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 $f(x)=a_{0}+\dots +a_{n}\,x^{n}$, $a_{n}\neq 0$, and $g(x)=b_{0}+\dots +b_{m}\,x^{m}$, $b_{m}\neq 0$. Thus $f$ is of degree $n$ and $g$ is of degree $m$.

The product $\left(f\,g\right)\left(x\right)$ is of degree $m+n$: $\left(f\,g\right)\left(x\right)=a_{0}\,b_{0}+(a_{1}\,b_{0}+a_{0}\,b_{1})\,x+\dots +a_{n}\,b_{m}\,x^{m+n}$. 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

$F(x)=\left\{{\frac {f(x)}{g(x)}}~\mid ~f,g\in F[x],\ g\neq 0\right\}$

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

Examples

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

Therefore evaluation homomorphism may not be onto.

Now let $\alpha =2$. Then $\phi _{2}:\mathbb {Q} [x]\to \mathbb {R}$ takes form $\phi _{2}(a_{0}+a_{1}\,x+\dots +a_{n}\,x^{n})=a_{0}+2a_{1}+\dots +2^{n}a_{n}$.

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

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

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

How to Plug Values into a Function

Definition 22.10.

Let $F\leq E$ be fields, $\alpha \in E$, and $f(x)=a_{0}+a_{1}\,x+\dots +a_{n}\,x^{n}\in F[x]$. If we have $\phi _{\alpha }:F[x]\to E$, we define

$f(\alpha )=\phi _{\alpha }\left(f(x)\right)=a_{0}+a_{1}\,\alpha +\dots +a_{n}\,\alpha ^{n}$

Section 23: Factorization of Polynomials over a Field

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

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

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 $a_{n},b_{m}\neq 0$ and $m\geq 0$ (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 $S=\left\{f(x)-g(x)\,s(x)~\mid ~s(x)\in F[x]\right\}$. 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

$r(x)=c_{t}\,x^{t}+c_{t-1}\,x^{t-1}+\dots +c_{0}\qquad c_{t}\neq 0$ 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 $\deg(s(x))<m$. In the more interesting case, if $t\geq m$, 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 $\deg(r(x))\leq t-1<r(x)$.

Let $s'(x)=s(x)+{\frac {c_{t}}{b_{m}}}\,x^{t-m}$, then $f(x)-s'(x)\,g(x)$ 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 $\deg(r_{1}),\deg(r_{2})<m$ and $q_{1}\neq q_{2}$. Then the product $g(x)\,\left(q_{1}(x)-q_{2}(x)\right)=r_{1}(x)-r_{2}(x)$ has LHS degree $\geq m$ and RHS degree $<m$ — contradiction.

quod erat demonstrandum

Factoring Examples

Let $f,g\in \mathbb {Z} _{5}[x]$ where ${\begin{aligned}f(x)&=x^{4}-3x^{3}+2x^{2}+4x-1\\g(x)&=x^{2}-2x+3\end{aligned}}$

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

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