Math4302 Modern Algebra (Lecture 29)

Rings

Polynomial Rings

R[x]=\{a_0+a_1x+\cdots+a_nx^n:a_0,a_1,\cdots,a_n\in R,n>1\}

Then $(R[x],+,\cdot )$ is a ring.

If $R$ has a unity $1$ , then $R[x]$ has a unity $1$ .

If $R$ is commutative, then $(R[x],+,\cdot )$ is commutative.

Definition of evaluation map

Let $F$ be a field, and $F[x]$ . Fix $\alpha\in F$ . $\phi_\alpha:F[x]\to F$ defined by $f(x)\mapsto f(\alpha)$ (the evaluation map).

Then $\phi_\alpha$ is a ring homomorphism. $\forall f,g\in F[x]$ ,

$(f+g)(\alpha)=f(\alpha)+g(\alpha)$
$(fg)(\alpha)=f(\alpha)g(\alpha)$ (use commutativity of $\cdot$ of $F$ , $f(\alpha)g(\alpha)=\sum_{k=0}^{n+m}c_k x^k$ , where $c_k=\sum_{i=0}^k a_ib_{k-i}$ )

Definition of roots

Let $\alpha\in F$ is zero (or root) of $f\in F[x]$ , if $f(\alpha)=0$ .

Example

$f(x)=x^3-x, F=\mathbb{Z}_3$

$f(0)=f(1)=0$ , $f(2)=8-2=2-2=0$

but note that $f(x)$ is not zero polynomial $f(x)=0$ , but all the evaluations are zero.

Factorization of polynomials

Division algorithm. Let $F$ be a field, $f(x),g(x)\in F[x]$ with $g(x)$ non-zero. Then there are unique polynomials $q(x),r(x)\in F[x]$ such that

$f(x)=q(x)g(x)+r(x)$

where $f(x)=a_0+a_1x+\cdots+a_nx^n$ and $g(x)=b_0+b_1x+\cdots+b_mx^m$ , $r(x)=c_0+c_1x+\cdots+c_tx^t$ , and $a^n,b^m,c^t\neq 0$ .

$r(x)$ is the zero polynomial or $\deg r(x)<\deg g(x)$ .

Proof

Uniqueness: exercise

Existence:

Let $S=\{f(x)-h(x)g(x):h(x)\in F[x]\}$ .

If $0\in S$ , then we are done. Suppose $0\notin S$ .

Let $r(x)$ be the polynomial with smallest degree in $S$ .

$f(x)-h(x)g(x)=r(x)$ implies that $f(x)=h(x)g(x)+r(x)$ .

If $\deg r(x)<\deg g(x)$ , then we are done; we set $q(x)=h(x)$ .

If $\deg r(x)\geq\deg g(x)$ , we get a contradiction, let $t=\deg r(x)$ .

$m=\deg g(x)$ . (so $m\leq t$ ) Look at $f(x)-(h(x)+\frac{c_t}{b_m}x^{t-m})g(x)$ .

then $f(x)-(h(x)+\frac{c_t}{b_m}x^{t-m})g(x)=f(x)-h(x)g(x)-\frac{c_t}{b_m}x^{t-m}g(x)$ .

And $f(x)-h(x)g(x)=r(x)=c_0+c_1x+\cdots+c_tx^t$ , $c_t\neq 0$ .

$\frac{c_t}{b_m}x^{t-m}g(x)=\frac{c_0c_t}{b_m}x^{t-m}+\cdots+c_t x^t$

That the largest terms cancel, so this gives a polynomial of degree $<t$ , which violates that $r(x)$ has smallest degree.

Example

$F=\mathbb{Z}_5=\{0,1,2,3,4\}$

Divide $3x^4+2x^3+x+2$ by $x^2+4$ in $\mathbb{Z}_5[x]$ .

3x^4+2x^3+x+2=(3x^3+2x-2)(x^2+4)+3x

So $q(x)=3x^3+2x-2$ , $r(x)=3x$ .

Some corollaries

$a\in F$ is a zero of $f(x)$ if and only if $(x-a)|f(x)$ .

That is, the remainder of $f(x)$ when divided by $(x-a)$ is zero.

Proof

If $(x-a)|f(x)$ , then $f(a)=0$ .

If $f(x)=(x-a)q(x)$ , then $f(a)=(a-a)q(a)=0$ .

If $a$ is a zero of $f(x)$ , then $f(x)$ is divisible by $(x-a)$ .

We divide $f(x)$ by $(x-a)$ .

$f(x)=q(x)(x-a)+r(x)$ , where $r(x)$ is a constant polynomial (by degree of division).

Evaluate at $f(a)=0=0+r$ , therefore $r=0$ .

Another corollary

If $f(x)\in F[x]$ and $\deg f(x)=0$ , then $f(x)$ has at most $n$ zeros.

Proof

We proceed by induction on $n$ , if $n=1$ , this is clear. $ax+b$ have only root $x=-\frac{b}{a}$ .

Suppose $n\geq 2$ .

If $f(x)$ has no zero, done.

If $f(x)$ has at least $1$ zero, then $f(x)=(x-a)q(x)$ (by our first corollary), where degree of $q(x)$ is $n-1$ .

So zeros of $f(x)=\{a\}\cup$ zeros of $q(x)$ , and such set has at most $n$ elements.

Done.

Preview: How to know if a polynomial is irreducible? (On Friday)