MATH 415 Lecture 19

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Review

Let $R$ be a ring. We construct a ring of polynomials in one indeterminate $R[x]$ with coefficients in $R$.

Exerise 24

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

Take $f(x),g(x)\in R[x]$ with $f(x)=a_0+\dots +a_n{x}^n$, $a_n\ne 0$, and $g(x)=b_0+\dots +b_m{x}^m$, $b_m\ne 0$. Thus $f$ is of degree $n$ and $g$ is of degree $m$.

The product $\left(fg\right)\left(x\right)$ is of degree $m+n$: $\left(fg\right)\left(x\right)=a_0{b}_0+(a_1{b}_0+a_0{b}_1)x+\dots +a_n{b}_m{x}^{m+n}$. The last coefficient $a_n{b}_n$ cannot be 0 since $R$ is an integral domain (i.e. it has no zero divisors).

Field of Quotients

The field of quotients of $F[x]$, denoted

$F(x)=\{\frac{f(x)}{g(x)}\mid f,g\in F[x],g\ne 0\}$

is a field of rational functions.

Note that in general, although $F$ is a field, $F[x]$ is a ring (In particular, an integral domain)

Evaluation Homomorphism

Let $F\le E$ be fields. Evaluating a polynomial $f\in F[x]$ at value $\alpha \in E$ (denoted ${\varphi }_{\alpha }(f)$) is a homomorphism from the ring $F[x]$ to field $E$.

Examples

Let $F=\mathbb{\mathbb{Q}}$, $E=\mathbb{ℝ}$, and $\alpha =0$. Then ${\varphi }_0:\mathbb{\mathbb{Q}}[x]\to \mathbb{ℝ}$ always has form ${\varphi }_0(a_0+a_{1}x+\dots +a_n{x}^n)=a_0$, which must be rational.

Therefore evaluation homomorphism may not be onto.

Now let $\alpha =2$. Then ${\varphi }_2:\mathbb{\mathbb{Q}}[x]\to \mathbb{ℝ}$ takes form ${\varphi }_2(a_0+a_{1}x+\dots +a_n{x}^n)=a_0+2a_1+\dots +2^n{a}_n$.

Note in particular that ${\varphi }_2(x^2+x-6)=2^2+2-4=0$, and therefore $x^2+x-6$ is a member of the kernel of ${\varphi }_2$.

Now let $E=\mathbb{ℂ}$ and $\alpha =i$, so ${\varphi }_i:\mathbb{\mathbb{Q}}[x]\to \mathbb{ℂ}$. Thus ${\varphi }_i(a_0+a_{1}x+\dots +a_n{x}^n)=a_0+a_{1}i+\dots +a_n{i}^n$. Note that $i^n\in \{\pm i,\pm 1\}$. Also note that ${\varphi }_i(x^2+1)=0$, so $i^2+1$ is in the kernel of ${\varphi }_i$.

However, ${\varphi }_{\pi }:\mathbb{\mathbb{Q}}[x]\to \mathbb{ℝ}$ is an isomorphism because $\pi$ is a transendental number (it is not a root of any polynomial with rational coefficients), hence the kernel of ${\varphi }_{\pi }$ is just $\left\{0\right\}$. Therefore ${\varphi }_{\pi }$ must be an isomorphism from $\mathbb{\mathbb{Q}}[x]$ onto $\mathbb{ℝ}$.

How to Plug Values into a Function

Definition 22.10.

Let $F\le E$ be fields, $\alpha \in E$, and $f(x)=a_0+a_{1}x+\dots +a_n{x}^n\in F[x]$. If we have ${\varphi }_{\alpha }:F[x]\to E$, we define

$f(\alpha )={\varphi }_{\alpha }(f(x))=a_0+a_1\alpha +\dots +a_n{\alpha }^n$

Section 23: Factorization of Polynomials over a Field

Let $F\le E$, $f(x)\in F[x]$ such that $f(x)=g(x)h(x)$ for $g(x),h(x)\in F[x]$. For $\alpha \in E$, we have

$f(\alpha )={\varphi }_{\alpha }(f(x))={\varphi }_{\alpha }(g(x)h(x))={\varphi }_{\alpha }(g(x))\cdot {\varphi }_{\alpha }(h(x))=g(\alpha )h(\alpha )$

Hence $f(\alpha )=0$ if and only if $g(\alpha )=0$ or $h(\alpha )=0$.

This factorization only makes sense when the degree of $g(x)$ and $h(x)$ is greater than 1. Hence we are concerned with nontrivial factorizations.

Theorem 23.1: Division in $F[x]$

Let $F$ be a field that forms the polynomial ring $F[x]$. Let $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\dots +a_0$ and $g(x)=b_m{x}^m+b_{m-1}{x}^{m-1}+\dots +b_0$ be elements of $F[x]$ with $a_n,b_m\ne 0$ and $m\ge 0$ (so $g(x)$ is not constant).

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 $r(x)=0$ or the degree of $r(x)$ is less than the degree $m$ of $g(x)$.

Proof. Consider $S=\{f(x)-g(x)s(x)\mid s(x)\in F[x]\}$. If $0\in S$, then there exists $s(x)\in F[x]$ such that $f(x)-g(x)s(x)=0$ Therefore $f(x)=g(x)s(x)+0$, hence $r(x)=0$.

Otherwise, let $r(x)\in S$ be of minimal degree . Then

$r(x)=c_t{x}^t+c_{t-1}{x}^{t-1}+\dots +c_0{c}_t\ne 0$ with $c_j\in F$

If $t<m$, then we're done: $f(x)-g(x)s(x)=r(x)$, so $f(x)=g(x)s(x)+r(x)$ with $\mathrm{deg}(s(x))<m$. In the more interesting case, if $t\ge m$, then

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} f(x) - s(x) \, g(x) - \left( \frac{c_t}{b_m} \right) x^{t-m} \, g(x) &= r(x) - \left( \frac{c}{b_m} \right) x^{t-m} \, g(x) \\ &= r(x) - (c_t \, x^t + \mbox{terms of lower degree}) \\ &= \cancel{c_t\,x^t + \dots - \cancel{c_t\,x^t} - \mbox{terms of lower degree}) \\ \end{align}}

Thus $\mathrm{deg}(r(x))\le t-1<r(x)$.

Let $s^{\prime }(x)=s(x)+\frac{c_t}{b_m}{x}^{t-m}$, then $f(x)-s^{\prime }(x)g(x)$ is of lesser degree than $r(x)$ — contradiction.

This proves the existence of $r(x)$. All that remains is to prove uniqueness.

Assume

$f(x)=g(x)q_1(x)+r_1(x)=g(x)q_2(x)+r_2(x)$

with

$\mathrm{deg}(r_1),\mathrm{deg}(r_2)<m$

and

$q_1\ne q_2$

. Then the product

$g(x)(q_1(x)-q_2(x))=r_1(x)-r_2(x)$

has LHS degree

$\ge m$

and RHS degree

$<m$

— contradiction.

quod erat demonstrandum

Factoring Examples

Let $f,g\in {\mathbb{\mathbb{Z}}}_5[x]$ where $\begin{array}{rl}f(x)& =x^4-3x^3+2x^2+4x-1\\ g(x)& =x^2-2x+3\end{array}$

Find the quotient $q(x)$ and remainder $r(x)$:

File:MATH 415 Factoring Example.jpg