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\title{Algebraic Geometry, PS 3}
\author{Bryden Cais}
\begin{document}
\maketitle

\begin{exer}

 \end{exer}

 \begin{exer}
    Let $A=\Z/(6).$  Consider the ideal $2A$.  If $x\in A\setminus 2A$ then $x$ is odd.  It follows that the ideal $(x,2A)$ is all of $A$
    (one way to see this is that $2\in 2A$ and since $x$ is odd, we have integers $a,b\mod 6$ such that $2a+x=1$).
    Thus $2A$ is maximal.  For the ideal $3A$, if $y\in A\setminus 3A$ then $y=1,2,4$ or 5.  If $y=1$ or 5, $y$ is a unit in
    $\Z/(6)$.  Otherwise, we have $3-2=4-3=1$ so that in any case $(3A,y)$ is the unit ideal.  Hence $3A$ is maximal.  Since
    if $x\not\in 2A,3A$ we have $x=1$ or $5$ (modulo 6) we see that any such $x$ is a unit.  It follows that $2A,3A$ ore the only maximal ideals
    of $A$.  Now consider the localization $A_{2A}$.  This consists of all fractions $a/b$ with $a,b\in A$ and $b\not\in 2A$ under the equivalence
    $a/b\sim c/d$ iff there exists $s\in A\setminus 2A$ with $s(ad-bc)=0$ in $A$.  Set $s=3$.  Then we have
    $a/b\sim c/d$ iff $ad\equiv bc\mod 2$.  Since $b,d$ are odd, we see that
    $$0\sim \frac{2}{*}\sim \frac{4}{*}$$ and
    $$\frac{1}{*}\sim \frac{3}{*}\sim \frac{5}{*},$$
    where $*$ denotes any element of $A\setminus 2A$.  Thus, $A_{2A}$ has two elements and is clearly isomorphic to $\Z/(2)$ since
    the numerator of $a/b+a/b$ is even and hence the sum is equivalent to $0$.  A similar analysis applies to the localization
    $A_{3A}$.  In this case, we choose $s=2$ and find that $a/b\sim c/d$ iff $$ad\equiv cd\mod (3).$$
    Now observe that $-a,b-a,2b-a$ are three consecutive terms in arithmetic progression so that one of them is divisible
    by $3$.  It follows that $a/b\sim 0,1$ or 2.  Since $3\sim 0$ we see that $A_{3A}\simeq \Z/(3)$.  Now let $P$ be a projective
    $A$ module of rank 1.  Then we know that $P_{2A},P_{3A}$ are free $A_{2A},A_{3A}$ (resp.) modules of rank 1, so pick $x\in P_{2A},y\in P_{3A}$
    such that
    \begin{align*}
        P_{2A}&=A_{2A}x\\
        P_{3A}&=A_{3A}y.
    \end{align*}
    By clearing denominators, we may assume that $x,y\in P$.  We now claim that $P=A(x+y)$ and that $x+y$ is not a zero divisor.
    It then follows that $P\simeq A$.  But how to show this claim??
 \end{exer}

 \begin{exer}
    Set $$A=\C[T_1,T_2]/(T_1^2-T_2(T_2-1)(T_2-2))$$ and let $t_1,t_2$ be the images of $T_1,T_2$ under the quotient map.  Let $I=(t_1,t_2)$.
    We show that for any maximal ideal $\M\subset A$ the localization $I_{\M}$ is a free $A_{\M}$ module of rank 1.  First,
    the maximal ideals of $A$ are in bijective correspondence to the maximal ideals of $\C[T_1,T_2]$ containing the principal ideal
    $T_1^2-T_2(T_2-1)(T_2-2).$  Since every maximal ideal of $\C[T_1,T_2]$ has the form $(T_1-\alpha,T_2-\beta),$ we see that
    the maximal ideals of $A$ are precisely the ideals
    $$\M_{\a,\b}=(t_1-\a,t_2-\b)$$ where $(T_1,T_2)=(\a,\b)$ is a root of $T_1^2-T_2(T_2-1)(T_2-2).$  Suppose $(\a,\b)\neq (0,0).$
    Then there exists some $f\in I$ with $f\not\in \M_{\a,\b}$ (since $I=\M_{0,0},\M_{\a,\b}$ are distinct maximal ideals).
    It follows that $I_{\M_{\a,\b}}$ contains $1$ and is therefore equal to $A_{\M_{\a,\b}}$, that is, is free of rank 1.
    Now consider $I_I$.  Since $I$ is the ideal of all $f\in A$ with no constant term, we find that any $f\in A_I$ with
    a nonzero constant term is invertible.  In particular, $(t_2-1)(t_2-2)$ is invertible.  It follows that
    $$\frac{t_1}{(t_2-1)(t_2-2)} t_1=t_2$$ with $\frac{t_1}{(t_2-1)(t_2-2)}\in A_I$.  From this, we see that
    $I_I=(t_1,t_2)I_=A_I t_1+A_I t_2=A_I t_1$ since $A_I t_2\subset A_I t_1$.  Now the elliptic curve $T_1^2-T_2(T_2-1)(T_2-2)$
    has distinct roots and is therefore irreducible.  It follows that $A$ is an integral domain and hence that $t_1$ in particular
    is not a zero divisor.  Therefore, $I_I=A_I t_1$ is free of rank one (sending $t_1\mapsto 1$ gives a surjective homomorphism
    from $I_I$ to $A_I$; injectivity follows from the fact that $t_1$ is not a zero divisor).  Therefore, since the localization
    of $I$ at every maximal ideal of $A$ is free of rank $1$, we see that $I$ is a projective $A$ module of rank 1.  Now we must show that
    $I$ is not free, or equivalently that $I$ is not a principal ideal.
 \end{exer}

