\magnification=\magstep1
\centerline{Mathematics 351 -- Abstract Algebra I}
\centerline{Solutions for Problem Set 1}
\centerline{September 8, 2007}
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\noindent
{\it Section 3.1}
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\noindent
8.  a)  $\overline{R} = \{(0,0),(1,0),(2,0)\}$ and $\overline{S} = \{(0,0),(0,1),(0,2),(0,3),(0,4)\}$.
\medskip\noindent
b)  (Note that the question says ``for all rings $R,S$'').  To 
save some writing, we will use the subring criterion from Theorem 
3.6 in the next section. $\overline{R}$ is nonempty since $R$ is.
 Let $(a,0_S), (b,0_S) \in \overline{R}$.  
Then $(a,0_S) - (b,0_S) = (a-b,0_S-0_S) = (a-b,0_S)$.  
Since $R$ is a ring, $a-b\in R$.  Hence $\overline{R}$ is closed
under differences.  Similarly, 
$(a,0_S)\cdot (b,0_S) = (a\cdot b, 0_S\cdot 0_S) = (a\cdot b, 0_S)$.
Since $R$ is a ring $a\cdot b\in R$.  Hence $\overline{R}$ is closed
under products.  It follows that $\overline{R}$ is a subring of 
$R\times S$.  
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c.  The proof is similar to the proof of b.
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\noindent
9.  Again we will use the criterion from Theorem 3.6.
Let $a + b\sqrt{2}, c + d\sqrt{2}$ be in ${\bf Z}[\sqrt{2}]$, which
is clearly nonempty.
Then 
$$(a + b\sqrt{2}) -  (c + d\sqrt{2}) = (a - c) + (b - d)\sqrt{2}$$
Since $a,c,b,d$ are integers, so are $a - c$ and $b - d$.  Thus
${\bf Z}[\sqrt{2}]$ is closed under differences.  Similarly,
$$(a + b\sqrt{2}) \cdot (c + d\sqrt{2}) = ac + ad\sqrt{2} + bc \sqrt{2} + 2bd = 
(ac + 2bd) + (ad + bc)\sqrt{2}$$
Since $a,c,b,d$ are integers, so are $ac + 2bd$ and $ad + bc$.  Thus
${\bf Z}[\sqrt{2}]$ is closed under products and a subring of ${\bf R}$.
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\noindent
11.  See the answer in the text.  {\it Note:}  When answers are given like
this, they are {\it only} to help you check your work.  To get full credit, you must show more details than given in the textbook.
For calculational problems like this, include
full details for at least some parts.
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\noindent
16.  Closure under sums and products follows by constructing the operation
tables, or by noting that the sum and product of integers divisible by 
$3$ is divisible by $3$, and the same is true when we take classes mod $18$, since
$18$ is also divisible by $3$ (if $3|n$ then $3|r$ where $n = 18q + r$ by division).  
$0 \in R$ by construction, and each element of $R$ has an additive
inverse in $R$ ($-0 = 0$, $-3 = 15$, $-6 = 12$, $-9 =9$, $-12 = 6$, and $-15 = 3$).  
Thus $R$ is a subring of ${\bf Z}_{18}$ by the criterion from Theorem 3.2 in 
the text.
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\noindent
{\it Section 3.2}
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\noindent
1 a) In a general ring $R$, we have distributive laws, but we 
do not assume $R$ is commutative, so
$$(a+b)(a - b) = a^2 + ba - ab - b^2$$
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b)  Similarly,
$$(a+b)^3 = (a+b)(a^2 + ab + ba + b^2) = a^3 + a^2b + aba + ab^2 + ba^2 + bab + b^2a + b^3.$$
\medskip\noindent
c)  If $R$ is commutative, then we simplify to the usual forms
$(a+b)(a-b) = a^2 - b^2$ and $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.
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\noindent
2.  a) There are infinitely many different idempotent $2\times 2$ matrices.  Four of 
them are the matrices:
$$\pmatrix{0&0\cr 0&0\cr}, \pmatrix{1&0\cr 0&1\cr}, \pmatrix{1&0\cr 0&0\cr}, 
\pmatrix{0&0\cr 0&1\cr}$$
\medskip\noindent
b)  The classes of $0,1,4,9$ are all of the idempotents in ${\bf Z}_{12}$. 
For example,
$$9^2 = 81 \equiv 9 \pmod{12},$$ 
since $81 - 9 = 72 = 6\cdot 12$, which shows that the class of 9 is an 
idempotent mod 12.
