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\begin{center}
\textbf{Mathematics 262} \\
\textbf{Homework 1 solutions}
\end{center}
\vspace{-0.05in}

1. Can you find a field with three elements?  With 4 elements?
[\textbf{Extra credit}: for which integers $n > 1$ can you find a
field with N elements?]

\textsc{Answer}: Yes, there is a field with three elements; there is
also a field with 4 elements.  If you want to know the answer to the
extra credit problem, let me know.  I won't give the answer here, in
case any of you want to keep working on it.

The integers modulo 3 form a field with three elements.  I'll give the
addition and multiplication tables.

\begin{tabular}{|c||c|c|c|}
\hline
$+$ & 0 & 1 & 2 \\ \hline \hline 
0 & 0 & 1 & 2 \\ \hline 
1 & 1 & 2 & 0 \\ \hline 
2 & 2 & 0 & 1 \\ \hline 
\end{tabular}
\hspace{0.6in}
\begin{tabular}{|c||c|c|c|}
\hline
$\times$ & 0 & 1 & 2 \\ \hline \hline 
0 & 0 & 0 & 0 \\ \hline 
1 & 0 & 1 & 2 \\ \hline 
2 & 0 & 2 & 1 \\ \hline 
\end{tabular}

Since addition and multiplication of integers satisfy most of the
appropriate properties (commutativity, associativity, distributivity),
then the same is true here.  The only thing you have to check is that
every nonzero element has a multiplicative inverse, and that's clear
from the multiplication table.

It may be harder to find a field with 4 elements, because the integers
modulo 4 don't work, and also because there are more possibilities for
addition and multiplication.  In any case, here are the tables:

\begin{tabular}{|c||c|c|c|c|}
\hline
$+$ & 0 & 1 & a & b \\ \hline \hline 
0 & 0 & 1 & a & b \\ \hline 
1 & 1 & 0 & b & a \\ \hline 
a & a & b & 0 & 1 \\ \hline 
b & b & a & 1 & 0 \\ \hline 
\end{tabular}
\hspace{0.6in}
\begin{tabular}{|c||c|c|c|c|}
\hline
$\times$ & 0 & 1 & a & b \\ \hline \hline 
0 & 0 & 0 & 0 & 0 \\ \hline 
1 & 0 & 1 & a & b \\ \hline 
a & 0 & a & b & 1 \\ \hline 
b & 0 & b & 1 & a \\ \hline 
\end{tabular}

To verify that these operations define a field, you have to check
commutativity for $+$ and $\times$ (clear since both tables are
symmetric), associativity for $+$ and $\times$, and distributivity.
(It is clear from the tables that $0$ and $1$ work as they should,
that every element has an additive inverse, and that every nonzero
element has a multiplicative inverse.)

Associativity for $+$: you have to see if $x+(y+z) = (x+y)+z$.  If any
of $x$, $y$, or $z$ is 0, then this is clear, so assume that they are
all nonzero.  If any two of them are equal, say $x=y$, then since
$x+x=0$ for every $x$ under consideration, this is also not too hard
to check.  This leaves the case in which they are all different.
Because of commutativity, this reduces to checking that $1+(a+b) =
(1+a)+b$.

Associativity for $\times$: check that $x(yz) = (xy)z$.  If any of
$x$, $y$, or $z$ is either $0$ or $1$, this is easy.  It is also easy
if $x=y=z$.  So you only have to check 
\begin{gather*}
a(ab) = (aa)b, \\
a(bb) = (ab)b, \\
a(ba) = (ab)a, \\
b(ab) = (ba)b, \\
b(aa) = (ba)a, \\
b(ba) = (ba)b.
\end{gather*}
Most of these follow from commutativity.  For example, the third one
holds because
\[
a(ba) = a(ab) = (ab)a.
\]
You only have to use the tables to check the first two, and these are
easy to verify.

Distributivity: This is similarly tedious, and I'm not going to work
it out.

\myline

2. Let $\mathbf{R}$ denote the field of real numbers.  Let
$\mathbf{R}^{\infty \times \infty}$ denote the set of all ``infinite
matrices'' with entries in $\mathbf{R}$; this is the set of all arrays
of real numbers of the form:
\[
\left[ 
\begin{array}{cccc}
a_{11} & a_{12} & a_{13} & \dots \\
a_{21} & a_{22} & a_{23} & \dots \\
a_{31} & a_{32} & a_{33} & \dots \\
\vdots & \vdots & \vdots
\end{array}
\right]
\]
You can add these the way you would expect, but I claim that the
standard matrix multiplication law has problems in this setting.
Explain.  Can you find a subset of $\mathbf{R}^{\infty \times \infty
}$ that forms a ring?