 \begin{exer}
    Suppose that $I$ is a homogenous ideal of $k[T]$ such that $m^s_+\subset I$ for some $s$.  Let $F\in k[T]$.  Then
    for any $G\in m^s_+$ we have $GF\in m^s_+\subset I$ so that $F\in I^{sat}$.  Thus, $I^{sat}=k[T]$.  For the second part
    of the question, I am convinced that there is a typo and that it cannot be ``Proposition 1.''  But then what proposition
    is it supposed to be?
 \end{exer}

  \begin{exer}
    Let $I=(T_0^2,T_0T_1)\subset k[T_0,T_1].$  Then since $m_+=(T_0,T_1)$ we have $T_0G\in I$ for any $G\in m_+$.
    Thus, $T_0\in I^{sat}$ so that $I\subset (T_0)\subset I^{sat}$.  We claim that $I^{sat}=(T_0)$.  For suppose that $FG\in (T_0)$ for all $G\in m_+^s$
    for some $s$.  In particular, $T_1^{s} F\in (T_0)$ so that $F$ is divisible by $T_0$ and hence $F\in (T_0)$.  It follows that $(T_0)$ is saturated
    and that $I^{sat}=(T_0).$
 \end{exer}

  \begin{exer}
    We first homogenize the equations to find that the projective closure is given by the equations
    $$Z_2Z_0-Z_1^2,Z_3Z_0^2-Z_1^3.$$
    The solution set can be parameterized.  Indeed, if $Z_0=0$ then we must have $Z_1=0$ also, whence we obtain the points
    $$(0,0,1,t)$$ and $$(0,0,0,1)$$
    for any $t\in k$.  Otherwise, $Z_0\neq 0$ whence we obtain the parameterization
    $$(1,t,t^2,t^3)$$ for any $t\in k$.  Thus, the projective closure is
    $$\left\{(1,t,t^2,t^3):t\in k\right\}\cup \left\{(0,0,1,t):t\in k\right\}\cup (0,0,0,1).$$
 \end{exer}

  \begin{exer}
	If $F$ is a homogenous polynomail free of multiple factors then $(F)=(F)^{sat}$.  We know that
	the set of solutions of $F$ in $\P_k^n(K)$ is irreducible if and only if the corresponding coordinate ring
is an integral domain.  But the coordinate ring is
$$A=k[T]/(F)^{sat}=k[T]/(F),$$
and is therefore an integral domain if and only f $F$ is irreducible.
 \end{exer}

