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\title[101 Midterm Exam]{Math 101: Algebra I \\ Midterm Exam solutions}
\date{11 October 2016}

\maketitle

\begin{problab}{1}
Let $G$ be a group.  Indicate if the following statements are true or false.  If true, give a proof; if false, give an explicit counterexample.

\begin{enumalph}
\item If $H,H' \trianglelefteq G$ and $G/H \simeq G/H'$, then $H \simeq H'$.

\item If $H,H' \trianglelefteq G$ and $H \simeq H'$, then $G/H \simeq G/H'$.

\item If $K,K'$ are groups and $G \times K \simeq G \times K'$, then $K \simeq K'$.
\end{enumalph}
\end{problab}

\begin{solu}
Part (a) is false.  Take $G=\Z/4\Z \times \Z/2\Z$, $H=\langle (0,1) \rangle \simeq \Z/2\Z$ and $H'=\langle (2,0) \rangle \simeq \Z/2\Z$.  Then $G/H \simeq \Z/4\Z$ and $G/H' \simeq \Z/2\Z \times \Z/2\Z$.  

Part (b) is also false. Take $G=\Z$ and $H=\Z$ and $H'=2\Z \simeq \Z$; then $G/H = \{1\} \not\simeq \Z/2\Z \simeq G/H'$.

Part (c) is also also false.  Take $G$ to be a countable product of copies of $\Z/2\Z$, and $K=\Z/2\Z$ and $K'=\Z/2\Z \times \Z/2\Z$.  Then $K \not\simeq K'$, but $G \simeq G \times K \simeq G \times K'$ are all isomorphic.  (It turns out that if $G,K,K'$ are finite, then the result becomes true, but this is not easy to prove.)
\end{solu}

\begin{problab}{2}
Let $R$ be a Euclidean domain with norm $N$.
\begin{enumalph}
\item Let 
\[ m=\min(\{ N(a) : a \in R, a \neq 0\}). \]
Show that every nonzero $a \in R$ with $N(a)=m$ is a unit in $R$.
\item Deduce that a nonzero element of norm zero in $R$ is a unit; show by an example that the converse of this statement is false.
\item Let $F$ be a field and let $R=F[[x]]$.  Show that $R$ is Euclidean.  What does part (a) tell you about $R^\times$?  What are the irreducibles in $R$, up to associates?
\end{enumalph}
\end{problab}

\begin{solu}
Let $a$ be a nonzero element of norm $m$.  Then we can write $1=qa+r$ with $r=0$ or $N(r)<N(a)=m$.  We cannot have the latter, since $m$ is the smallest such, hence $r=0$ so $1=qa$ and hence $a \in R^*$, which proves (a).  For (b), if there is an element of norm zero then $m=0$ so by (a) every nonzero element of norm zero is a unit.  The converse of this statement is false, namely, that every unit has norm zero: the ring $\Z[i]$ is a Euclidean domain and $a \in \Z[i]$ is a unit with respect to the complex norm if and only it has norm $1$.

Finally, part (c).  For $\alpha = a_nx^n+\dots \in F[[x]]$ with $a_n \neq 0$, we define the norm $N(\alpha)=n \geq 0$.  Then $R$ is Euclidean under this norm as follows.  Let $\alpha,\beta \in R$ with $\beta \neq 0$.  If $N(\alpha)<N(\beta)$, then we can write $\alpha=0\beta+\alpha$.  Otherwise $N(\alpha) \geq N(\beta)$, and we claim $\beta \mid \alpha$, i.e., $\alpha=(\alpha/\beta)\alpha+0$ with $\alpha/\beta \in F[[x]]$.  Indeed, write $\alpha=x^{N(\alpha)} \alpha_0(x)$ and $\beta=x^{N(\beta)} \beta_0(x)$ with $\beta_0(x)=b_0+\dots$ and $b_0 \neq 0$; we showed in class that $\beta_0(x) \in F[[x]]^\times$ by solving linear equations, so $\alpha/\beta = x^{n-m}\alpha_0(x)\beta_0(x)^{-1} \in F[[x]]$.  Therefore $F[[x]]$ is Euclidean under this norm.  Then part (a) reminds us that $R^\times=F[[x]]^\times$ consists of the elements with nonzero constant term, reading off the definition of the norm.  The only irreducible, up to associates, is $x$.  Indeed, we know that $F[[x]]$ is a UFD so irreducibles are the same as primes, and $F[[x]]/(x) \simeq F$ so $x$ is irreducible; and any $\alpha(x)=x^{N(\alpha)} \alpha_0(x) \neq 0$ with $\alpha_0(x) \in F[[x]]^\times$ is then a factorization of $\alpha(x)$ as a power of the irreducible $x$ (times a unit).  
\end{solu}

\begin{problab}{3}
Let $F$ be a field and let $V=\Mat_{2\times 3}(F)$ be the $F$-vector space of $2\times 3$-matrices.  
\begin{enumalph}
\item The group $\GL_2(F)$ acts on $V$ by left multiplication.  For $M,M' \in V$, the relation $M \sim M'$ if and only if $M'=AM$ for some $A \in \GL_2(F)$ defines an equivalence relation on $V$.  

What are the equivalence classes (i.e., the \emph{orbits} of the action)?

