 Algebra Notes - Printable Version +- OMath! (http://math.elinkage.net) +-- Forum: Math Forums (/forumdisplay.php?fid=4) +--- Forum: Algebra (/forumdisplay.php?fid=5) +--- Thread: Algebra Notes (/showthread.php?tid=716) Pages: 1 2 Algebra Notes - elim - 04-02-2015 11:29 AM 1 Background 1.1 Rings Ring 1.1 Definition A Ring $R\;(R,+,\cdot)$ is always assumed commutative with $1$: $\quad\bullet \;(R,+)$ is an abelian group (write $0$ for the additive identity), $\quad\bullet \;(R,\cdot)$ is associative, commutative, and $\exists 1\in R\,\forall a\in R\;(1\cdot a = a = a\cdot 1)$ $\quad\bullet \;\forall a,b,c\in R\;(a\cdot (b+c) = a\cdot b + a\cdot c)$ (Distributive law) $\quad\bullet \;0\ne 1$ (additive and multiplicative identities are different) Write $ab$ for $a\cdot b$. Integral domain 1.1.2 Definition An integral domain is a ring s.t. $\small(ab = 0)\implies (a = 0)\vee (b=0)$. Field 1.1.3 Definition A field is a ring $F$ such that $\forall b\in F^{\times}\;\exists b^{-1}\in F^{\times}\;(bb^{-1} = 1)$ $\qquad F^{\times}:= F-\{0\}$ is the multiplicative group of $F.$ (Write $a/b$ or $\dfrac{a}{b}$ for $ab^{-1}$). $\qquad$We reserve letter $F$ for a field unless otherwise stated. 1.1.4 Remark A field is an integral domain. 1.1.5 Examples $\mathbb{Z}$ is an integral domain while $\mathbb{Q,\,R,\,C}$ are fields. Subring, Subfield and Ideal 1.1.6 Definitions $S$ is a subring of a ring $R\;$ iff $\;(1\in S\cdot S = S\le_+ R)$. $\quad S\,$is a subfield of a field $F\,$ iff $(S\subset_R F)\wedge(S-\{0\}\le_{\LARGE\mathbf{\cdot}} F^{\times})$. $\quad I\,$is an ideal in a ring $R$ iff $(1\not\in I\le_+ R)\wedge (RI\subset I)$. 1.1.7 Lemma If $I$ is an ideal in a ring $R$, then $R/I = \{r+I\mid r\in R\}$ $\quad\color{grey}{\small{(r+I = \{r+i\mid i\in I\})}}\quad$form a (quotient) ring with $\;\; +,\,\cdot$ given by $\quad (r+I) + (s+I) = (r+s) +I,\;(r+I)(s+I) = rs + I.\;$ 1.1.8 Lemma $\quad\mathbb{Z}_n :=\mathbb{Z}/n\mathbb{Z}$ is the ring of integers modulo $n$ $\quad\small(\in\mathbb{N}-\{1\},\;\mathbb{Z}_0 = \mathbb{Z})$and $\mathbb{Z}_n$ is a field iff $n$ is a prime number. $\quad$The elements of $\mathbb{Z}_n$ will be denoted by $0,1,\ldots,n-1$ or $\bar{0},\,\bar{1},\ldots,\overline{n-1}$ $\quad$instead of $...,k + n\mathbb{Z},..$ 1.1.9 Definition $\phi:R\to S$ is a homomorphism of rings if $\qquad(\phi(1) = 1)\wedge \forall a,b\in R\;(\phi(a+b) = \phi(a) +\phi(b),\,\phi(ab) = \phi(a)\phi(b))$ $\qquad\phi$ is monomorphism,epimorphism, or isomorphism if $\phi$ is in additionally $\qquad$injective, surjective,or bijective, respectively. 