  \begin{exer}
    Clearly, $P$ has only simple roots in $\bar{k}$ iff $P,\ P^{\prime}$ have
    no common roots.  By exercise \ref{discprod} this is the case iff $\res(P^{\prime},P)\neq 0$
    (since the resultant is a product over all pairwise differences of the roots of $P$ and $P^{'}$).
    Let us now compute $\res(P,P^{'})$ in the case where $P$ is of degree less than $4$.
    \begin{enumerate}
        \item $\deg(P)=2:$  Set $P(Z)=a_0Z^2+a_1Z+a_2$ so that $P^{'}(Z)=2a_0Z+a_1$.  We then have
        \begin{align*}
            \res(P,P^{'})&=\begin{vmatrix}a_0 & a_1 & a_2 \\ 2a_0 & a_1 & 0 \\ 0 & 2a_0 & a_1\end{vmatrix}\\
            &=-a_0(a_1^2-4a_2a_0).
        \end{align*}
        \item $\deg(P)=3:$ Set $P(Z)=a_0Z^3+a_1Z^2+a_2Z+a_3$ so that $P^{'}(Z)=3a_0Z^2+2a_1Z+a_2$.  We then have
        \begin{align*}
            \res(P,P^{'})&=\begin{vmatrix}a_0 & a_1 & a_2 & a_3 & 0\\ 0 & a_0 & a_1 & a_2 & a_3  \\ 3a_0 & 2a_1 & a_2 & 0 & 0 \\
            0& 3a_0 & 2a_1 & a_2 & 0\\ 0& 0& 3a_0 & 2a_1 & a_2\end{vmatrix}\\
            &=27a_3^2a_0^3 + (-18a_3a_2a_1 + 4a_2^3)a_0^2 + (4a_3a_1^3 - a_2^2a_1^2)a_0.
        \end{align*}

        \item $\deg(P)=4:$ Set $P(Z)=a_0Z^4+a_1Z^3+a_2Z^2+a_3Z+a_4$ so that $P^{'}(Z)=4a_0Z^3+3a_1Z^2+2a_2Z+a_3$.  Proceeding as above (and using
        \texttt{PARI}), we have
        \begin{align*}
            \res(P,P^{'})&=\begin{vmatrix}a_0 & a_1 & a_2 & a_3 & a_4 & 0 & 0\\ 0 & a_0 & a_1 & a_2 & a_3 & a_4 & 0\\0& 0 &a_0 & a_1 & a_2 & a_3 & a_4\\
            4a_0 & 3a_1 & 2a_2 & a_3 & 0 & 0 & 0\\0 &4a_0 & 3a_1 & 2a_2 & a_3 & 0 & 0 \\0 & 0 & 4a_0 & 3a_1 & 2a_2 & a_3 & 0 \\
            0 & 0 & 0 &4a_0 & 3a_1 & 2a_2 & a_3\end{vmatrix}\\
            &=256 a_4^3 a_0^4 + (-192 a_4^2 a_3 a_1 + (-128 a_4^2 a_2^2 + 144 a_4 a_3^2 a_2 - 27 a_3^4)) a_0^3 + ((144 a_4^2 a_2 - 6 a_4 a_3^2) a_1^2\\
            & + (-80 a_4 a_3 a_2^2 + 18 a_3^3 a_2) a_1 + (16 a_4 a_2^4 - 4 a_3^2 a_2^3)) a_0^2 + (-27 a_4^2 a_1^4 + (18 a_4 a_3 a_2 - 4 a_3^3) a_1^3\\
            &+ (-4 a_4 a_2^3 + a_3^2 a_2^2) a_1^2) a_0.
        \end{align*}

