NoteNextra-origin/content/Math4302/Math4302_L29.md

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

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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.

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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)