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\title{Algebraic Geometry, PS 2}
\author{Bryden Cais}
\begin{document}
\maketitle

\begin{exer}
    Let $X$ be the subvariety of $\A_k^2$ defined by the equation $T_2^2-T_1^2-T_1^3$ and let $f:\A_k^1\rightarrow X$
    be the morphism given by
    $$T_1\mapsto T^2-1,T_2\mapsto T(T^2-1).$$
    We then have a map $f^*:\O(X)\rightarrow k[T]$ defined
    as above on the generators $T_1,T_2$ of $\O(X)$.  Observe that
    $$g(T)=f^*(p(T_1,T_2))=p(T^2-1,T(T^2-1))\in k[T]$$
    satisfies
    $g(1)=g(-1)=p(0,0)\in k$.  Conversely, suppose that $g(T)$ satisfies $g(1)=g(-1)$ and $\cha(k)\neq 2$.
    Then the polynomial $G(T)=g(T)-g(1)$ is divisible by $T\pm 1$ and hence by $T^2-1$.  Now observe
    that $T^{2n}=((T^2-1)+1)^n$ is a polynomial in $(T^2-1)$ and $T^{2n+1}=T\cdot T^{2n}$ is $T$ times a polynomial in $T^2-1$.  Thus, any polynomial
    may be written as a $k$ linear combination of $(T^2-1)^i$ and $T(T^2-1)^j$ with $i,j\geq 0$.  It thus follows that
    $g(T)\in f^*(\O(X))$.  Thus $f^*(\O(X))$ is the subring of $k[T]$ consisting of polynomials $g(T)$ such that $g(1)=g(-1)$.

    Now suppose that $\cha(k)=2$.  Then $T^2-1=(T-1)^2$ and if $g(T)$ is in the image of $f^*$ then $g(T)-g(1)$ has a double
    root at $T=1$ (since it is divisible by $(T-1)^2$ and therefore $g^{'}(1)=0$.  Conversely, suppose that $g^{'}(1)=0$ for
    $g\in k[T]$.  Then $g^{'}(T)$ is a polynomial in $T-1$ with no constant term.  It follows that $g(T)$ is a polynomial
    in $T-1$ with no linear term.  We can therefore write $g(T)=(T-1)^2 h(T)+g(0)$.
    Now observe that $T^{2n}=((T-1)^2+1)^n$ is a polynomial in $(T-1)^2$ and
    $T^{2j+1}=T((T-1)^2+1)^j$ is a $k$ linear combination of terms of the form $(T-1)^{2i}$ and $T(T-1)^{2j}$
    for $i,j\geq 0$.  Therefore, $h$ is a $k$ linear combination of such terms so that $g(T)$ is in the image of
    $\O(X)$ under $f^*$.

    When $\cha(K)=2$, the variety $X$ is isomorphic to the variety $Y$ given by the equation $U_1^2-U_2^3=0$.  To see this,
    define
    $$f^*:\O(Y)\longrightarrow \O(X)$$ by $$f^*(U_1)=T_2-T_1, f^*(U_2)=T_1.$$ This is clearly well defined since
    $f(U_1^2-U_2^3)=T_2^2-T_1^2-T_1^3$.  Moreover, the morphism $g$ given by $$g^*(T_1)=U_2,g^*(T_2)=U_1+U_2$$ is inverse to
    $f^*$ (i.e. $f^* g^*:\O(X)\rightarrow \O(X)=\id_{\O(X)}$ and $ g^* f^*:\O(Y)\rightarrow \O(Y)=\id_{\O(Y)}$).  It therefore
    follows that $\O(X),\O(Y)$ are isomorphic and hence that $X,Y$ are isomorphic.
 \end{exer}

 \begin{exer}
    Let $Y$ be the variety given by $T_1T_2-1=0$.  If $Y$ were isomorphic to $\A_k^1$ then we would have an isomorphism
    $$f^*:k[T_1,T_2]/(T_1,T_2)\longrightarrow k[U].$$  So suppose that $f^*(T_1)=g(U)$ and $f^*(T_2)=h(U)$.
    Then $f^*(T_1T_2)=gh=f^*(1)=1$.  Therefore $g,h$ are invertible elements of $k[U]$ and hence are both elements of $k$.
    It follows that the image of $f^*$ is contained in $k$ and hence $f^*$ is not an isomorphism of $\O(Y)$ with $\O(\A_k^1)$.
 \end{exer}