    \end{enumerate}
 \end{exer}

 \begin{exer}\label{discprod}
         Let
        \begin{alignat*}{3}
        P&=a_0(Z-\alpha_1)\cdots (Z-\alpha_n)&=a_0Z^n+a_1Z^{n-1}+\cdots+a_n,\\
        Q&=b_0(Z-\beta_1)\cdots (Z-\beta_m)&=b_0Z^m+b_1Z^{m-1}+\cdots+b_m.\\
        \end{alignat*}
        Let $V(X_1,X_2,\ldots,X_r)$ denote the matrix
        \begin{eqnarray*}
        \begin{pmatrix}
        1 & X_1 & \cdots & X_1^r \\
        1 & X_2 & \cdots & X_2^r \\
        \vdots & \vdots & \cdots & \vdots \\
        1 & X_r & \cdots & X_r^r \\
        \end{pmatrix},
        \end{eqnarray*}
        set $M=V(\beta_m,\cdots,\beta_1,\alpha_n,\cdots,\alpha_1)$ and
        \begin{eqnarray*}
        R\ =
        \begin{pmatrix}
            a_0 & & & b_0 & & \\
            a_1 & \ddots & & \vdots & \ddots & \\
            \vdots & \ddots & a_0 & b_{m-1} & \ddots & b_0 \\
            a_n & \ddots & a_1 & b_m & \ddots & \vdots \\
             & \ddots & \vdots & & \ddots & b_{m-1} \\
             & & a_n & & & b_m \\
        \end{pmatrix}.
        \end{eqnarray*}
        We first show that $\det(V(X_1,\ldots,X_r))=\prod_{1\leq i<j\leq r}(X_i-X_j).$
        To see this, for $1\leq j\leq r-1$, subtract $X_1$ times the $j^{\text{th}}$ column
        from the $(j+1)^{\text{st}}$ column of the matrix $V(X_1,\ldots,X_r).$
        These operations do not alter the determinant, and we find that
        \begin{eqnarray*}
        \det(V(X_1,\ldots,X_r))=\prod_{2\leq i\leq r}(X_i-X_1)\,
        \det(V(X_2,\ldots,X_r)).
        \end{eqnarray*}
        Induction then gives the proposed formula.  We now compute the determinant of $MR$ in
        two different ways.  It is not difficult to see that
        \begin{eqnarray*}
        \lefteqn{
        \begin{pmatrix}
        1 & \beta_m & \cdots & \beta_m^{m+n-1}\\
        \vdots & \vdots & \cdots & \vdots \\
        1 & \beta_1 & \cdots & \beta_1^{m+n-1}\\
        1 & \alpha_n & \cdots & \alpha_n^{m+n-1}\\
        \vdots & \vdots & \cdots & \vdots \\
        1 & \alpha_1 & \cdots & \alpha_1^{m+n-1}\\
        \end{pmatrix}
        \,
        \begin{pmatrix}
            a_0 & & & b_0 & & \\
            a_1 & \ddots & & \vdots & \ddots & \\
            \vdots & \ddots & a_0 & b_{m-1} & \ddots & b_0 \\
            a_n & \ddots & a_1 & b_m & \ddots & \vdots \\
             & \ddots & \vdots & & \ddots & b_{m-1} \\
             & & a_n & & & b_m \\
        \end{pmatrix}
        }\\\\
        &=&
        \begin{pmatrix}
            P(\beta_m) & \cdots & P(\beta_m)\beta_m^{m-1}  & & &\\
            \vdots & \cdots & \vdots & & &\\
            P(\beta_1) & \cdots & P(\beta_1)\beta_1^{m-1}  & & &\\
            & & & Q(\alpha_n) & \cdots & Q(\alpha_n)\alpha_n^{n-1} \\
            & & & \vdots & \cdots & \vdots \\
            & & & Q(\alpha_1) & \cdots & Q(\alpha_1)\alpha_1^{n-1} \\
        \end{pmatrix},
        \end{eqnarray*}
        and hence that
        \begin{eqnarray*}
            \lefteqn{\det(MR)=}\\
            &&\det(V(\beta_m,\ldots,\beta_1)V(\alpha_n,\ldots,\alpha_1))
        \prod_{1\leq j\leq m}P(\beta_j)\,\prod_{1\leq i\leq n}Q(\alpha_i).
        \end{eqnarray*}
        On the other hand,
        \begin{eqnarray*}
        \det(MR)&=&\det(M)\,\det(R^T)\\
        &=&\det(V(\beta_m,\ldots,\beta_1,\alpha_n,\ldots,\alpha_1))\res(P,Q),
        \end{eqnarray*}
        where $R^T$ denotes the transpose of $R$.
        Now, using our formula for the determinant of $V(X_1,\ldots,X_r)$, we see that
        \begin{eqnarray*}
        \lefteqn{\det(V(\beta_m,\ldots,\beta_1,\alpha_n,\ldots,\alpha_1))}\\
        &=&\det(V(\beta_m,\ldots,\beta_1))\det(V(\alpha_n,\ldots,\alpha_1))
        \prod_{1\leq i\leq n}\prod_{1\leq j\leq m}(\alpha_i-\beta_j)\\
        &=&\det(V(\beta_m,\ldots,\beta_1))\det(V(\alpha_n,\ldots,\alpha_1))
        \frac{1}{b_0^n}\,\prod_{1\leq i\leq n}Q(\alpha_i).
        \end{eqnarray*}
        It thus follows that
        \begin{eqnarray*}
        \res(P,Q)=b_0^n\,P(\beta_1)\cdots P(\beta_m).
        \end{eqnarray*}
        Notice that $\res(P,Q)=(-1)^{mn}\res(Q,P)$ since to swap the roles
        of the $a_i$'s and the $b_j$'s in the defining determinant of the resultant requires
        a bubble sort involving $mn$ flips of columns.  Thus, we also have
        \begin{eqnarray*}
        \res(P,Q)=(-1)^{mn}a_0^m\,Q(\alpha_1)\cdots Q(\alpha_n).
        \end{eqnarray*}
 \end{exer}