1.1.10 Proposition If $\phi:R\to S$ is a ring homomorphism, then $\text{Im}(\phi)$ is $\qquad$a subring of $S$ and $\ker(\phi)$ is an ideal in $R$, and $\bar{\phi}:{\small{\dfrac{R}{\ker(\phi)}}}\to \text{Im}(\phi)$ $\qquad\color{grey}{\small{(\bar{\phi}(r+\ker(\phi)) = \phi( r))}}$ is a ring isomorphism. 1.2 Prime subfields - elim - 04-02-2015 02:02 PM 1.2.1 Definition Call $\displaystyle{\bigcap\{S\mid S\subset_{\small F} F\}}$ the prime subfield of field $F$ 1.2.2 Remark A prime subfield $\text{psub}(F)$ is a field. 1.2.3 Examples $\mathbb{Q},\;\mathbb{Z}_p\;$($p$ is prime) are their own prime subfields. 1.2.4 Proposition For any field $F,\;\text{psub}(F)$ is isomorphic to either $\mathbb{Z}_p$ $\qquad$for some prime $p$(in which case we say $F$ has characteristic $p$) $\qquad$or to $\mathbb{Q}$ ($F$ has characteristic $0$ then). 1.3 Fields of fractions - elim - 04-02-2015 04:40 PM Let $R$ be a ring, $\Omega_R = \{(r,s)\mid r,s\in R,\;s\ne 0\},$ relation $\sim$ on $\Omega_R$ by $\qquad (r,s)\sim (u,v)\iff rv = su$. Clearly $\sim$ is an equivalence relation, $\qquad\small[r,s] = \{(u,v)\in\Omega: (u,v)\sim (r,s)\}$ as the $\sim$ equivalence class of $\small(r,s)$. 1.3.1 Proposition If $R$ is an integral domain, then $\small F_R = \{[r,s]\mid (r,s)\in\Omega_R\}$ $\qquad$forms the field of fractions of $R$ with ring operations $\qquad [r,s]+[u,v] = [rv +su,\,sv],\;[r,s][u,v] = [ru,sv].$ $\qquad$We may identify $R$ with the subring $\{[r,1]\mid r\in R\}$ of $F_R.\;$ $\qquad$i.e. $\;i: R\to F_R\;(r\mapsto [r,1])$ is the natural monomorphism. 1.3.2 Proposition If $K$ is a field and $\phi: R\underset{\;}{\to} K$ is a monomorphism, $\qquad$then $\phi$ extends uniquely to $F_R$, i.e. there is a unique homomorphism $\qquad\tilde{\phi}:F_R\to K$ such that $\phi = \tilde{\phi}\circ i.$ 1.4 Polynomial rings - elim - 04-02-2015 05:11 PM 1.4.1 Definition A polynomial over a ring $R$ with the indeterminate $t$ is an expression $a_n t^n +a_{n-1}t^{n-1}+\cdots +a_1 t + a_0 \tag{1}$$\qquad$where $n\in\mathbb{N},\; a_j \in R\,(j = \overline{0,n})$ We call $a_j$ the coefficient of $t^j\;(a_j = 0\;(j > n))$ $\qquad$If $D_f:=\{j\mid a_j\ne 0\}\ne\varnothing$, we say polynomial $f$ in (1) has degree $\deg(f) = \max D_f$ $\qquad$otherwise $\deg(f) := -\infty$. $a_m$ is called the leading coefficient of $f$ if $m = \deg(f)\ge 0$. $\qquad f$ is said to be monic if its leading coefficient is $1$. $\qquad$Given two polynomials $\sum a_j t^j,\; \sum b_j t^j,$ we define their sum to be $\sum (a_j + b_j) t^j$. $\qquad$and their product to be $\quad\small{ \displaystyle{\sum c_j t^j\quad(c_j = \sum_{j = i+k} a_i b_k)}}.