 \begin{exer}
    Let $f:K^2\rightarrow K^2$ be given by $(x,y)\mapsto (x,xy).$  Then the image of $f$ is neither open nor closed and dense in $K^2$.
    We first show that it is dense.  Clearly, the image is
    $$U=\left\{(a,b)\in K^2:a\neq 0\quad\text{or}\quad a=b=0\right\}.$$
    Suppose now that $f\in k[T_1,T_2]$ vanishes on $U$.  Then in particular, $f$ vanishes on all points of the line
    $(t,b)$ for any fixed $b$ and $t\neq 0$.  However, on this line, $f$ is a polynomial of the variable $t$ alone with
    infinitely many roots.  It follows that $f\equiv 0$ and thus that $I(\bar{U})=(0),$ where $\bar{U}$ denotes the closure
    of $U$ in the Zariski topology.  Therefore, the image of $f$ is dense.  Since the image is dense and not all of $K^2$,
    it cannot be closed.  Now suppose that $U$ is open.  Then the complement is closed.  Let $f(T_1,T_2)$ be any polynomial vanishing
    on $T_1=0,T_2\neq 0$.  Then by the same argument as before, $f$ must also vanish at $(0,0)$, a contradiction.  Hence
    the image is not open.
 \end{exer}

  \begin{exer}
    Suppose that $f:\A_k^1\rightarrow A_k^1$ is an isomorphism.  then we have an isomorphism
    $$f^*:k[T]\longrightarrow k[U].$$
    Let $g^*$ be inverse to $f^*$.
    Suppose that $f^*(T)=P(U)$ and that $g^*(U)=Q(T)$.  Then $Q\circ P=P\circ Q=\id$.  It follows that
    $P,Q$ have.  Conversely, and morphism $f^*:k[T]\rightarrow k[U]$ given by
    $f^*(T)=aU+b$ for $a,b\in k$ and $a\neq 0$ is an isomorphism, with inverse given by $g^*(U)=a^{-1}(U-b)$.
 \end{exer}

  \begin{exer}
	Recall that we have the identification
	$$X(K)=\hom{\O(X)}{K}.$$  Therefore, we want to show that
	$$\hom{\O(X)}{K}\times \hom{\O(Y)}{K}= \hom{\O(X)\otimes \O(Y)}{K}$$
(that is, we have a bijection between the two sets).  Given any homomorphism 
	$$(\alpha,\beta)\in\hom{\O(X)}{K}\times\hom{\O(Y)}{K}$$ we get a bilinear
	map $$\alpha\beta:\O(X)\times\O(Y)\longrightarrow K$$ given by
	$\alpha\beta(x,y):\alpha(x)\beta(y)$.  By the universal property
	of the tensor product, we get a unique linear map $L:\O(X)\otimes \O(Y)\rightarrow K$
	such that the usual diagram commutes.  To give the correspondence in the reverse direction,
	let $L\in \hom{\O(X)\otimes\O(Y)}{K}$.  Then by composition with the natural
	map $N:\O(X)\otimes\O(Y)\rightarrow \O(X)\otimes\O(Y)$ we get a bilinear map
	$B:\O(X)\times\O(Y)\rightarrow K.$  This gives elements $\alpha=B(*,1)\in\hom{\O(X)}{K}$
	and $\beta=B(1,*)\in \hom{\O(Y)}{K}$.  It is not difficult to see that $B(x,y)=\alpha(x)\beta(y)$
	and hence, (recalling how multiplication is defined in the tensor product)
	that this is the reverse correspondence required.  Thus we have a bijection and we are done.

