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\begin{document}

\begin{center}
{{\large\bf ALGEBRA I}\\

\bigskip

{\large\bf HOMEWORK ASSIGNMENT I}\\
\bigskip
{\bf (Sample Solutions.)}}
\end{center}

\vspace{.25in}

{\bf 
\underline{2.}
Suppose that a finite ordered set $\langle P;\leq\rangle$
is a semilattice with a top element.
Show that it is a lattice. Give an example
to show that this statement is not true for 
infinite ordered sets.
}

\medskip
{\bf Solution.}
Assume that $\lb A;\wedge\rb$ is a finite
(meet) semilattice with a top element $\top$.
To show that {\rm lub}'s exist, arbitrarily 
choose a pair
of elements $a, b\in A$. We will find
their {\rm lub}.

Let $U$ be the set
of common upper bounds of $a$ and $b$:
\[U = \{x\in A\;|\; a\leq x\;\mbox{\rm and}\;b\leq x\}.\]
$U$ is nonempty because it contains $\top$.
It is not to hard to prove that any finite nonempty
subset
of an ordered set has minimal elements (how?),
so $U$ has a minimal element $m$.\footnote{Here is the
definition of ``minimal'': If $\lb A;\leq \rb$ is an ordered set and 
$U\subseteq A$, then an element $m\in A$ is 
{\it minimal in $U$} if
\begin{enumerate}
\item[(i)] $m\in U$; and 
\item[(ii)] if $m'\in U$ and $m'\leq m$, then $m' = m$.
\end{enumerate}}
We will argue that $m = {\rm lub}(a,b)$.

We chose $m\in U$, so $m$ satisfies the first part of the
definition of ${\rm lub}(a,b)$:
\begin{enumerate}
\item[(i)] $a\leq m$, $b\leq m$.
\end{enumerate}
To show that $m$ satisfies the second part,
\begin{enumerate}
\item[(ii)] if $a\leq d$, $b\leq d$, then $m\leq d$,
\end{enumerate}
choose any $d\in A$ such that $a\leq d$ and $b\leq d$.
Our goal is to prove that $m\leq d$, or equivalently
that $m = m\wedge d$.

Since $a\leq d$ and $b\leq d$ we get that $d\in U$.
We claim that 
$m\wedge d\in U$ too.
To see this, observe that since $a\leq m$
and $a\leq d$ we get that
$a\leq m\wedge d$. Similarly $b\leq m\wedge d$. 
Thus $m\wedge d$ satisfies the condition defining
membership in $U$.

Since $m$ is minimal in $U$,
$m\wedge d\in U$, and $m\wedge d\leq m$, 
we conclude that $m\wedge d = m$.
Hence $m\leq d$. This completes the proof
that $m$ is the {\rm lub} of $a$ and $b$.

\vfill
\break

To give an example of an infinite meet semilattice
that is not a lattice, we can start building
our set with a pair
of incomparable elements $a$ and $b$, add their meet,
and then add an infinite descending chain above 
all elements of the set $\{a, b, a\wedge b\}$:

\bigskip

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%\put(5,17.5){\circle*{1}}
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\put(5,0){\line(1,1){5}}
\put(5,0){\line(-1,1){5}}
\put(5,25){\line(0,-1){11}}
\put(4.6,10){$\vdots$}
\put(3.5,28){$\top$}
\put(-4,4){$a$}
\put(12,4){$b$}
\put(1,-4){$a\wedge b$}
\end{picture}
\end{center}

\bigskip\bigskip

The infinite descending chain is chosen
so that it has no least element.
The elements $a$ and $b$ are the only incomparable 
elements in this order. Since they have a meet,
this IS a meet semilattice. If they had a join it would have
to be the least element in the descending chain, and
we built this ordered set so that there is no such element.

