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 $gh$.

Let $f\in \mathbb{\mathbb{Q}}[x]$, where $f=\frac{m_0}{n_0}+\frac{m_1}{n_1}x+\dots +\frac{m_k}{n_k}{x}^k$. Let $M=\mathrm{l}\mathrm{c}\mathrm{m}\left(n_i\right)$.

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

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 $u_i\ne 0$ 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

$x^4+3x^3+2x+4=(x-1{)}^3(x+1)\in {\mathbb{\mathbb{Z}}}_5[x]$. At the same time, $x^4+3x^3+2x+4=(x-1{)}^2(2x-2)(3x+3)$

Review for Exam 2

• Groups
• Homomorphisms ($\varphi :G\to H$)
  • $\varphi (e_G)=e_H$
  • $\varphi (a^{-1})=\varphi (a{)}^{-1}$
  • Kernel of $\varphi$ is normal subgroup of $G$, and $\varphi [G]\simeq G/\mathrm{K}\mathrm{e}\mathrm{r}\varphi$
  • Image of $\varphi$ is a subgroup of $H$
  • Trivial homomorphism: $\varphi (a)=e_H$
  • page 134, exercises 33–42
  • exercise 33: $\varphi :{\mathbb{\mathbb{Z}}}_{12}\to {\mathbb{\mathbb{Z}}}_5$ must be trivial since $k=|\varphi [{\mathbb{\mathbb{Z}}}_{12}]|$ must divide 12 and 5, so $k=1$.
  • excrcise 44: $\left|G\right|<\mathrm{\infty }$, then $\varphi [G]$ is also finite and $|\varphi [G]|$ is divisor of $\left|G\right|$.
  • exercise 45: $\left|H\right|<\mathrm{\infty }$, then $\varphi [G]$ also finite and divisor of $\left|H\right|$ (because it is subgroup of $H$; use lagrange's theorem
• Factor Groups.
  • Find order of $g=(c,d)+⟨(a,b)⟩\in G\oplus H/⟨(a,b)⟩$.
  • 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 $ax\equiv b(\mathrm{mod}n)$
• Describe all units in a ring of form ${\mathbb{\mathbb{Z}}}_n$
• Find characteristic of ring (least $n$ such that $nx=0$ for all $x\in R$)
• Find Zeroes of polynomial in finite ring

Back to Material

As stated at beginning,

$f(x)\in \mathbb{\mathbb{Q}}[x]$ is reducible if and only if $Mf(x)\in \mathbb{\mathbb{Z}}[x]$ is reducible, where $M=\mathrm{l}\mathrm{c}\mathrm{m}\left(n_i\right)$ for coefficients $a_i=\frac{m_i}{n_i}$ of $f$. If $f(x)\in \mathbb{\mathbb{Q}}[x]$ and $Mf(x)\in \mathbb{\mathbb{Z}}[x]$ are reducible, then $f(x)=g(x)h(x)$ for $g(x),h(x)\in \mathbb{\mathbb{Q}}[x]$, and $Mf(x)=g_1(x)h_1(x)$ for $g_1(x),h_1(x)\in \mathbb{\mathbb{Z}}[x]$. In particular, $\mathrm{deg}g=\mathrm{deg}{g}_1$ and $\mathrm{deg}h=\mathrm{deg}{h}_1$.

Corollary. If $f(x)=1\cdot x^n+a_{n-1}{x}^{n-1}+\dots +a_{1}x+a_0$ is in $\mathbb{\mathbb{Z}}[x]$ with $a_0\ne 0$, and if $f(x)$ has a zero in $\mathbb{\mathbb{Q}}$, then it has a zero $m$ in $\mathbb{\mathbb{Z}}$, and $m$ must divide $a_0$.

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