Skip to main content
Contents
Search Book
Search Results:
No results.
Embed Readability settings Prev Up Next \(\require{mathtools}\require{color} \setcounter{MaxMatrixCols}{15}
\newcommand{\mbu}[1]{\mathbf{\mathrm{#1}}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\cC}{\mathcal{C}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\GF}{\mathrm{GF}}
\DeclareMathOperator{\Span}{Span}
\DeclareMathOperator{\rank}{rank}
\DeclareMathOperator{\lcm}{lcm}
\DeclareMathOperator{\rk}{rk}
\DeclareMathOperator{\wt}{wt}
\DeclareMathOperator{\im}{im}
\makeatletter
\@ifundefined{char}{
\DeclareMathOperator{\char}{char}
}{}
\makeatother
\DeclareMathOperator{\GRS}{GRS}
\DeclareMathOperator{\RS}{RS}
\DeclareMathOperator{\BCH}{BCH}
\newcommand{\transpose}[1]{#1^{\top}}
\newcommand{\by}{\mathbf{y}}
\newcommand{\bc}{\mathbf{c}}
\newcommand{\bx}{\mathbf{x}}
\newcommand{\bm}{\mathbf{m}}
\newcommand{\bs}{\mathbf{s}}
\newcommand{\be}{\mathbf{e}}
\newcommand{\bu}{\mathbf{u}}
\newcommand{\bv}{\mathbf{v}}
\newcommand{\bh}{\mathbf{h}}
\newcommand{\br}{\mathbf{r}}
\newcommand{\bzero}{\mathbf{0}}
\newcommand{\balpha}{\boldsymbol{\alpha}}
\newcommand{\subgroup}[1]{\langle{#1}\rangle}
\newcommand{\ideal}[1]{\langle{#1}\rangle}
\newcommand{\isom}{\cong}
\DeclareMathOperator\Aut{Aut}
\DeclareMathOperator\Char{char}
\DeclareMathOperator\cl{cl}
\DeclareMathOperator\Conj{Conj}
\DeclareMathOperator\Inn{Inn}
\DeclareMathOperator\Gal{Gal}
\DeclareMathOperator\Out{Out}
\DeclareMathOperator\orb{orb}
\DeclareMathOperator\Orb{Orb}
\DeclareMathOperator\Perm{Perm}
\DeclareMathOperator\stab{stab}
\DeclareMathOperator\Stab{Stab}
\DeclareMathOperator\fix{fix}
\DeclareMathOperator\Fix{Fix}
\DeclareMathOperator\Syl{Syl}
\DeclareMathOperator\Sym{Sym}
\DeclareMathOperator\soc{soc}
\DeclareMathOperator{\nil}{nil}
\DeclareMathOperator{\Nil}{Nil}
\DeclareMathOperator\jac{jac}
\DeclareMathOperator\Jac{Jac}
\DeclareMathOperator\Eq{Eq}
\DeclareMathOperator\Hol{Hol}
\DeclareMathOperator\Frac{Frac}
\DeclareMathOperator\Ann{Ann}
\DeclareMathOperator\GL{GL}
\DeclareMathOperator\SL{SL}
\DeclareMathOperator\So{SO}
\DeclareMathOperator\SU{SU}
\DeclareMathOperator\PGL{PGL}
\DeclareMathOperator\PSL{PSL}
\DeclareMathOperator\PSU{PSU}
\DeclareMathOperator\PSP{PsP}
\DeclareMathOperator\AGL{AGL}
\DeclareMathOperator\Heis{Heis}
\DeclareMathOperator\Dic{Dic}
\DeclareMathOperator\SA{SA}
\DeclareMathOperator\SD{SD}
\DeclareMathOperator\Fr{Fr}
\DeclareMathOperator\Mod{Mod}
\DeclareMathOperator\DQ{DQ}
\DeclareMathOperator\QD{QD}
\DeclareMathOperator\OD{OD}
\DeclareMathOperator\Cl{Cl}
\DeclareMathOperator\BinTet{2T}
\DeclareMathOperator\BinOct{2O}
\DeclareMathOperator\BinIcos{2I}
\DeclareMathOperator\Aff{Aff}
\DeclareMathOperator\BS{BS}
\newcommand{\normal}{\lhd}
\newcommand{\normaleq}{\unlhd}
\newcommand{\nnormal}{\ntriangleleft}
\newcommand{\nnormaleq}{\ntrianglelefteq}
\def\longto{\longrightarrow}
\def\into{\hookrightarrow}
\def\longinto{\longhookrightarrow}
\def\onto{\twoheadrightarrow}
\DeclareRobustCommand\longonto{\relbar\joinrel\twoheadrightarrow}
\DeclareMathOperator{\Image}{Im}
\DeclareMathOperator\Ker{Ker}
\DeclareMathOperator{\Id}{Id}
\newcommand{\ceil}[1] {\left\lceil #1 \right\rceil}
\newcommand{\floor}[1] {\left\lfloor #1 \right\rfloor}
\definecolor{xRed}{RGB}{229, 31, 58}
\definecolor{xBlue}{RGB}{68, 119, 170}
\definecolor{xGreen}{RGB}{33, 135, 51}
\definecolor{xPurple}{RGB}{170, 51, 119}
\definecolor{xOrange}{RGB}{ 197, 83, 17}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\newcommand{\fillinmath}[1]{\mathchoice{\underline{\displaystyle \phantom{\ \,#1\ \,}}}{\underline{\textstyle \phantom{\ \,#1\ \,}}}{\underline{\scriptstyle \phantom{\ \,#1\ \,}}}{\underline{\scriptscriptstyle\phantom{\ \,#1\ \,}}}}
\)
Section Introduction to Rings
Worksheet Part 1: Basic Ring Definitions
The goal of these problems is to practice with the basic definitions of ring theory.