\bigskip\bigskip

{\bf 
\underline{4.}
Show that a lattice that satisfies one distributive law:
\[x\wedge (y\vee z) = (x\wedge y)\vee (x\wedge z)
\]
also satisfies the other one:
\[x\vee (y\wedge z) = (x\vee y)\wedge (x\vee z).\]
Thus, the dual of a distributive lattice is distributive.
}

\medskip
{\bf Solution.} Assume that $\m l$ satisfies 
the first distributive law. Then 
\[
{\arraycolsep=0pt
\begin{array}{rlll}
\underline{(x\vee y)}\wedge (\underline{x}\vee \underline{z})
&=(\underline{(x\vee y)}\wedge \underline{x})\vee 
(\underline{(x\vee y)}\wedge \underline{z})&
\hspace{.5in}&\mbox{\rm Dist. Law $\#1$.}\\
&=(x\wedge (x\vee y))\vee 
(z\wedge (x\vee y))&
\hspace{.5in}&\mbox{\rm Comm. Law.}\\
&=x\vee 
(z\wedge (x\vee y))&&\mbox{\rm Absorp. Law.}\\
&=x\vee ((z\wedge x)\vee (z\wedge y))&&\mbox{\rm Dist. Law $\#1$.}\\
&=x\vee ((x\wedge z)\vee (y\wedge z))&&\mbox{\rm Comm. Law.}\\
&=(x\vee (x\wedge z))\vee (y\wedge z))&&\mbox{\rm Assoc. Law.}\\
&=x\vee (y\wedge z))&&\mbox{\rm Absorp. Law.}
\end{array}}
\]

\vfill
\break

{\bf 
\underline{5.} Let ${\bf L}$ be a lattice. For a fixed $a\in L$
define a function 
\[J_a(x):L\rightarrow L: x\mapsto a\vee x.\]
Show that if ${\bf L}$ is distributive, then 
$J_a(x\vee y) = J_a(x)\vee J_a(y)$ and
$J_a(x\wedge y) = J_a(x)\wedge J_a(y)$.
}

\medskip
{\bf Solution.}
\[
{\arraycolsep=0pt
\begin{array}{rl}
J_a(x\vee y) &= a\vee (x\vee y)\\
&= (a\vee a) \vee (x\vee y)\\
&= a\vee (a \vee (x\vee y))\\
&= a\vee ((a \vee x)\vee y)\\
&= a\vee (y\vee (a \vee x))\\
&= (a\vee y)\vee (a \vee x)\\
&= (a\vee x)\vee (a \vee y)\\
&= J_a(x)\vee J_a(y).
\end{array}}
\]

\[
{\arraycolsep=0pt
\begin{array}{rl}
J_a(x\wedge y) &= a\vee (x\wedge y)\\
&= (a\vee x) \wedge (a\vee y)\\
&= J_a(x)\wedge J_a(y).
\end{array}}
\]

\bigskip\bigskip

{\bf 
\underline{7.} Show that the lattice of subspaces of a 
vector space is modular.
}

\medskip
{\bf Solution.}
Assume instead that a certain vector space has subspaces
$U, V$ and $W$ such that 

\bigskip\bigskip

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\put(22,19){$U$}
\put(-5,14){$W$}
\put(-2,-5){$U\cap W = V\cap W$}
\put(-2,32){$U + W = V + W$}
\end{picture}
\end{center}

\bigskip\bigskip

Choose a vector $a\in U-V$. Since 
$a\in U\subseteq V+W$ we can express $a$
as $a = v+w$ for some $v\in V\;(\subseteq U)$ and some $w\in W$.
Now $w = a-v$, and both $a$ and $v$ belong to $U$, so
$w = a-v\in U$. This shows that $w\in U\cap W = V\cap W\subseteq
V$.
But if $v, w\in V$, then $a = v+w\in V$. This contradicts our
choice of $a$. We conclude that there is no pentagon
in the lattice of subspaces.

\vfill
\break

{\bf 
\underline{8.} Show that the lattice of equivalence relations
on $X = \{1, 2, 3, 4\}$ is not modular.
}

\medskip
{\bf Solution.}

\bigskip

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\put(7,32){$1234$}
\end{picture}
\end{center}

\end{document}

