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\def\sol{\bigskip\item{}{\it Solution:} }
\nopagenumbers
\centerline{Mathematics 351 -- Abstract Algebra I}
\centerline{Solutions for Midterm Exam 1 -- October 1, 2007}
\bigskip
\noindent
I.  {\it Terminology.}
\item{A)} (5) Give an example of a ring $R$ that is {\it not} a {\it commutative ring}.
\sol
The ring of $2\times 2$ matrices with real entries, $M_{2\times 2}({\bf R})$, is
one example.
\bigskip
\item{B)} (5) What does it mean to say that $R$ is a {\it ring with identity}?
\sol
$R$ is a ring with identity if there is a {\it multiplicative identity}
in $R$: an element $1_R$ satisfying $1_R\cdot a = a\cdot 1_R = a$ for all $a \in R$.
\bigskip
\item{C)} (5) What does it mean to say that $u \in R$ is a {\it unit}?
\sol  $u$ is a unit if $u$ has a multiplicative inverse in $R$ (an element $x$
such that $u\cdot x = 1_R = x \cdot u$.  Note that this only makes sense
if $R$ is a ring with identity(!) 
\bigskip
\item{D)} (5) Is every commutative ring with identity an {\it integral domain}?
If so, say why.  If not, give a counterexample.
\sol 
The answer is {\it no}.  The ring ${\bf Z}_6$ is a commutative ring with identity,
but it is not an integral domain, since for instance, $[2][3] = 0$, but $[2] \ne [0]$
and $[3] \ne [0]$.
\bigskip
\noindent
II.  Let $R$ be a ring.  Let $b$ be any one particular element of $R$, and
let
$$C_b = \{a \in R : a\cdot b = b\cdot a\}.$$
\item{A)}  (10)  Show that $C_b$ is a subring of $R$.  
\sol
We use the criterion from Theorem 3.6 in Hungerford.  First notice that 
$C_b$ is not empty, since $0_R\cdot b = b\cdot 0_R = 0_R$.  Hence
$0_R \in C_b$.  Let $a, a' \in C_b$. Then by the distributive law
$$
(a - a')\cdot b = a\cdot b - a'\cdot b = b\cdot a - b\cdot a' = b\cdot(a - a').
$$ 
This shows $a - a' \in C_b$ so $C_b$ is closed under differences.  Similarly
by associativity of multiplication,
$$
(a \cdot a')\cdot b = a \cdot (a'\cdot b) = a\cdot (b \cdot a') = 
(a\cdot b) \cdot a' = (b \cdot a) \cdot a' = b \cdot (a \cdot a').
$$
Hence $C_b$ is also closed under products, and hence is a subring of $R$.  
\bigskip
\item{B)}  (10)  Let $S$ be a second ring, and let
$\varphi : R \to S$ be a ring homomorphism.  Show that if
$b \in R$, then $\varphi(C_b) \subseteq C_{\varphi(b)} \subseteq S$.  
\sol
Let $a \in C_b$, so $a\cdot b = b \cdot a$.
By the definition of a ring homomorphism, 
$$
\varphi(a)\cdot \varphi(b) = \varphi(a\cdot b) = \varphi(b\cdot a) = \varphi(b)\cdot \varphi(a).
$$
This shows $\varphi(a) \in C_{\varphi(b)}$.  Hence $\varphi(C_b) \subseteq C_{\varphi(b)}$.
($\varphi(C_b) \subseteq S$ by definition, so it is not necessary to prove that part).
\bigskip
\noindent
III.  Let $F[x]$ be the ring of polynomials with coefficients in a field
$F$.
\item{A)} (10) Show that the quotient and remainder under the division 
of $f(x)$ by $g(x)$ in $F[x]$ are unique.
\sol
Suppose $f(x) = q(x) g(x) + r(x)$ and $f(x) = q_1(x) g(x) + r_1(x)$, where
both $r(x)$ and $r_1(x)$ are either zero or have degree less than the degree
of $g(x)$.  Then subtracting these equations shows
$0 = (q(x) - q_1(x))g(x) + r(x) - r_1(x)$, so $r(x) - r_1(x) = (q_1(x) - q(x))g(x)$.
If $r(x) - r_1(x) \ne 0$, then it is a polynomial of degree $\le \deg(g)$.  However,
$\deg((q_1(x) - q(x))g(x)) = \deg(q_1(x) - q(x)) + \deg(g(x)) \ge \deg(g(x))$.
This gives a contradiction, so $r(x) = r_1(x)$ and $q(x) = q_1(x)$.  
\bigskip
\item{B)} (5)  {\it Define:}  $f(x)$ is {\it irreducible} in $F[x]$.
\sol
$f(x)$ is irreducible if in every factorization $f(x) = g(x) h(x)$ 
in $F[x]$, one of $g(x),h(x)$ is a unit (a nonzero constant polynomial),
and the other is an associate of $f(x)$.  