\item Show that this action $\GL_2(F) \circlearrowright V$ induces an injective group homomorphism 
\[ \phi:\GL_2(F) \hookrightarrow \Aut_F(V).  \]

\item Under the isomorphism $\Aut_F(V) \simeq \GL_6(F)$ given by the basis of matrix units, describe $\phi$ explicitly.
\end{enumalph}
\end{problab}

\begin{solu}
For (a), a matrix $A \in \GL_2(F)$ acts on the left by row operations.  So every $M \in V$ can be put into reduced row echelon form by this action.  By linear algebra, the reduced row echelon form is unique.  The possible forms (choosing pivots) are
\[ \begin{pmatrix} 1 & 0 & * \\ 0 & 1 & * \end{pmatrix},  
\begin{pmatrix} 1 & * & 0 \\ 0 & 0 & 1 \end{pmatrix}, 
\begin{pmatrix} 1 & * & * \\ 0 & 0 & 0 \end{pmatrix}, 
\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}, 
\begin{pmatrix} 0 & 1 & * \\ 0 & 0 & 0 \end{pmatrix}, 
\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix},  
\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]
where $*$ denotes an arbitrary element of $F$.

For (b), we get a map $\phi:\GL_2(F) \to \End_F(V)$ because matrix multiplication map is $F$-linear: 
\[ A(M+cM')=AM+cAM' \quad\text{for all $A \in \GL_2(F)$, $M,M' \in V$, and $c \in F$}. \]  
The map is a homomorphism because this holds for matrix multiplication: 
\[ \phi(AB)(M)=(AB)M=A(B(M))=(\phi(A) \circ \phi(B))(M) \]
for all $A,B \in \GL_2(F)$ and $M \in V$.  In a group action, we always have the image landing in the symmetric group on the set (acting bijectively), and indeed the inverse to $A$ is $A^{-1}$, so the image lands in $\Aut_F(V)$.  Finally, the map is injective: take $M=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$ to see that $AM=M$ implies $A=1$.

For (c), taking the basis $e_{11},e_{21},e_{12},e_{22},e_{13},e_{23}$, and noting that matrix multiplication acts independently on column vectors, we compute that $\phi(A)$ is the block diagonal matrix with three copies of $A$ down the diagonal, for each $A \in \GL_2(F)$.  
\end{solu}

\begin{problab}{4}
For the purposes of this exercise, we say that an isomorphism of $F$-vector spaces is \emph{natural} if it does not depend on a choice of basis.  

Let $F$ be a field and let $V,W$ be finite-dimensional vector spaces over $F$.  Show that there is a (well-defined) natural isomorphism of $F$-vector spaces
\[ \phi:V^* \otimes_F W \xrightarrow{\sim} \Hom_F(V,W). \]
\end{problab}

\begin{solu}
To start, note that 
\[ \dim_F(V \otimes_F W^*) = \dim_F(V)\dim_F(W^*)=\dim_F(V)\dim_F(W)=\dim_F \Hom_F(V,W) \]
so there certainly is an isomorphism.  By this dimension count, it is enough to exhibit a natural injective $F$-linear map.

There is really only one thing we could write down: given a simple tensor $f \otimes w \in V^* \otimes W$, we define $\phi(f \otimes w) \in \Hom_F(V,W)$ by $\phi(f \otimes w)(v) = f(v)w$, and we extend the map to a sum of simple tensors by linearity.  The map $\phi(f \otimes w)$ is indeed $F$-linear, since 
\[ \phi(f \otimes w)(v+cv') = f(v+cv')w = f(v)w+cf(v')w = \phi(f \otimes w)(v) + c\phi(f \otimes w)(v'). \]
To show tjat $\phi$ is well-defined, we observe first that
\[ \phi((f+cf') \otimes w)(v)=(f+cf')(v)w = f(v)w + cf'(v)w = \phi(f \otimes w)(v) + c\phi(f' \otimes w)(v) \]
for all $f,f' \in V^*$, $c \in F$, $v \in V$, and $w \in W$, so we conclude that 
\[ \phi((f+cf') \otimes w) = \phi(f \otimes w) + c\phi(f' \otimes w). \]
In a similar fashion, one can show that
\[ \phi(f \otimes (w+cw')) = \phi(f \otimes w) + c\phi(f \otimes w'). \]
and immediately we see that the map $\phi$ is $F$-linear.

To show that $\phi$ is injective, we may choose a basis $v_1,\dots,v_n$ of $V$ and $w_1,\dots,w_m$ of $W$.  Let $v_i^*$ be the dual basis of $V^*$.  Then $v_i^* \otimes w_j$ is an $F$-basis of $V^* \otimes W$.  Let $\sum_{i,j} c_{ij} v_i^* \otimes w_j \in \ker \phi$.  Then for all $v \in V$, we have
\[ \phi\biggl(\sum_{i,j} c_{ij} v_i^* \otimes w_j\biggr)(v) = \sum_{i,j} c_{ij} v_i^*(v)w_j = 
\sum_j \biggl(\sum_i c_{ij} v_i^*(v)\biggr) w_j = 0. \]
Since the $w_j$ are linearly independent, we have $\sum_i c_{ij} v_i^*(v) = 0$ for all $j$ and all $v \in V$, which means $\sum_i c_{ij} v_i^*=0$; but the $v_i^*$ are linearly independent, so $c_{ij}=0$ for all $i,j$.  
\end{solu}

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