\textsc{Answer}: The problem with the multiplication is convergence.
If I try to multiply using ordinary matrix multiplication, I have to
take the dot product of two vectors from $\mathbf{R}^{\infty}$.  If
the $i$th row of the infinite matrix $A$ is $(1, 1, 1, \dots)$, and
the $j$th column of the infinite matrix $B$ is the same, then the
$(i,j)$-entry of the product $AB$ is the dot product of this vector
with itself, $\displaystyle \sum_{i=1}^{\infty} 1$, which doesn't
converge.

There are various ways to avoid this.  One of them is to restrict your
attention to infinite matrices in which every row and every column has
only finitely many nonzero entries.  Then when you take the dot
product of a row $(a_{i1}, a_{i2}, \dots)$ with a column $(b_{1j},
b_{2j}, \dots)$, the answer
\[
\sum_{k=1}^{\infty} a_{ik} b_{kj}
\]
has only finitely many nonzero terms, and therefore converges.  (You
have to check that this set of matrices is closed under addition and
multiplication, but this is not very hard.  Neither are associativity,
distributivity, etc.)

As a special case, you can just use diagonal matrices.  There are
other options, too.  If you want to get fancy, then you can consider
all infinite matrices whose rows and columns are the terms of
absolutely convergent series.  These form a ring; I'll let you check
the details.  (The main thing you have to check is that the
term-by-term product of absolutely convergence series is absolutely
convergent.  In other words, if $\sum a_{k}$ and $\sum b_{k}$ converge
absolutely, then so does $\sum a_{k} b_{k}$.)  Come to think of it,
you can probably get away with matrices whose rows are the terms of an
absolutely convergent series, and whose columns are the terms of any
convergent series.  (Or you can switch the roles of row and column:
look at all matrices whose columns give absolutely convergent series,
and whose rows give convergence series.)

\myline

3. Write down addition and multiplication tables for the ring
$\mathbf{Z}/6\mathbf{Z}$.

\textsc{Answer}: You perform ordinary addition and multiplication, and
then take remainders upon division by 6.  Here's the answer:

\begin{tabular}{|c||c|c|c|c|c|c|}
\hline
$+$ & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \hline 
0 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline
1 & 1 & 2 & 3 & 4 & 5 & 0 \\ \hline
2 & 2 & 3 & 4 & 5 & 0 & 1 \\ \hline
3 & 3 & 4 & 5 & 0 & 1 & 2 \\ \hline
4 & 4 & 5 & 0 & 1 & 2 & 3 \\ \hline
5 & 5 & 0 & 1 & 2 & 3 & 4 \\ \hline
\end{tabular}
\hspace{0.6in}
\begin{tabular}{|c||c|c|c|c|c|c|}
\hline
$\times$ & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \hline 
0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline
1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline
2 & 0 & 2 & 4 & 0 & 2 & 4 \\ \hline
3 & 0 & 3 & 0 & 3 & 0 & 3 \\ \hline
4 & 0 & 4 & 2 & 0 & 4 & 2 \\ \hline
5 & 0 & 5 & 4 & 3 & 2 & 1 \\ \hline
\end{tabular}

Note that this is not a field: the element 2 does not have a
multiplicative inverse (and neither do 3 or 4).

\myline

From Section 5.2 in the book:

3. I'm going to get tired of writing subscripts, so instead I'll write
the matrix $A$ as $\left( \begin{smallmatrix} a & b \\ c & d
\end{smallmatrix} \right)$.  Then the adjoint of $A$ is $\adj A = \left(
\begin{smallmatrix} \ d & -b \\ -c & \ a \end{smallmatrix} \right)$.

3(a) Multiply $\adj A$ by $A$ using ordinary matrix
multiplication:
\[
(\adj A) A = 
\begin{pmatrix}
\ d & -b \\ -c & \ a 
\end{pmatrix}
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
= 
\begin{pmatrix}
ad-bc & 0 \\ 0 & ad-bc
\end{pmatrix}
= 
(\det A) I.
\]
The same sort of computation works for the product $A (\adj A)$.

3(b) Compute the determinant of $\adj A$ using the standard formula
for a $2 \times 2$ determinant to get $da - (-b)(-c) = ad-bc = \det
A$.

3(c) Since $A^{t} = \left( \begin{smallmatrix} a & c \\ b & d
\end{smallmatrix} \right)$, then 
\[
\adj (A^{t}) = 
\begin{pmatrix}
\ d & -c \\ -b & \ a
\end{pmatrix}
= (\adj A)^{t}.
\]

\myline

12. I will assume that $D(A)$ is not 0 for some $A$, and prove that
$D(I) = 1$.  Indeed, suppose that $D(A) = c \neq 0$.  Then $c = D(A) =
D(A I) = D(A) D(I) = c D(I)$.  Now multiply both sides by $c^{-1}$: $1
= D(I)$.

Now, if $A$ is invertible with inverse $B$, then 
\[
1 = D(I) = D(AB) = D(A) D(B).
\]
Therefore $D(A)$ is nonzero (if $D(A) = 0$, then $D(A) D(B) = 0$ no
matter what $D(B)$ is).

\end{document}