$ $\qquad$In this way $\displaystyle{R[t] = \{{\small{\sum_{j=0}^n}} a_j t^j\mid n\in\mathbb{N},\; a_j \in R\}}$ becomes a polynomial ring over $\qquad R$ in the indeterminate $t$, which contains subring $R$ (the set of degree-0 polynomials) 1.4.2 Definition If $f\in R[t]$ is as in (1) and $\alpha\in R$, then the evaluation of $f$ at $\alpha$ is$f(\alpha) = a_n\alpha^n + a_{n-1}\alpha^{n-1}+\cdots +a_1\alpha + a_0\in R$$\qquad$Any fixed $\alpha\in R$ determines an evaluatin homomorphism $R[t]\to R\;(f\mapsto f(\alpha))$. 1.4.3 Lemma If $R$ is an integral domain, so is $R[t].$ 1.4.4 Remark Since $R[t]$ is an integral domain whenever $R$ is, it determines a field of fractions $\qquad R(t)$ called the field of rational functions over $R$ in the indeterminate $t$, which is expressible $\qquad$as the quotient of polynomials. $\qquad$If $\phi:R\to S$ is a ring homomorphism, then $\sum a_j t^j \mapsto \sum \phi(a_j)a^j$ determines an induced map $\qquad R[t]\to S[t]$ which is clearly a ring homomorphism; we often use the same notation $\phi$ for it. 1.5 Polynomial rings over fields - elim - 04-03-2015 04:40 PM Fix a field $K$ in this section 1.5.1 Proposition (Division Algorithm for polynomials over a field). $\qquad\forall f,g\in F[t]\;((f\ne 0)\implies \exists !\,q\in F[t]\;(g = qf + r\;(\deg(r ) < \deg(f))))$ 1.5.2 Definition Let $f,g\in F[t]$. We say that $f$ divides $g$ (write $f\mid g$) if $\exists h\in F[t]\;(g = fh)$. $\qquad$We write $f\not\mid g$ if $f$ does not divide $g\underset{\,}{.}$ $\qquad$A member of $\underset{\,}{\,}D(f,g) = \{ h\in R[t]: (h\mid f)\wedge(h\mid g)\}$ is called a common divisor of $f$ and $g$. $\qquad d(\in D(f,g))$ is a highest common divisor (hcd) of $f,\;g$ if $h\in D(f,g)\implies h\mid d\underset{\;}{.}$ $\qquad$we say $f,\;g$ are relatively prime.If $1$ is a hcd of $f$ and $g$. 1.5.3 Proposition hcd of $f,\,g$ exists hence $\forall f,g\in F[t]\;(D(f,g)\ne \varnothing)\underset{\,}{.}$ Proof WLOG, assume $\{f,g\}\ne \{0\}.$ then $C = \{h\mid h = af + bg\ne 0, \; a,b\in F[t] \}\ne \varnothing\underset{\,}{.}$ $\qquad \therefore\;\exists d\in C\,\forall h\in C\;(\deg(d)\le \deg(h)).\;\;\{\lambda d\mid \lambda\in K-\{0\}\}$ is clearly the set of all hcd's. 1.5.4 Definition $f\in R[t] - R$ is irreducible over a ring $R$ if $\forall g,h\in R[t]\;(f = gh\implies \{g,h\}\cap R \ne\varnothing)$ 1.5.5 Examples All polynomials of degree $1$ are irreducible. $t^2 - 2$ is irreducible over $\mathbb{Q}$ $\qquad$but reducible over $\mathbb{R}$. 1.5.6 Remark Clearly each $f\in R[t] - R$ is expressible as the product of irreducible polynomials. $\qquad$If $f$ is monic, then it is a product of monic irreducibles. We return our attention to the case where $R = F$ is a field. 