1.
Find an integer \(n\) that shows that the rings \(\Z_n\) need not have the following properties enjoyed by \(\Z\text{.}\) In each part, \(a,b,c\in \Z_n\) are arbitrary elements of the ring.
\(a^2=a\) implies
\(a=0\) or
\(a=1\)
\(ab=0\) implies
\(a=0\) or
\(b=0\)
\(ab=ac\) and
\(a\neq 0\) implies
\(b=c\)
Solution .
There are many possible solutions. The smallest \(n\) where all three properties fail is \(n=6\text{.}\) Then we have
\(3^2=3\) but
\(3\neq 0\) and
\(3\neq 1\)
\(2\cdot 3=0\) but
\(2\neq 0\) and
\(3\neq 0\)
\(2\cdot 1=2\cdot 4\) but
\(1\neq 4\)
Definition 145 . Division in a Ring.
Let
\(R\) be a commutative ring and
\(a,b\in R\text{.}\) We say that
\(a\) divides \(b\) or
\(a\) is a
factor of
\(b\) and write
\(a\mid b\) if there exists a ring element
\(c\) such that
\(b=ac. \)
2. Show that
\((x-3)\) divides
\(x^2-9\) in
\(\Z[x]\text{,}\) the ring of polynomials with integer coefficients.
Solution .
We have
\(x^2-9=(x-3)(x+3)\text{,}\) so
\(x-3\) divides
\(x^2-9\) in
\(\Z[x]\text{.}\)
3. Show that no linear polynomial
\(x-a\) divides
\(x^2+1\) in
\(\Z[x]\text{.}\)
Solution .
Suppose we have \(x^2+1=(x-a)g(x)\) for some \(g(x)\in \Z[x]\text{.}\) Then from the degree we see \(g(x)\) must be a linear polynomial, say \(g(x)=bx+c\text{.}\) Then we have
\begin{equation*}
x^2+1=bx^2+(c-ab)-ac
\end{equation*}
so we have the system
\begin{equation*}
b=1, \quad c-ab=0, \quad 1=-ac\text{.}
\end{equation*}
From this we derive simultaneously \(c=a\) and \(1=-a^2\text{,}\) which is impossible in \(\Z\text{.}\) Therefore no linear polynomial divides \(x^2+1\) in \(\Z[x]\text{.}\)
4.
Prove as many of the following properties of rings as you can. Let \(a,b,c\) be elements of a ring \(R\) with identity \(1\) (not necessarily commutative).
\(\displaystyle a0=0a=0\)
\(\displaystyle a(-b)=(-a)b=-(ab)\)
\(\displaystyle (-a)(-b)=ab\)
\(a(b-c)=ab-ac\) and
\((b-c)a=ba-ca\)
\(\displaystyle (-1)a=-a\)
\(\displaystyle (-1)(-1)=1\)
\(0\text{,}\) \(-a\text{,}\) \(1\text{,}\) and
\(a^{-1}\) are unique, if they exist.
Solution .
Let \(a,b,c\) be elements of a ring \(R\) with identity \(1\) (not necessarily commutative).
