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\centerline{Mathematics 351 -- Abstract Algebra I}
\centerline{Solutions for Midterm Exam 2 -- November 14, 2007}
\vskip 10pt
\noindent
I.  {\it Terminology.}
\vskip 10pt
\begin{enumerate}
\item[A)]  Let $G$ be a group and $n$ be an integer.
What does it mean to say that $|g| = n$ for $g \in G$?
\sol $|g|$ is the {\it order} of the element $g$.  This
is, by definition, the smallest positive integer 
such that $g^n = e$ (the identity element in $G$).
\vskip 10pt
\item[B)] Let $I$ be an ideal in a ring $R$.  What does it mean to
say that $a \equiv b \bmod I$?
\sol This equivalent to saying $a - b \in I$.
\vskip 10pt
\item[C)] Give an example of an ideal in a ring $R$ that is prime
but not maximal.
\sol  One such example is the ideal $P = \{(a,0) : a \in {\mathbb Z}\}$
in $R = {\mathbb Z}\times {\mathbb Z}$.  The quotient ring $R/P$ is 
isomorphic to ${\mathbb Z}$, which is an integral domain but
not a field.  Hence $P$ is a prime ideal, but not a maximal ideal. 
\end{enumerate}
\vskip 10pt
\noindent
II. Let $R$ be the ring of functions $f : {\mathbb R} \to {\mathbb R}$
that have derivatives of all orders.  Show that
$I = \{f \in R : f(0) = f'(0) = 0\}$ is an ideal in $R$.
\begin{enumerate}
\sol $I$ is clearly nonempty since it contains the zero function, $f(x) = x^2$, etc.
Let $f,g \in I$ so $f(0) = f'(0) = 0$ and $g(0) = g'(0) = 0$.  
Then $(f - g)(0) = f(0) - g(0) = 0$ and $(f - g)'(0) = f'(0) - g'(0) = 0 - 0 = 0$ (by the sum rule for derivatives).  Hence $f - g \in I$.  Next, 
let $f \in I$, and let $g$ be any function in $R$.  Since $R$
is a commutative ring, we only need to check that $fg \in I$.  But
$(fg)(0) = f(0) g(0) = 0\cdot g(0) = 0$.   Moreover, by the
product rule for derivatives, 
$$(fg)'(0) = f'(0)g(0) + f(0)g'(0) = 0\cdot g(0) + 0\cdot g'(0) = 0.$$
Hence $fg \in I$.
\end{enumerate}
\vskip 10pt
\noindent
III. All parts of this problem refer to $p(x) = x^2 + x$ in ${\mathbb Z}_2[x]$.
\vskip 10pt
\begin{enumerate}
\item[A)] How many distinct cosets of the ideal $(p(x))$
are there in ${\mathbb Z}_2[x]$?
\sol Since ${\mathbb Z}_2$ is a field, we have a division algorithm
in the polynomial ring ${\mathbb Z}_2[x]$, and by our general results
in this situation, the distinct cosets of $(p(x))$ are in one-to-one
correspondence with the possible remainders on division by 
$p(x)$.  Since $p(x)$ has degree 2, these remainders are all 
the polynomials of degree 1 or less in ${\mathbb Z}_2[x]$ (and the
zero polynomial): $0,1,x,x+1$.  There are 4 distinct cosets:
$$0 + (p(x)), 1 + (p(x)), x + (p(x)), x + 1 + (p(x)).$$
\item[B)] Construct the coset addition and multiplication tables for the
quotient ring  ${\mathbb Z}_2[x]/(p(x)).$
\sol  The addition table (omitting the $+ (p(x))$ in the description
of the cosets for simplicity):
\[
\begin{array}{cccccc}
          + &\vline & 0 & 1 & x & x+1\\ \hline
          0 &\vline & 0 & 1 & x & x+1\\
          1 &\vline & 1 & 0 & x + 1 & x\\
          x &\vline & x & x + 1 & 0 & 1\\
 x+1 & \vline & x + 1 & x & 1 & 0 \\
\end{array}
\]
The multiplication table is:
\[
\begin{array}{cccccc}
          \cdot &\vline & 0 & 1 & x & x+1\\ \hline
          0 &\vline & 0 & 0 & 0 & 0\\
          1 &\vline & 0 & 1 & x & x + 1\\
          x &\vline & 0 & x & x & 0\\
          x+1 & \vline & 0 & x + 1 & 0 & x + 1 \\
\end{array}
\]

\item[C)] State and prove the First Isomorphism Theorem for rings.
