Update Math4302_L29.md

content/Math4302/Math4302_L29.md

@@ -44,6 +44,8 @@ Division algorithm. Let $F$ be a field, $f(x),g(x)\in F[x]$ with $g(x)$ non-zero
 
 $f(x)=q(x)g(x)+r(x)$
 
+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>
@@ -57,4 +59,87 @@ 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)