1.5.7 Lemma If $f,g,h\in F[t]$ and $f$ is irreducible, then $(f\mid gh)\implies (f\mid g)\vee (f\mid h)$ Proof. A hcd of $f,\;g$ divides irreducible $f$ hence must be either $\lambda$ or $\lambda f\;(\lambda\in F^{\times}).$ The latter case $\qquad$implies $f\mid g$. Otherwise $f,\;g$ are relatively prime so Proposition 1.5.3 shows that $af + bg = 1$ $\qquad$for some $a,b\in F[t]$ hence $h = afh + bgh$ and so $f\mid h.\quad\square$ 1.5.8 Theorem Every $f\in F[t] - F$ is a product of irreducible polynomials, unique up to order $\qquad$and multiplication by constants. If $f$ is monic, it is a product of monic irreducible polynomials, $\qquad$unique up to order. 1.5.9 Definition If $R$ is a ring, $\alpha\in R,\;f\in R[t],\;\; f(\alpha) = 0$ then $\alpha$ is a zero or root of $f$. 1.5.10 Lemma $\alpha \in F$ is a zero of $f\in F[t]$ if and only if $(t -\alpha) \mid f$. 1.5.11 Example If $\underset{\,}{\,}f\in F[t],\;\deg(f).$ Then $f$ is irreducible if and only if $\forall \alpha\in F\; (f(\alpha) \ne 0)$. $\qquad$For instance $f = t^3 + t^2 + 2 \in \mathbb{Z}_3[t]$ is irreducible since $f(0)=2 = f(2),\,f(1) = 1.$ 1.6 Irreducibility over the rational field - elim - 04-04-2015 06:00 PM Proposition 1.6.1 (Gauss's Lemma) Irreducible polynomials over $\mathbb{Z}$ remain irreducible over $\mathbb{Q}\underset{\,}{.}$ Proof. Let $g,\,h\in\mathbb{Q}[t],\underset{\,}{\;} f = gh\in\mathbb{Z}[t],\; \deg(g)\le\deg(h) < \deg(f)$. We'll see that $f$ is reducible. $\qquad$Take the smallest $c = c_g c_h\in\mathbb{N}^+$ such that $g' = c_g g, \,h' = c_h h\in\mathbb{Z}[t]$ thus $cf = g'h'\underset{\,}{.}$ $\qquad$If $c > 1$, there is a prime $p\mid c$. Suppose $g' = \sum a_j t^j,\underset{\,}{\;} h = \sum b_j t^j$. Since $c$ is the minimum, $\qquad$both $\frac{1}{p}g',\;\frac{1}{p}h'\not\in\mathbb{Z}[t]$. So there are smallest $k,\,l\in\mathbb{N}$ such that $\frac{a_k}{p}\not\in\mathbb{Z},\;\frac{b_l}{p}\not\in\mathbb{Z}$. $\qquad$Now consider $cf$'s coefficient of $t^{k+l}$, that's $\small{\displaystyle{c_{k+l} = \sum_{j=0}^{k+l} a_j b_{k+l -j}}}.$ But $p\mid a_j b_{k+l-j}\iff j\ne k$ $\qquad$and so $p\nmid c_{k+l}\implies \frac{c}{p}f\not\in\mathbb{Z}[t]\implies f\not\in\mathbb{Z}[t]$. Such contradiction shows that $c = 1$ hence $\qquad f = g'h'$ is not irreducible. 