\(a0=0a=0:\) Write
\begin{equation*}
a^2=a\cdot a = a(a+0)=a^2+a0\text{.}
\end{equation*}
Then by group cancellation for addition we have \(0=a0\text{.}\) By rewriting the left factor of \(a\) instead we can derive \(0a=0\text{.}\)
\(a(-b)=(-a)b=-(ab)\text{:}\) Write
\begin{equation*}
0=a0=a(b+(-b))=ab+a(-b)\text{.}
\end{equation*}
By uniqueness of additive inverses we have \(-(ab)=a(-b)\text{.}\) By swapping the role of a and b we can derive \(-(ab)=(-a)b\text{.}\)
\((-a)(-b)=ab\text{:}\) Since arbitrary elements
\(a\in R\) are in bijection with
\(-a\in R\text{,}\) we can replace the
\(a\) s in the previous property with
\(-a\) and the result follows at once since
\(-(-a)=a\text{.}\)
\(a(b-c)=ab-ac\) and \((b-c)a=ba-ca\text{:}\) Write
\begin{equation*}
a(b-c)=a(b+(-c))=ab+a(-c)=ab-ac
\end{equation*}
by distributivity and property 2. The second equality follows similarly.
\((-1)a=-a\text{:}\) Choose
\(a=1\) and
\(b=a\) in property 2.
\((-1)(-1)=1\text{:}\) Choose
\(a=b=1\) in property 3.
\(0\text{,}\) \(-a\text{,}\) \(1\text{,}\) and
\(a^{-1}\) are unique, if they exist: The first two follow from the properties of
\((R,+)\) as a group.
Now suppose that there are elements \(1_a\) and \(1_b\) in \(R\) that both act as a multiplicative identity. Then we have
\begin{equation*}
1_a=1_a\cdot 1_b=1_b\text{,}
\end{equation*}
where the first equality follows from the fact that \(1_b\) is a multiplicative identity and the second equality follows from the fact that \(1_a\) is a multiplicative identity. Therefore \(1_a=1_b\) and the multiplicative identity is unique.
Finally, suppose that there are two elements \(b,c\) in \(R\) such that \(ab=ba=1=ac=bc\text{.}\) Then we get
\begin{align*}
b \amp =b(1)\\
\amp =b(ac)\\
\amp =(ba)c\\
\amp =(1)c\\
\amp =c\text{.}
\end{align*}
Worksheet Part 2: Subring Practice
The goal of these problems is to practice determining whether a subset of a ring is a subring of that ring.
1.
Decide which of the following are subrings of
\(\Q\text{.}\)
(a) the set of all rational numbers with odd denominators (when written in lowest terms)
Solution .
This is a subring of \(\Q\text{.}\) Suppose we have \(a/b,c/d\) with \(b,d\) odd and \(\gcd(a,b)=\gcd(c,d)=1.\) Then we have
\begin{equation*}
(a/b)-(c/d)=\frac{ad-bc}{cd}
\end{equation*}
and \(cd\) is odd so \(\gcd(ad-bc,cd)\) is odd and so even after reducing to lowest terms this fraction is in the set. Similarly,
\begin{equation*}
(a/b)(c/d)=(ab)(cd)
\end{equation*}
is in the set because \(cd\) odd forces \(\gcd(ab,cd)\) odd. Therefore by the Subring Test this is a subring of \(\Q\text{.}\)
(b) the set of all rational numbers with even denominators (when written in lowest terms)
Solution .
This is not a subring of
\(\Q\) for various reasons. In particular,
\(0\) is not in the set because
\(0/1\) has an odd denominator.
(c) the set of nonnegative rational numbers
Solution .
This is not a subring of
\(\Q\) because it has no additive inverses for any nonzero element.
(d) the set of squares of rational numbers
Solution .
This is not a subring of
\(\Q\) because it has no additive inverses for any nonzero element (this is a subset of the previous set).
(e) the set of all rational numbers with odd numerators (when written in lowest terms)
Solution .
This is not a subring of
\(\Q\) because it does not contain
\(0\) (which has an even numerator).
(f) the set of all rational numbers with even numerators (when written in lowest terms)
Solution .
This is a subring of \(\Q\text{.}\) Let \(a/b,c/d\) be elements of the set with \(a,c\) even and \(\gcd(a,b)=\gcd(c,d)=1.\) Then
\begin{equation*}
(a/b)-(c/d)=\frac{ad-bc}{bd}
\end{equation*}
Since \(a,c\) are even and \(\gcd(a,b)=\gcd(c,d)=1\text{,}\) both \(b\) and \(d\) are odd. So \(bd\) is odd and thus \(\gcd(ad-bc,bd)\) is odd, so since \(ad-bc\) is even, the fraction reduces to lowest terms with an even numerator. Similarly,
\begin{equation*}
(a/b)(c/d)=(ac)/(bd)
\end{equation*}
will have an even numerator when written in lowest terms because \(ac\) is even and \(bd\) is odd. Therefore by the Subring Test this is a subring of \(\Q\text{.}\)