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c)  Let $e$ be an idempotent in an integral domain $R$.  Then 
$e^2 = e$, which implies $e^2 - e = 0_R$.  Factoring on the left,
this shows $e(e - 1_R) = 0_R$.  Since we assume $R$ is an integral
domain, this shows either $e = 0_R$, or $e - 1_R = 0_R$.  The second
case is equivalent to $e = 1_R$.  
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\noindent
8.  a)  Almost any randomly chosen matrix will fail to commute
with some matrix.  Matrices that do commute with every matrix 
are any matrix of the form 
$\pmatrix{r&0\cr 0&r}$ for $r \in {\bf R}$.
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b)  We must show that the given $C$ is a subring of $R$.
We will use the criterion of Theorem 3.6.  First note that
$C$ is always nonempty since it contains at least $0_R$
(part 1 of Theorem 3.5 in the text).  Let $a,b \in C$.
Then $ar = ra$ and $br = rb$ for all $r \in R$.  Then
using other parts of Theorem 3.5 and the above,
$$(a - b)r = ar - br = ra - rb = r(a - b)$$
for all $r\in R$.  It follows that $a - b \in C$.
Similarly, using associativity and the equations $ar = ra$, $br = rb$,
$$(ab)r = a(br) = a(rb) = (ar)b = (ra)b = r(ab)$$
for all $r \in R$.  Hence $ab \in C$, and $C$ is a subring of $R$.
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\noindent
10.  The statement is {\it false}, which can be seen by considering
the counterexample $R = {\bf Z}$.  The set of units in ${\bf Z}$ is 
easily seen to be $U = \{1, -1\}$ (no other nonzero integer has
a multiplicative inverse in ${\bf Z}$).  $U$ is not closed under
sums:  $1 + -1 = 0$.  Hence $U$ is not a subring of ${\bf Z}$.
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\noindent
12.  For this problem, we need to recall a number of facts about 
integers, divisibility, and gcds that 
were covered in Algebraic Structures (MATH 243), or Chapters 1 and 
2 of Hungerford (see Theorem 1.3 on page 9 in particular).  First, the 
notation $(a,n)$ is shorthand for the {\it greatest common 
divisor} of $a,n$ in ${\bf Z}$.  Second, for all integers $a,b$
not both zero, $(a,b)$ is the smallest positive integer in the
set $\{ua + vb | u,v \in {\bf Z}\}$.  Hence, in particular, the
gcd itself can be written in the form $(a,b) = ua + vb$ for 
some integers $u,v$.  Finally note that each part is an ``if and only if''
statement, so we must prove both implications.
\medskip\noindent
a)  $\Rightarrow:$  Let $[a]$ be a unit in ${\bf Z}_n$.  This means
that there is some $[u]$ in ${\bf Z}_n$ such that $[u][a] = [1]$ in 
${\bf Z}_n$ (since $[1]$ is the multiplicative identity in this
ring).  But this says $ua \equiv 1\!\!\!\pmod n$, so as an equation in
${\bf Z}$, 
$au - 1 = (-v)n$
for some integer $v$.  This last equation shows $ua + vn = 1$.
By the fact about the gcd mentioned above, this implies that 
$(a,n) = 1$. 
\medskip\noindent
$\Leftarrow:$  Assume that $(a,n) = 1$.  Then there are 
integers $u,v$ such that $ua + vn = 1$.  This can be 
rearranged to say $ua \equiv 1\!\!\!\pmod n$, so $[u][a] = [1]$
in ${\bf Z}_n$.  It follows that $[a]$ is a unit in 
${\bf Z}_n$.
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\noindent
b) $\Rightarrow$:  Assume that $[a]$ is a nonunit in ${\bf Z}_n$.  By 
part a, this means that $(a,n) = d > 1$ in ${\bf Z}$.  By the definition
of the gcd, we have $a = dp$ and $n = dq$ for some integers $p,q$. 
We may assume $0 < a < n-1$ if we like.  From $n = dq$, since $d > 1$, it
follows that $[q] \ne [0]$ in ${\bf Z}_n$.  Multiplying by $q$ on both
sides of the equation $a = dp$, we get:
$$qa = q(dp) = (dq)p = np.$$
Hence $[q][a] = [0]$ in ${\bf Z}_n$. This implies $[a]$ is a zero 
divisor in ${\bf Z}_n$ (if $[a] \ne [0]$). 
