients in $R$ to be an infinite formal sum \[\sum_{j=0}^\infty a_j x^j = a_0 + a_1 x+ a_2 x^2 + \cdots + a_n x^n + \cdots\] where $a_j\in R$ for all $j$ and $a_j =0$ for all but finitely many values of $j$.

We denote the set of such polynomials as $R[x]$.

Since the $a_j=0$ for all but finitely many values of $j$, by the well-ordering principle, there is a largest value of $j$ for which $a_j \neq 0$. We call this largest value of $j$ the degree of the polynomial. If all of the coefficients are $0$, we say the polynomial is degree zero.

So the polynomial $p(x)=3x^4-2x^2+1$ is a degree $4$ polynomial.

So we can define the function “$\mathrm{degree}$” that maps the non-zero polynomials to the natural numbers by \[\mathrm{degree}\left(\sum_{j=0}^\infty a_j x^j\right) \mbox{ is the largest value of $j$ for which $a_j$ is non-zero}.\] We call the $a_j$ that corresponds to this value of $j$ the leading coefficient.

Related Content Standards

(HSA.APR.1) Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials.

If $a(x)=\sum_{j=0}^n a_j x^j$ and $b(x)=\sum_{j=0}^m b_j x^j$ are two elements of $R[x]$, with $n\leq m$, then we define \[a(x)+b(x) := \sum_{j=0}^m (a_j+b_j) x^j\] where $a_j:=0$ for $n<j\leq m$. From this definition, we see that addition is closed. We also see that addition is associative from the associativity of addition on $R$. Since $R$ is a ring, addition is commutative on $R$ and so addition is commutative on $R[x]$. We also see that the zero polynomial, $0(x)=0$, is an additive identity for the set. So $(R[x],+)$ is an Abelian group.

Multiplication on $R[x]$ is more challenging to write in a compact form. We will assume that the indeterminate, $x$, commutes multiplicatively with every element in $R$, i.e. $ax=xa, \: \forall a\in R$ (the set of all elements of a ring that commute with every element of the ring is called the center of the ring). We can then define multiplication as

\[\begin{align*} a(x) \cdot b(x) &:= \left( \sum_{j=0}^n a_j x^j \right) \cdot \left(\sum_{k=0}^m b_k x^k\right) = \sum_{j=0}^n \left( a_j x^j \left(\sum_{k=0}^m b_k x^k\right) \right) \\ &= \sum_{j=0}^n \sum_{k=0}^m \left(a_j x^j b_k x^k\right) = \sum_{j=0}^n \sum_{k=0}^m \left(a_j b_k\right) x^{(j+k)} \end{align*}\] and we see that $R[x]$ is closed under multiplication.

Unless $R$ is a commutative ring, we must maintain the order of the coefficients. The associativity of multiplication and distributive properties are inherited from $R[x]$. If the ring $R$ has a multiplicative identity, then $R[x]$ also has a multiplicative identity of the polynomial of the multiplicative identity. Finally, if $R$ is a commutative ring, then multiplication on $R[x]$ is commutative.

Theorem 8.7 The set $R[x]$ of polynomials with coefficients from a ring $R$ with the binary operations of polynomial addition and multiplication is a ring.

One of the main activities that we do with polynomials in the secondary curriculum is to treat the polynomials as functions and evaluate them for various values.

The following theorem gives us the mathematical justification for such evaluations.

In this theorem we use a field $F$ and a subfield $E$, where subfied signifies that $E$ is a subset of $F$ that also satisfies the field properties inherited from $F$. The reason that we use the field $E$ and the subfield $E$ is that we often have polynomials whose coefficients are from one field and the point at which we are evaluating from a larger field. For instance, we often look at quadratic polynomials with real coefficients and determine that they have complex zeros. In this instance, the subfield would be $\mathbb{R}$ and the larger field would be $\mathbb{C}$.

Theorem 8.8 Let $F$ be a subfield of a field $E$ and let $\alpha$ be any element of $E$. The map $\phi_\alpha: F[x] \rightarrow E$ defined by \[\phi_\alpha(a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n) = a_0 + a_1 \alpha + a_2 \alpha^2 + \cdots + a_n \alpha^n\] is a ring homomorphism from $F[x]$ into $E$.

We go ahead and include the proof of this theorem below to demonstrate how these evaluation homomorphisms work and to demonstrate certain properties of polynomial functions.

Proof. Let $F$ be a subfield of a field $E$ and let $\alpha$ be any element of $E$. We also let $a(x)$ and $b(x)$ be generic elements of $F[x]$ and label them as \[a(x) = \sum_{j=0}^\infty a_j x^j \mbox{ and } b(x)=\sum_{j=0}^\infty b_j x^j\] where the $a_j$ and the $b_j$ are zero for all but finitely many values of $j$.

