# 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 $f(\alpha)=0$.

<details>
<summary>Example</summary>

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

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

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

</details>

#### 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 $\deg r(x)<\deg g(x)$.

<details>
<summary>Proof</summary>

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.

</details>

<details>
<summary>Example</summary>

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

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

</details>

#### 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.

<details>
<summary>Proof</summary>

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 $r=0$.

</details>

#### Another corollary

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

<details>
<summary>Proof</summary>

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.

</details>

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