\medskip\noindent
$\Leftarrow:$  If $[a] \ne [0]$ is a zero divisor in ${\bf Z}_n$, there
is some $[b] \ne [0]$ such that $[a][b] = [0]$ in ${\bf Z}_n$.
If $[a]$ was a unit, then we could multiply both sides of this equation
by $[u] = [a]^{-1}$ and we would obtain the contradiction $[b] = [0]$.  
Hence $[a]$ cannot be a unit.
({\it Comment:}  This last reasoning also shows that in any ring
a zero divisor cannot also be a unit.)
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\noindent
18.  (A different method from the hint.)  $0_R \in S$ because $S$ is 
a subring and is closed under differences.  The element $0_R$ satisfies
$0_R + a = a$ for all $a \in S$, so it is an additive identity for $S$.
Then $0_R = 0_S$ by the uniqueness of the additive identity in a 
ring (see Theorem we proved in class on Wednesday, September 5).  
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\noindent
19.  a) Consider $R = {\bf Z}_6$ and $S = \{0,2,4\}$.  Then it is easy to
see $S$ is a subring of $R$.  Since $0\cdot 4 = 0$, $2\cdot 4 = 2$,
and $4\cdot 4 = 4$ in ${\bf Z}_6$, $4$ is a multiplicative identity for $S$ 
which is different from the identity $1 \in R$.  
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\noindent
b)  Let $a \ne 0$ be an element of the subfield $S$ (there must be such an 
element by the definition of a field).  Then we have
$a\cdot 1_S = a$ and $a \cdot 1_R = a$, so $a\cdot (1_S - 1_R) = 0_R$.
Multiply both sides by the multiplicative inverse $a^{-1}$ from $R$.
We get $1_S - 1_R = 0_R$, so $1_S = 1_R$.    
({\it Comment:} Actually, we only used the hypothesis that $R$ is a field.)
\medskip
\noindent
c)  Because $1_S$ is a multiplicative identity
for $S$, we have $1_S\cdot 1_S = 1_S$.  Similarly, since
$1_R$ is a multiplicative identity for $R$ and $1_S \in R$,
$1_S\cdot 1_R = 1_S$.  Subtracting these equations gives 
$1_S\cdot(1_S - 1_R) = 0_R$.  This equation shows that if
$1_S \ne 1_R$, then $1_S$ is a zero-divisor in $R$.  
({\it Comment:}  the statement ``if $p$, then $q$ or $r$''
is logically equivalent to ``if $p$ and not $q$, then $r$''.)
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\noindent
31.  {\it Comment:}  For this question and the next one, you need to 
look at the definition of $n\cdot a$ for $n\in {\bf Z}$ and $a\in R$
from page 59 of the text.  This is {\it not} a product using the
multiplication in the ring $R$.  It is a shorthand notation 
the {\it sum} $a + a + \cdots + a$ ($n$ terms). 
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\noindent
a)  In ${\bf Z}$, we never have $1 + 1 + \cdots + 1 = 0$, if there are
$n \ge 1$ terms.  Hence ${\bf Z}$ has characteristic zero.  
In ${\bf Z}_n$, we have $1 + 1 + \cdots + 1 = 0$ ($n$ $1$'s), and this
is the smallest number of 1's that adds up to 0.  Hence the 
characteristic is $n$.  
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\noindent
b)  The additive and multiplicative identities are $(0,0),(1,1)$ 
respectively.  Adding $n$ $(1,1)$'s together will only give $(0,0)$
in ${\bf Z}_4 \times {\bf Z}_6$ 
when $n$ is a multiple of $4$ and a multiple of $6$ simultaneously.
Hence the characteristic of ${\bf Z}_4 \times {\bf Z}_6$ is ${\rm lcm}(4,6) = 12$.
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\noindent
32.  Let $R$ be a finite ring with identity, and consider the 
elements $n\cdot 1_R$ for all $n\ge 1$.  Since there are infinitely
many integers $n$ but only finitely many different elements in $R$,
if we compute the $n\cdot 1_R$ for each $n\ge 1$ in turn, 
at some point, we must obtain a ``repeat'' of an element previously
seen:
$m\cdot 1_R = k\cdot 1_R$ for some $1\le k < m$.  From the definition,
we have $(m - k)\cdot 1_R = 0_R$, and $m - k > 0$.  Hence
$$S = \{n > 0\ |\ n \cdot 1_R = 0_R\}$$
is a nonempty set of natural numbers.  By the Well-Ordering Property
it has a smallest element, which is the characteristic of $R$.      
\bye