Then $\phi_\alpha \left(a(x)\right) = \sum_{j=0}^\infty a_j \alpha^j$ and $\phi_\alpha\left(\left(b(x)\right)\right) = \sum_{j=0}^\infty b_j \alpha^j$, which are elements of $E$ since addition and multiplication on $E$ are closed and only finitely many elements of each of the sums are non-zero.

Using the definition of addition of two polynomials, we have that $a(x)+b(x)$ is a polynomial that is written as \[a(x)+b(x) = \sum_{j=0}^\infty (a_j+b_j) x^j.\] Therefore, \[\phi_\alpha \left(a(x)+b(x)\right) = \sum_{j=0}^\infty (a_j+b_j) \alpha^j\] and by rearranging the finite number of terms using the distributive property and that both addition and multiplication are commutative in $E$ we have that \[\phi_\alpha \left(a(x)+b(x)\right) = \sum_{j=0}^\infty a_j \alpha^j + \sum_{j=0}^\infty b_j \alpha^j = \phi_\alpha \left(a(x)\right) + \phi_\alpha \left(b(x)\right). \]

Similarly, one finds that \[\phi_\alpha \left(a(x) \cdot b(x)\right) = \phi_\alpha (a(x)) \cdot \phi_\alpha(b(x)).\]

You should keep in mind that the operations are using the same symbol for operations in two different rings. For the case of $a(x)+b(x)$, the addition is in the polynomial ring $F[x]$, while $\phi_\alpha \left(a(x)\right) + \phi_\alpha \left(b(x)\right)$ is addition in $E$.

We say that an element $a \in E$ is a zero of a polynomial $p(x)\in F[x]$ if \[\phi_a ( p(x))=0.\]

We would really like to have the property that for all $a,b \in R$, $a$ and $b$ are the only zeros of the polynomial $(x-a)(x-b)$. In order for that to be true, we need $R[x]$ to be an integral domain. However, we first need some properties related to the degree of polynomials.

Theorem 8.9 Let $R$ be an integral domain and let $a(x)$ and $b(x)$ be elements of $R[x]$. Then \[\mathrm{degree}(a(x)+b(x))\leq \mathrm{max}\left( \mathrm{degree}(a(x)), \mathrm{degree}(b(x))\right) \mbox{, and}\] \[\mathrm{degree}\left( a(x) b(x)\right) = \mathrm{degree}(a(x))+\mathrm{degree}(b(x)).\]

Proof. Let $a(x)=\sum_{j=0}^n a_j x^j \in R[x]$ with degree $n$ and $b(x)=\sum_{j=0}^m b_j x^j\in R[x]$ with degree $m$. We can assume without any loss of generality that $n\leq m$.

If $n<m$, $a(x)+b(x) = \sum_{j=0}^m (a_j+b_j) x^j$, where $a_j=0$ for $n<j\leq m$, we see that $\mathrm{degree} (a(x)+b(x)) =m$. If $n=m$ the sum also has degree $m$, unless the two leading coefficients are additive inverses. Therefore, \[\mathrm{degree}(a(x)+b(x))\leq \mathrm{max}\left( \mathrm{degree}(a(x)), \mathrm{degree}(b(x))\right).\]

Since \[a(x) \cdot b(x) = \sum_{j=0}^n \sum_{k=0}^m \left(a_j b_k\right) x^{(j+k)}\] we see that the leading term of $a(x)\cdot b(x)$ is $a_n b_m$. This is non-zero because $R$ is an integral domain. So the degree of $a(x) \cdot b(x)$ is $m+n$. And we have that \[\mathrm{degree}\left( a(x) b(x)\right) = \mathrm{degree}(a(x))+\mathrm{degree}(b(x)).\]

We are now ready to prove that $R$ being an integral domain implies the $R[x]$ is an integral domain.

Theorem 8.10 If $R$ is an integral domain, then $R[x]$ is an integral domain.

Proof. Assume that there are two polynomials, $a(x)=\sum_{j=0}^n a_j x^j$ and $b(x)=\sum_{j=0}^m b_j x^j$ in $R[x]$, such that $a_n\neq 0$ and $b_m\neq 0$. Then the product of the polynomials has degree of $n+m$. If $n+m>0$, then the product cannot be the zero polynomial. If $n+m=0$, then $n$ and $m$ are both zero and $a(x)=a_0$, $b(x)=b_0$, and $a(x)\cdot b(x)=a_0 b_0$. Since $a_0\neq 0$ and $b_0\neq 0$, since $R$ is an integral domain, the product is non-zero.

8.2.1 Exercises

In $\mathbb{Z}_{12}$, what are the zeros of the polynomial $6-5x+x^2$?
Use the various models of multiplication presented in Section 4.2 to add and multiply the following pairs of polynomials:
1. $a(x)=-2x+3$ and $b(x)=x-1$
2. $a(x)=2x^2-4x+1$ and $b(x)=x^3-2$
3. Which models transition well to polynomials and which do not?