1.6.2 Theorem (Eisenstein's criterion) $\qquad$If $f = a_n t^n +\cdots +a_0\in\mathbb{Z}[t] -\mathbb{Z}$ and $p\nmid a_n, \;p^2\nmid a_0,\; p\mid a_j \,(j < n)$ for some prime $p$, $\qquad$Then $f$ is irreducible over $\mathbb{Z}$ hence irreducible over $\mathbb{Q}$ (Lemma 1.6.1) Proof Suppose that $f = gh$ with $g = b_r t^r+\cdots +b_0\in\mathbb{Z}[t]-\mathbb{Z},\; h = c_s t^s + \cdots c_0\in\mathbb{Z}[T]-\mathbb{Z}$ $\qquad\because\;p\nmid a_n\implies (p\nmid b_r)\wedge (p\nmid c_s),$ there is a smallest $i(\le r < r+s =n)$ such that $p\nmid b_i$. $\qquad$So by assumption we have $p\mid a_i = \sum_{0\le j\le i} b_{i -j} c_j \implies p\mid c_0$. Symmetrically, $p\mid b_0$ $\qquad$hence $p^2 \mid a_0 = b_0 c_0$, A contradiction! 1.6.3 Examples $\qquad$(1) $f = \frac{2}{9}t^5 +\frac{5}{3}t^4 + t^3 +\frac{1}{3}$ is irreducible over $\mathbb{Q}$ if and only if $9f = 2t^5 + 15t^4+9t^3 + 3$ $\qquad\quad$is irreducible over $\mathbb{Z}$, and this is the case by Eisenstein's criterion with $p = 3$. $\qquad$(2) $t^3 +t^2 -5t +1$ is irreducible over $\mathbb{Q}$ if and only if $(t+1)^3 +(t+1)^2 -5(t+1) +1$ $\qquad\quad\; = t^3 + 4t^2 -2$ is irreducible over $\mathbb{Q}$. Which is true by Eisenstein's criteron for $p = 2$. 1.6.4 Remark For any prime $p$, $\overline{\quad}:\mathbb{Z}\to \mathbb{Z}_p\;(n\longmapsto \bar{n} = n +p\mathbb{Z})\underset{\,}{\,}$ (The natural quotient $\qquad$homomorphism) extends uniquely to $\overline{\quad}:\mathbb{Z}[t]\to \mathbb{Z}_p[t]\;(\sum a_j t^j = f\longmapsto \bar{f} = \sum \bar{a}_j t^j)$ 1.6.5 Lemma If $f\in \mathbb{Z}[t]$ whose leading coefficient is not divisible by prime $p$ and $\bar{f}\in \mathbb{Z}_p[t]$ is $\qquad$irreducible, then $f$ is irreducible over $\mathbb{Q}$. Proof. Suppose $f = gh\;(g,\,h\in \mathbb{Z}[t]-\mathbb{Z})$, then $\bar{f} = \bar{g}\bar{h}$ and $\deg(\bar{g})\le \deg(g) < \deg(f) = \deg(\bar{f}).$ $\qquad$Similarly $\deg(\bar{h}) < \deg{\bar{f}}$ and so $\bar{f}$ is not irreducibla.$\quad\square$ 1.6.6 Example Let $f = t^4+15 t^3 + 7\in\mathbb{Z}[t],\; p = 5$. Then $\bar{f} = t^4 + 2$ which has no zeroes in $\mathbb{Z}$. $\qquad$So if $\bar{f}$ is reducible, then $t^4 + 2 = (t^2 + at +b)(t^2 + ct +d)\;(a,b,c,d\in \mathbb{Z}_5)$. $\qquad$thus $a+c = 0 = ac + b+d,\; bd = 2$. Then $b+d = -ac = a(-c +0) = a(-c + a+c) = a^2$ $\qquad$and so $b(a^2 -b) = 2.$ But this is imposible for $b\in\mathbb{Z}_5,\; a^2\in\{0,1,4\}$. Therefore $\bar{f}$ is irreducible $\qquad$over $\mathbb{Z}_5$ and so $f$ is irreducible over $\mathbb{Z}$ (Lemma 1.6.5) $\qquad$Note that over $\mathbb{Z}_3$ we get $t^4 +1 = (t^2+t-1)(t^2-t-1)$. so choose prime $p$ with care. RE: Algebra Notes - elim - 04-04-2015 08:22 PM ............ Notes - elim - 04-05-2015 07:51 AM ...... Notes - elim - 04-05-2015 08:02 AM ........ Notes - elim - 04-06-2015 07:05 AM ...space...