\bigskip
\item{C)} (10)  In what sense are factorizations of polynomials into 
irreducibles unique?  (State the theorem that answers this question; you
do not need to prove it.)
\sol
The sense in which irreducible factorizations are unique in $F[x]$ is that if 
$$f(x) = p_1(x) \cdots p_s(x) \qquad {\rm and} \qquad q_1(x) \cdots q_s(x),
$$
where all the $p_i(x)$ and $q_j(x)$ are irreducible in $F[x]$, then 
$r = s$, and for each $1 \le i \le s$, there exists some $j$ such that 
$p_i(x)$ and $q_j(x)$ are associates.  
\bigskip
\noindent
IV. 
\item{A)} (10)  Let $F$ be a field.  
Show that $a \in F$ is a root of $f(x) \in F[x]$ if and 
only if $(x - a)|f(x)$ in $F[x]$.
\sol
Use the division algorithm in $F[x]$, dividing $x - a$ 
into $f(x)$ to write $f(x) = q(x)(x - a) + r(x)$,
where either $r$ is zero, or else $\deg(r) < \deg(x - a) = 1$. 
If $a\in F$ is a root of $f(x)$, then $0 = f(a) = q(a)(a - a) + r$, so 
$r = 0$.  This shows $(x - a)|f(x)$.  Conversely, if $(x - a)|f(x)$, 
then $r = 0$, so $f(a) = q(a)(a - a) = 0$, so $a$ is a root of $f(x)$.  
\bigskip 
\item{B)} (10) Let $F$ be a field.  If $a_1,\ldots,a_n$ are {\it distinct}
roots of $f(x) \in F[x]$ in $F$, show that 
$\left((x - a_1)(x - a_2)\cdots (x - a_n)\right)|f(x)$.
(Hint:  induction on $n$).  
\sol
Following the hint, part A is the base case for the induction.  So assume
that whenever a polynomial $g(x)$ in $F[x]$ has $k$ distinct roots
$a_1, \ldots, a_k$, then $((x - a_1)\cdot\cdots\cdot (x - a_k))|g(x)$.
Assume now that $f(x)$ is a polynomial with $k + 1$ distinct
roots in $F$, say $a_1, \ldots, a_{k+1}$.  By part A, we know that $(x - a_{k+1})|f(x)$
so write 
$$f(x) = (x - a_{k+1})q(x)\leqno(1)$$
for some $q(x) \in F[x]$.  If we substitute $x = a_i$, for any $i$, $1 \le i \le k$, then 
$0 = f(a_i) = (a_i - a_{k+1})q(a_i)$.  The first factor cannot be zero since the 
$a_j$'s are distinct.  Hence $q(x)$ must have $a_1, \ldots, a_k$ as roots, 
and by induction $((x - a_1)\cdot\cdots\cdot (x - a_k))|q(x)$.  This 
means 
$$q(x) = ((x - a_1)\cdot\cdots\cdot (x - a_k)) s(x)\leqno(2)$$ 
for some $s(x) \in F[x]$.
Hence combining (1) and (2),
$$f(x) = (x - a_{k+1}) ((x - a_1)\cdot\cdots\cdot (x - a_k)) s(x)$$
which shows what we wanted to prove.
\bigskip
\item{C)} (5)  What are the roots of $x^2 + 2x \in {\bf Z}_8[x]$ in ${\bf Z}_8$?  
Why is your result {\it not a contradiction} to part B?
\sol
$x = 0, 2, 4, 6$ all satisfy $x^2 + 2x = 0$ in ${\bf Z}_8$.  This is not
a contradiction to part B since ${\bf Z}_8$ is {\it not a field}.  The 
statement in part B does not apply.
\bigskip
\noindent
V.  All parts of this question refer to 
$$f(x) = 5x^5 - 12 x^3 + 36 x^2 + 18\in {\bf Q}[x].$$
\item{A)} (5) List all rational numbers that could be roots of $f$ according 
to the Rational Root Test.  
\sol 
The possible rational roots of $f(x)$ are all rational numbers in lowest
terms of the form $p/q$, where $p|18$ and $q | 5$, so: 
$$x = \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18, \pm {1\over 5}, 
\pm {2\over 5},\pm {3\over 5},\pm {6\over 5},\pm {12\over 5},\pm {18\over 5}.$$
\bigskip
\item{B)} (5) Use any applicable method to decide whether $f$
is irreducible in ${\bf Q}[x]$, and state which criterion or criteria you are using.  
\sol
By the Eisenstein Criterion with $p = 2$, $f(x)$ is irreducible in 
${\bf Q}[x]$:  $2\not|\ 5$, $2|(-12), 2|36, 2|18$, but $2^2 = 4\not|\ 18$.  
\bye
