MATH 415 Lecture 21

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End Exam 2 content

Exam Next Tuesday

Extra office hours Monday 15:00–16:00

Review

A function $f(x)\in F[x]$ is called irreducible if and only if $f$ cannot be expressed as product of two nonconstant polynomials $g\,h$.

Let $f\in \mathbb {Q} [x]$, where $f={\frac {m_{0}}{n_{0}}}+{\frac {m_{1}}{n_{1}}}\,x+\dots +{\frac {m_{k}}{n_{k}}}\,x^{k}$. Let $M=\mathrm {lcm} \left(n_{i}\right)$.

$f\in \mathbb {Q} [x]$ is irreducible over $\mathbb {Q} [x]$ if and only if $M\,f\in \mathbb {Z} [x]$ is irreducible over $\mathbb {Z}$

Uniqueness of Factorization

Divisibility

Theorem. Let $p(x)\in F[x]$ be irreducible. If $p(x)$ divides $r(x)\,s(x)$ for $r,s\in F[x]$, then $p(x)$ divides either $r(x)$ or $s(x)$.

Corollary. If $p(x)$ divides $\prod _{i}r_{i}(x)$, then $p(x)$ divides $r_{i}(x)$ for some $i$. (by induction of previous theorem).

Theorem 23.20

If $F$ is a field, then every nonconstant polynomial $f(x)\in F[x]$ can be factored into a product of irreducible polynomials, the irreducible polynomials being unique up to the ordering and unit (i.e. nonzero constant) factors in $F$.

Uniqueness

Suppose $f(x)=q_{1}(x)\,q_{2}(x)\,\dots \,q_{s}(x)=p_{1}(x)\,p_{2}(x)\,\dots \,p_{r}(x)$. Then

1. $s=r$
2. there is a permutation $\sigma \in S_{r}$, and
3. $q_{i}(x)=u_{i}p_{\sigma (i)}(x)$, where $u_{i}\in F$ and $u_{i}\neq 0$ are constants (units).

Proof. If $p_{1}\mid f$, then $p_{1}\mid q_{i}$ for some $i$. Since $q_{i}$ is irreducible, we have $p_{i}=u_{i}\,q_{i}$ for some nonzero constant $u_{i}$.

Example

$x^{4}+3x^{3}+2x+4=(x-1)^{3}(x+1)\in \mathbb {Z} _{5}[x]$. At the same time, $x^{4}+3x^{3}+2x+4=(x-1)^{2}(2x-2)(3x+3)$

Review for Exam 2

• Groups
• Homomorphisms ($\phi :G\to H$)
  • $\phi (e_{G})=e_{H}$
  • $\phi (a^{-1})=\phi (a)^{-1}$
  • Kernel of $\phi$ is normal subgroup of $G$, and $\phi [G]\simeq G/\mathrm {Ker} \,\phi$
  • Image of $\phi$ is a subgroup of $H$
  • Trivial homomorphism: $\phi (a)=e_{H}$
  • page 134, exercises 33–42
  • exercise 33: $\phi :\mathbb {Z} _{12}\to \mathbb {Z} _{5}$ must be trivial since $k=\left|\phi [\mathbb {Z} _{12}]\right|$ must divide 12 and 5, so $k=1$.
  • excrcise 44: $\left|G\right|<\infty$, then $\phi [G]$ is also finite and $\left|\phi [G]\right|$ is divisor of $\left|G\right|$.
  • exercise 45: $\left|H\right|<\infty$, then $\phi [G]$ also finite and divisor of $\left|H\right|$ (because it is subgroup of $H$; use lagrange's theorem
• Factor Groups.
  • Find order of $g=(c,d)+\left\langle (a,b)\right\rangle \in G\oplus H/\left\langle (a,b)\right\rangle$.
  • Classify given group according to Fundamental Theorem of Finitely Generated Abelian Groups (examples 15.10–15.12 and exercises 15.1–15.12)
• Fermat's Little Theorem
• Find all solutions to congruence $a\,x\equiv b{\pmod {n}}$
• Describe all units in a ring of form $\mathbb {Z} _{n}$
• Find characteristic of ring (least $n$ such that $n\,x=0$ for all $x\in R$)
• Find Zeroes of polynomial in finite ring

Back to Material

As stated at beginning,

$f(x)\in \mathbb {Q} [x]$ is reducible if and only if $M\,f(x)\in \mathbb {Z} [x]$ is reducible, where $M=\mathrm {lcm} \left(n_{i}\right)$ for coefficients $a_{i}={\frac {m_{i}}{n_{i}}}$ of $f$. If $f(x)\in \mathbb {Q} [x]$ and $M\,f(x)\in \mathbb {Z} [x]$ are reducible, then $f(x)=g(x)\,h(x)$ for $g(x),h(x)\in \mathbb {Q} [x]$, and $M\,f(x)=g_{1}(x)\,h_{1}(x)$ for $g_{1}(x),h_{1}(x)\in \mathbb {Z} [x]$. In particular, $\deg {g}=\deg {g_{1}}$ and $\deg {h}=\deg {h_{1}}$.

Corollary. If $f(x)=1\cdot x^{n}+a_{n-1}\,x^{n-1}+\dots +a_{1}\,x+a_{0}$ is in $\mathbb {Z} [x]$ with $a_{0}\neq 0$, and if $f(x)$ has a zero in $\mathbb {Q}$, then it has a zero $m$ in $\mathbb {Z}$, and $m$ must divide $a_{0}$.

Proof. $f(x)=(a\,x+b)\,(c\,x^{n-1}+\dots +d)=a\,c\,x^{n}+\dots =(x+b)\,(x^{n-1}+\dots +d)$. Hence $-b$ is a zero of $f$.