 \end{exer}

  \begin{exer}
    To show that the correspondence $k\rightarrow O(n,K)$ is an abstract affine algebraic $k$ variety, it is enough to construct a finitely
    generated $k$ algebra $A$ such that for each $K$ there is a bijection $X(K)\leftrightarrow \hom{A}{K}$.  We therefore
    let
    $$A=k[T_{11},\ldots,T_{nn}]/I,$$
    where $I$ is the ideal generated by
    $$\sum_{k=1}^n T_{ik}T_{jk}-\delta_{ij}$$ for $1\leq i,j\leq n$,
    where
    $$\delta_{ij}=\begin{cases}1 &\text{if $i=j$}\\ 0 & \text{otherwise}\end{cases}.$$
    Any $\varphi\in\hom{A}{K}$ is given by its values on generators, so let $\varphi(T_{ij})=b_{ij}\in K$.
    Then the required bijection is given by $$\varphi \mapsto (b_{ij})=M\in O(n,K).$$  It is clear that
    this satisfies all the required properties of an abstract affine algebraic variety.
 \end{exer}

  \begin{exer}
    Let $S_1=\Z\subset \C$.  Consider the map $f:S_1\rightarrow S_1$ given by
    $$f(x)=\begin{cases}x &\text{if $x\neq 0,1$}\\ 0 &\text{if $x=1$}\\ 1 &\text{if $x=0$} \end{cases}.$$
    Now the Zariski topology on $\Z\subset \C$ is just that the closed sets are the finite sets.  Clearly
    the preimage of any closed ($=$finite) set of $S_1$ under $f$ is again a (finite$=$) closed set, so that
    $f$ is continuous in the Zariski topology.  On the other hand, $f$ is not regular: if it were given by a polynomial
    then $f-x$ is a polynomial map from $S_1\subset \C$ to itself with infinitely many zeroes, and is therefore identically zero,
    which is a contradiction.
 \end{exer}

  \begin{exer}
    Suppose that $k$ is a field with $\cha(k)\neq 2$.  Let $X$ be the affine algebraic $K$ set defined by the
    equations $T_1^2+T_2^2+T_3^2=0,T_1^2-T_2^2-t_3^2+1=0$.  It is not difficult to see that this system of equations is
    equivalent to the system
    $$T_1^2+1/2,T_2^2+T_3^2-1/2.$$
    If $k$ does not contain a square root of $-1/2$ then this system has no solutions in $k$ and hence is empty.  Otherwise,
    $$X=U_1\cup U_2,$$
    where $U_i$ is the variety given by the system of equations
    $$\left(T_1+(-1)^i\sqrt{-\frac{1}{2}},T_2^2+T_3^2-\frac{1}{2}\right),$$ each of which is obviously irreducible.
 \end{exer}

  \begin{exer}
    It is not difficult to see that the system of equations
    $$T_2^2-T_1T_3=0,T_1^2-T_2^3$$ defining a variety $X$ has one solution set
    $$T_1=T_2=0,T_3\in \A_k^1.$$  Let $V_1$ be the corresponding variety. Then $V_1$ is clearly one irreducible component of $X$.
    Now we may suppose that $T_1,T_2\neq 0$ so that
    our system of equations is equivalent to
    $$T_1T_2^2-T_2^3T_3=0,T_1^2-T_2^3=0,$$ which is equivalent to
    $$T_1-T_2T_3=0,T_1^2-T_2^3=0.$$  Since we are working over a field, this is equivalent to
    $$T_2=T_3^2,T_1=T_3^3.$$  Thus, if we let $V_2$ be the variety corresponding to
    $$T_2=T_3^2,T_1=T_3^3,T+3\in \A_k^1,$$ we see that $V_2$ is irreducible and that
    $$X=V_1\cup V_2.$$  Moreover, it is easy to see that $V_1,V_2$ are birationally isomorphic to $\A_k^1$.
    One simply takes the maps
    \begin{align*}
    \phi_1:&R(V_1)\longrightarrow k(T)\\
    \phi_2:&R(V_2)\longrightarrow k(T)
    \end{align*}
    given by
    \begin{align*}
    \phi_1(T_3)=T,\phi_1(T_2)=\phi_1(T_1)=0,\\
    \phi_2(T_3)=T,\phi_2(T_2)=T^2,\phi_2(T_1)=T^3.\\
    \end{align*}
    It is clear that these maps are isomorphisms.
 \end{exer}