\sol  The First Isomorphism states that if $f : R \to S$ is 
a ring homomorphism, then ${\rm im}(f) \simeq R/\ker(f)$.  To prove this,
we write $K = \ker(f)$ and consider the mapping $\phi : R/K \to {\rm im}(f)$
defined by $\phi(a + K) = f(a)$.  This is well-defined since
if $a + K = b + K$, then $a - b \in K$, so $f(a - b) = 0$, which
shows $f(a) = f(b)$.  Hence $\phi(a + K) = \phi(b + K)$.  Next,
we show that $\phi$ is a ring homomorphism.  Let $a + K$ and $b + K$
be elements of $R/K$.  Then $\phi((a + K) + (b + K)) = 
\phi((a + b) + K) = f(a + b) = f(a) + f(b) = 
\phi(a + K) + \phi(b + K)$.
Similarly, $\phi((a + K) \cdot (b + K)) =
\phi((a\cdot b) + K) = f(a\cdot b) = f(a)\cdot f(b) = \phi(a + K)\cdot \phi(b + K).$ (We are using the definition of the 
coset sum and product operations and the fact that $f$ is a ring homomorphism here.  Now, $\phi$ is onto ${\rm im}(f)$ by definition.
Every element of ${\rm im}(f)$ is of the form $f(a)$ for some
$a \in R$.  Hence $f(a) = \phi(a + K)$.  Finally, $\phi$ is 
one-to-one since if $\phi(a + K) = \phi(b + K)$, then
$f(a) = f(b)$ which implies $f(a - b) = 0$ so $a - b \in K$, and
hence $a + K = b + K$.  
\vskip 10pt
\item[D)] Let ${\mathbb Z}_2 \times {\mathbb Z}_2 = \{(a,b) : a,b \in {\mathbb Z}_2\}$,
a ring under the component-wise sum and product operations.
Show that $f : {\mathbb Z}_2[x] \to {\mathbb Z}_2\times {\mathbb Z}_2$ defined by
$f(g(x)) = (g(0), g(1))$ is a ring homomorphism and determine $\ker(f)$.
\sol $f$ is a ring homomorphism because
$$f(g(x) + h(x)) = (g(0) + h(0), g(1) + h(1)) = (g(0),g(1)) + (h(0),h(1)) = f(g(x)) + f(h(x))$$
and 
$$f(g(x) \cdot h(x)) = (g(0) \cdot h(0), g(1) \cdot h(1)) = (g(0),g(1)) \cdot (h(0),h(1)) = f(g(x)) \cdot f(h(x)).$$
The kernel of $f$ consists of all polynomials $g(x) \in {\mathbb Z}_2[x]$
such that $g(0) = 0$ and $g(1) = 0$.  This means that 
$x$ and $(x + 1)$ must divide $g$, so $g(x) \in (p(x)) = (x(x+1))$.
On the other hand every polynomial in $(p(x))$ is clearly 
in $\ker(f)$ since such polynomials are zero at $x = 0$ and $x = 1$.
Hence $\ker(f) = (p(x))$.
\item[E)] Deduce that ${\mathbb Z}_2[x]/(p(x))$ is isomorphic
as a ring to ${\mathbb Z}_2 \times {\mathbb Z}_2$.
\sol  This follows from part D by the First Isomorphism Theorem.
Let $f$ be as in part D.
