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\author{Bryden R. Cais}
\title{Math 593 Assignment \# 9 Solutions}
\begin{document}

\maketitle
\begin{enumerate}
    \item Let $R=\C[x,y]$ and $I=(x,y)\subset R$.  Define $\phi:I\otimes_R I\rightarrow \C$ by
    $$\phi(f\otimes g)=f_x(0,0)g_y(0,0)-f_y(0,0)g_x(0,0).$$  We claim that $\psi$ is an $R$ module
    homomorphism, where $\C$ is considered as an $R$ module by
    $$f\cdot z:=f(0,0)z$$ for any $f\in R$ and $z\in \C$.  We first show that the map
    $\psi:I\times I\rightarrow \C$ given by
    $$\psi(f,g)=f_x(0,0)g_y(0,0)-f_y(0,0)g_x(0,0)$$ is $R$ bilinear.  This is essentially a sequence
    of straightforward calculations.  For example, for any $r\in R$ and $f,g,h\in I$ we have
    \begin{eqnarray*}
        \lefteqn{\psi(rf+g,h)}\\
        &=&(rf+g)_x(0,0) h_y(0,0)-h_x(0,0)(rf+g)_y(0,0)\\
        &=&(r_x(0,0)f(0,0)+f_x(0,0)r(0,0)+g_x(0,0))h_y(0,0)-h_x(0,0)(r_y(0,0)f(0,0)+f_y(0,0)r(0,0)+g_y(0,0))\\
        &=&(f_x(0,0)r(0,0)+g_x(0,0))h_y(0,0)-h_x(0,0)(f_y(0,0)r(0,0)+g_y(0,0))\\
        \lefteqn{\text{since $f\in I$ and hence $f(0,0)=0$,}}\\
        &=&r(0,0)(f_x(0,0)h_y(0,0)-h_x(0,0)f_y(0,0))+g_x(0,0)h_y(0,0)-h_x(0,0)g_y(0,0)\\
        &=&r\cdot\psi(f,h)+\psi(g,h).
    \end{eqnarray*}
    The calculations to show $R$ linearity in the second component are similar, so we omit them.  By the universal
    property of the tensor product, we have that $\phi$ is $R$ linear, as required.

    Now if $x\otimes y-y\otimes x=0$ then since $\phi$ is an $R$ module homomorphism we have
    \begin{align*}
        0&=\phi(0)\\
        &=\phi(x\otimes y-y\otimes x)\\
        &=\phi(x\otimes y)-\phi(y\otimes x)\\
        &=(1\cdot 1-0\cdot 0)-(0\cdot 0-1\cdot 1)\\
        &=2,
    \end{align*}
    a contradiction.

    \item For $p,q\in \Q$ write $p=a/b,\ q=c/d$ for integers $a,b,c,d$.  Then for $p\otimes q\in \Q\otimes_{\Z} \Q$ we have
    \begin{align*}
        p\otimes q &= \frac{a}{b}\otimes \frac{c}{d}\\
        &=\left(\frac{acd}{bd}\otimes \frac{1}{d}\right)\\
        &=d \left(\frac{ac}{bd}\otimes \frac{1}{d}\right)\\
        &= \left(\frac{ac}{bd}\otimes \frac{d}{d}\right)\\
        &= \left(\frac{ac}{bd}\otimes 1\right)\\
        &=(pq\otimes 1),
    \end{align*}
    where we have only used $\Z$-bilinearity of the tensor.  This suggests that we consider the map $\phi:Q\otimes_{\Z}\Q\rightarrow \Q$
    given by $p\otimes q\mapsto pq$.  Observe that from our calculations above, it is clear that this is a $\Q$ module homomorphism.
    Indeed, we have
    \begin{align*}
        \phi(p\otimes q+r\otimes s) &= \phi((pq\otimes 1)+(rs\otimes 1))\\
        &=\phi(((pq+rs)\otimes 1))\\
        &=pq+rs\\
        &=\phi(p\otimes q)+\phi(r\otimes s),
    \end{align*}
    and
    \begin{align*}
        \phi(\frac{x}{y} (p\otimes q))&=\phi(\frac{x}{y}(\frac{y}{y} pq\otimes 1))\\
        &=\phi((\frac{x}{y} pq\otimes 1))\\
        &=\frac{x}{y} pq\\
        &=\frac{x}{y}\phi(p\otimes q),
    \end{align*}
    for any $x/y\in\Q$.
    Surjectivity of $\phi$ is clear since $\phi(p\otimes 1)=p$ for any $p\in \Q$.  Injectivity is evident from the above calculation
    since if $\phi(p\otimes q)=\phi(r\otimes s)$ then $pq=rs$ so that
    $$p\otimes q=pq\otimes 1=rs\otimes 1=r\otimes s.$$
    It follows that $\phi$ is an isomorphism of $\Q$ modules, as required.