  \begin{exer}
    Let $X$ be the variety defined by $T_2^2-T_1^2-T_1^3=0$ and let $f:A_k^1\rightarrow X$
    be the morphism given by $T_1\mapsto T^2-1,T_2\mapsto T(T^2-1).$  We claim that $f$ is birational.
    We must show that it induces an isomorphism of the fields $R(\O(X))$ and $k(T)$.  So define
    $g:k(T)\rightarrow R(\O(X))$ by $T\mapsto T_2/T_1.$  Then we easily check
    \begin{align*}
    g\circ f (T_1)&=(T_2^2-T_1^2)/T_1^2=T_1^3/T_1^2=T_1\\
    g\circ f (T_2)&=T_2(T_2^2-T_1^2)/T_1^3=T_2T_1^3/T_1^3=T_2\\
    f\circ g (T)&=f(T_2/T_1)=T(T^2-1)/(T^2-1)=T,
    \end{align*}
    which shows that $f$ and $g$ are inverses and hence that $f$ is birational.
 \end{exer}

  \begin{exer}
    Let $F(T_1,\ldots,T_n)=G(T_1,,\ldots,T_n)+H(T_1,,\ldots,T_n)$ be irreducible, where $G$ is homogenous of degree $d-1$ and
    $H$ is homogenous of degree $d$.  Let $X$ be the algebraic set $V(F)$ and let
    \begin{align*}
    \varphi:&R(X)\longrightarrow k(U_1,\ldots,U_{n-1})\\
    \intertext{be given by}
    \varphi(T_1)&=-\frac{G(1,U_1,\ldots,U_{n-1})}{H(1,U_1,\ldots,U_{n-1})}\\
    \intertext{and}\\
    \varphi(T_i)&=-U_{i-1}\frac{G(1,U_1,\ldots,U_{n-1})}{H(1,U_1,\ldots,U_{n-1})}\\
    \intertext{if $i>1$.}
    \end{align*}
    Writing $\lambda$ for $-G(1,U_1,\ldots,U_{n-1})/H(1,U_1,\ldots,U_{n-1})$,
    we find that
    \begin{align*}
    \varphi(F)&=G(\lambda,\lambda U_1,\ldots,\lambda U_{n-1})+H(\lambda,\lambda U_1,\ldots,\lambda U_{n-1})\\
    &=\lambda^{d-1}\left(G(1,U_1,\ldots,U_{n-1})+\lambda H(1,U_1,\ldots,U_{n-1})\right)\\
    &=\lambda^{d-1}\left(G(1,U_1,\ldots,U_{n-1})-G(1,U_1,\ldots,U_{n-1})\right)\\
    &=0,
    \end{align*}
    so that $\varphi$ is well defined.  Now let $\psi:k(U_1,\ldots,U_{n-1})\longrightarrow R(X)$ be given by
    \begin{align*}
    \psi(U_i)&=\frac{T_{i+1}}{T_1}.
    \end{align*}
    Then it is not difficult to check that $\psi\circ\varphi:R(X)\rightarrow R(X)$ and $\varphi\circ \psi:k(U_1,\ldots,U_{n-1})$
    are both the respective identity mappings.  It follows that $\varphi$ is an isomorphism and hence that $X$ is rational.
 \end{exer}

  \begin{exer}
    We now show that the affine algebraic sets given by $T_1^3+T_2^3-1=0$ and $Y^2-X^3/3+1/12=0$ are
    birationally isomorphic.  Explicitly, let
    \begin{align*}
    f(T_1)&=\frac{Y+1/2}{X}\\
    f(T_2)&=\frac{-Y+1/2}{X}\\
    g(X)&=\frac{1}{T_1+T_2}\\
    g(Y)&=\frac{1}{2}\frac{T_1-T_2}{T_1+T_2}.
    \end{align*}
    Then one readily checks that $f,g$ give maps
    \begin{align*}
        f:k[T_1,T_2]/(T_1^3+T_2^3-1)&\longrightarrow k[X,Y]/(Y^2-X^3/3+1/12)\\
        g:k[X,Y]/(Y^2-X^3/3+1/12)&\longrightarrow k[T_1,T_2]/(T_1^3+T_2^3-1)
    \end{align*}
    that are inverse to eachother (this is just a computation: we have, for example,
    $f(T_1^3+T_2^3-1)=(3Y^2+1/4)/X^3-1=3/X^3 (Y^2-X^3/3+1/12)$, as required) and hence $f,g$ are isomorphisms.
 \end{exer}


\end{document}