The image of $f$ is all of ${\mathbb Z}_2\times {\mathbb Z}_2$, so 
that is isomorphic to ${\mathbb Z}_2[x]/\ker(f)$
and $\ker(f) = (p(x))$.  {\it Comment}:  It is also possible to 
see this by comparing the addition and multiplication tables for
the two rings.  But it is necessary then to be somewhat careful in
defining the correspondence between the elements.  For example,
the ``obvious'' mapping 
\begin{eqnarray*}
0 + (p(x)) & \mapsto & (0,0)\\
1 + (p(x)) & \mapsto & (1,0)\\
x + (p(x)) & \mapsto & (0,1)\\
x + 1 + (p(x)) & \mapsto & (1,1)
\end{eqnarray*}
is an isomorphism of the additive groups of the two rings, but 
{\it not an isomorphism of rings} -- note that the multiplicative
identity in $\mathbb{Z}_2[x]/(p(x))$ is $1 + (p(x))$, but the 
multiplicative identity in $\mathbb{Z}_2 \times \mathbb{Z}_2$ is $(1,1)$. 
A ring isomorphism would have to map the multiplicative identity
to the multiplicative identity, so a correct mapping to get a ring
isomorphism here is:
\begin{eqnarray*}
0 + (p(x)) & \mapsto & (0,0)\\
1 + (p(x)) & \mapsto & (1,1)\\
x + (p(x)) & \mapsto & (0,1)\\
x + 1 + (p(x)) & \mapsto & (1,0).
\end{eqnarray*}
This is the mapping obtained from $f$ from part D by the proof
of the First Isomorphism Theorem(!)  There is also {\it another}
correct way to do this by mapping 
\begin{eqnarray*}
0 + (p(x)) & \mapsto & (0,0)\\
1 + (p(x)) & \mapsto & (1,1)\\
x + (p(x)) & \mapsto & (1,0)\\
x + 1 + (p(x)) & \mapsto & (0,1).
\end{eqnarray*}
\end{enumerate}
\noindent
IV.  Let $G = \{4,8,12,16\} \subset {\mathbb Z}_{20}$.  Is $G$ a group under
{\it multiplication} mod 20?  If so, say why and determine if $G$ is cyclic.
If not, identify which of the group axioms fail.
\begin{enumerate}
\sol  $G$ is a group under multiplication mod 20, as can be seen from 
the operation table
\[
\begin{array}{cccccc}
          \cdot &\vline & 4 & 8 & 12 & 16\\ \hline
          4 &\vline & 16 & 12 & 8 & 4\\
          8 &\vline & 12 & 4 & 16 & 8\\
          12 &\vline & 8 & 16 & 4 & 12\\
          16 & \vline & 4 & 8 & 12 & 16 \\
\end{array}
\]
We see $16$ is an identity for multiplication, every
element has an inverse (that is, for each $u$, there is
some $v$ such that $uv = 16$), and associativity follows
from associativity of multiplication in $\mathbb{Z}_{20}$.
The group $G$ {\it is cyclic}, since the powers of $8$ 
generate all the elements: 
$8^1 = 8$, $8^2 = 4$, $8^3 = 12$, $8^4 = 16$.
\end{enumerate}
\bigskip
\noindent
V.
\begin{enumerate}
\item[A)] State Lagrange's Theorem for groups.
\sol Lagrange's Theorem says that if $G$ is a finite
group and $H$ is a subgroup of $G$, then $|H|$ divides
$|G|$.  In particular, $|G| = [G:H]|H|$.  
\item[B)] Let $G$ be a group of order 60.  Is it possible for $G$
to contain elements $a,b$ with $|a| = 3$ and $aba^{-1} = b^4$?  Why or why not? 
\sol  There certainly can be such elements.  Note for instance that
a cyclic group of order 60 generated by $c$ will contain
elements $a = c^{20}$ and $b = c^{40}$, both of order 3.
Then $aba^{-1} = b a a^{-1} = b = b^4$ since $a$ and $b$ commute and 
$b$ also has order 3.
\end{enumerate}

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