    \item Define
    $$\phi:V^*\times W\rightarrow \Hom(V,W)$$ by
    $$(f,w)\mapsto \left(v \mapsto f(v)\cdot w \right).$$
    It is not difficult to see that $\phi$ is bilinear.  For every $v\in V$ we have
    \begin{align*}
        \phi(af+g,bw+u)(v)&=(af+g)(v) (bw+u)\\
        &=(af(v)+g(v))(bw+u)\\
        &=ab f(v)w+a f(v)u+bg(v)w+bg(v)u\\
        &=ab\phi(f,w)(v)+a\phi(f,u)(v)+b\phi(g,w)(v)+b\phi(g,u)(v).
    \end{align*}
    Thus we obtain a homomorphism $\tilde{\phi}:V^*\otimes W\rightarrow \Hom(V,W)$ given
    by $\phi(f,w)(v)=f(v)w$.  To check that this is an isomorphism, pick any basis
    $\{v_i\}$ of $V$ and $\{w_j\}$ of $W$.  Let $\{v_i^*\}$ be the dual basis to $\{v_i\}$.  Then
    we know that $\{v_i^*\otimes w_j\}$ is a basis for $V^*\otimes W$.  Suppose that
    $f\in \Hom(V,W)$ is given by
    $$f(v_i)=\sum_j a_{ij}w_j.$$  Let
    $$x=\sum_{k,j} a_{kj}(v_k^*\otimes w_j)\in V^*\otimes W.$$
    Then clearly we have
    $$\phi(x)(v_i)=\sum_{k,j} a_{kj}(v_k^*(v_i) w_j)=\sum_{j} a_{ij}w_j=f(v_i),$$
    so that $\phi$ is surjective.  Since $\dim \Hom(V,W)=\dim V^*\otimes W$, we must have that $\phi$ is an isomorphism.

    Observe that although we chose a basis to show that $\phi$ was an isomorphism, the definition of $\phi$ is independent of any choice
    of basis.  This is what it means for an isomorphism of vector spaces to be ``canonical.''

    \item Let $M$ be an $R$ module and $I\subset R$ an ideal.  Recall that we have an exact sequence
    $$\begin{CD}0@>>>I@>u>>R@>v>>R/I@>>>0\end{CD}.$$
    By right exactness of the tensor product, this gives an exact sequence
    $$\begin{CD}I\otimes_R M@>u\otimes\id>>R\otimes_R M@>v\otimes\id>>R/I\otimes_R M@>>>0\end{CD}.$$
    We have
    \begin{align*}
        u\otimes \id \left(\sum_l i_l\otimes m_l\right)&= u\otimes \id \left(\sum_l 1\otimes i_lm_l\right)\\
        &=\sum_l u(1)\otimes i_lm_l\\
        &=u(1)\otimes \left(\sum_l i_lm_l\right)\\
    \end{align*}
    so that if $u\otimes\id (x)=0$ then since $u$ is injective and $u(1)\neq 0$ we have $x=0$ so that $u\otimes\id$ is injective.
    We claim that $I\otimes_R M\simeq IM$.  For we may define
    \begin{align*}
        f:&I\otimes M\rightarrow IM,\\
        g:&IM\rightarrow I\otimes M
    \end{align*}
    by
    \begin{align*}
        &f\left(\sum_l i_l\otimes m_l\right)=\sum_l i_lm_l,\\
        &g\left(x\right)=1\otimes x.
    \end{align*}
    A priori, it is unclear that the image of $g$ is contained in $IM$.  However, we have
    \begin{align*}
        g\circ f\left(\sum_l i_l\otimes m_l\right)&=1\otimes \sum_l i_lm_l\\
        &=\sum_l 1\otimes i_lm_l\\
        &=\sum_l i_l\otimes m_l
    \end{align*}
    since the tensor product is over $R$ and $I\subset R$.
    Moreover, for $x=\sum_l i_l m_l\in IM$ we have
    \begin{align*}
        f\circ g(x)&=f\left(1\otimes \sum_l i_lm_l\right)\\
        &=\sum_l  i_lm_l\\
        &=x,
    \end{align*}
    from which it follows that $f$ is an isomorphism, since it is obviously a homomorphism.
    Similarly, one shows that $R\otimes_R M\simeq M$ (the isomorphism is the same as above).
    It follows that
    $$M\otimes_R R/I\simeq R/I\otimes_R M\simeq M/IM,$$
    as required.

    \item Let $\{e_1,e_2\}$ be a basis of $\R^2$.  Let $v=ae_1+be_2$ and $w=ce_1+de_2$.  Then
    \begin{align*}
    v\otimes w&=ac e_1\otimes e_1+ad e_1\otimes e_2+bc e_2\otimes e_1+bd e_2\otimes e_2\\
    &=\alpha e_1\otimes e_1+\beta e_1\otimes e_2+\gamma e_2\otimes e_1+\delta e_2\otimes e_2,
    \end{align*}
    with
    $$\alpha\delta-\beta\gamma=0.$$
    This shows that any simple tensor corresponds to a point of the quadric hypersurface given by
    $$\alpha\delta-\beta\gamma=0.$$
    Conversely, suppose that we have a point $(\alpha,\beta,\gamma,\delta)$ of the above hypersurface.
    If $\alpha,\beta,\gamma,\delta$ are all 0, it is clear that this corresponds to the simple tensor $0\otimes 0$.
    Otherwise, without loss of generality, suppose that $\alpha\neq 0$ and that $\alpha\delta-\beta\gamma=0$.
    Then
    $$\delta=\frac{\beta\gamma}{\alpha}$$ and
    \begin{align*}
    \alpha e_1\otimes e_1+\beta e_1\otimes e_2+\gamma e_2\otimes e_1+\delta e_2\otimes e_2
    &=(\alpha e_1+  \gamma e_2)\otimes (e_1+\frac{\beta}{\alpha}e_2)
    \end{align*}
    is a simple tensor.
    
\end{enumerate}

\end{document}