  \begin{exer}
    Let $X$ be the cubic curve defined projectively by $T_1^2T_0-T_2^2-T_0^3=0.$  Except for the point at infinity, $(0,1,0),$
    all points on $X(\C)$ lie in $\A^1_{\C}$ (where we take the copy of $\A^1_{\C}\subset \P^1_{\C}$ given by $T_0=1$.
    We thus proceed to determine explicit formulae for the group law on the affine curve $X^{'}$ given by $y^2=x^3+1.$  Let $P_0=(x_0,y_0),P_1=(x_1,y_1)$
    be any two {\em distinct} points on $X^{'}.$  Then line through these points intersects $X$ at a third point whose $x$ coordinate
    is the third root of
    $$\left(\frac{y_1-y_0}{x_1-x_0}(x-x_0)+y_0\right)^2=x^3+1.$$
    Call this root $x_2$.  Then since $x_0,x_1$ are also roots, we know that
    $$\left(\frac{y_1-y_0}{x_1-x_0}\right)^2=x_0+x_1+x_2.$$  Similarly, $y_2$ is given by
    $$y_2=\frac{y_1-y_0}{x_1-x_0}(x_2-x_0)+y_0.$$  Taking the point $(0,1,0)$ as the zero element, we find that the sum
    $P_0+P_1$ on $X$ is the point $P_2=(x_2,-y_2).$  Observe that
    \begin{align*}
    x_2&=\left(\frac{y_1-y_0}{x_1-x_0}\right)^2-x_0-x_1\\
    &=\left(\frac{x_1^3-x_0^3}{(x_1-x_0)(y_1+y_0)}\right)^2-x_0-x_1\\
    &=\left(\frac{x_0^2+x_0x_1+x_1^2}{y_1+y_0}\right)^2-x_0-x_1,
    \end{align*}
    so that if $P_0=P_1$ we have $2P_0=(x_2,-y_2),$ where
    $$x_2=\left(\frac{3x_0^2}{2y_0}\right)^2-2x_0,$$ and
    $$y_2=\left(\frac{3x_0^2}{2y_0}\right)(x_2-x_0)+y_0.$$
    All of these formulae make sense except when $x_0=x_1$ but $P_0\neq P_1$ (in which case the sum $P_0+P_1$ is seen to be the point
    at infinity) and when $y_0=0,$ when $x_0$ is a cube root of $-1$ whence the sum $2P_0$ is again the point at infinity.
 \end{exer}
 \begin{exer}
    Using our formulae above, we can determine explicit formulae for $3P$ when $P$ is a point on $X$.  Let $P=(x,y)$.  It is not difficult
    to see from the formulae that the points of order 3 in $X(C)$ correspond to those points $P$ with $2P=-P,$ i.e. those points $P$ with
    \begin{align*}
        \left(\frac{3x^2}{2y}\right)^2-3x&=0.
    \end{align*}
    This gives
    \begin{align*}
        x^4+4x&=0,
    \end{align*}
    or $x=0,-\sqrt[3]{4},\rho\sqrt[3]{4},-\rho\sqrt[3]{4},$
    where $\rho=e^{\pi i/3}$ is a cube root of $-1$.  For each of these four values of $x$ there are two corresponding values of $y$,
    giving $8$ points of order exactly $3$.  Since the point at infinity $O$ also satisfies $3O=O,$ we see that there are nine points
    in $X(C)$ satisfying $3P=O$.  They are
    \begin{align*}
        O&=(0,1,0)\\
        \pm P_1&=(1,\pm 1,0)\\
        \pm P_2&=(1,\pm i\sqrt{3},-\sqrt[3]{4})\\
        \pm P_3&=(1,\pm i\sqrt{3},\rho\sqrt[3]{4})\\
        \pm P_4&=(1,\pm i\sqrt{3},-\rho\sqrt[3]{4}).
    \end{align*}
    Now let us determine the inflection points of $X$ in $\C.$  A point $(T_0,T_1,T_2)$ is an inflection point of $X$ if and only if
    \begin{alignat*}{3}
        \begin{vmatrix}-6T_0 & 2 T_1 & 0\\2T_1 & 2T_0 & 0\\0 & 0 & 6T_2\end{vmatrix}&=-24T_2(3T_0^2+T_1^2)&=0.
    \end{alignat*}
    These are the points on $X$ with $T_2=0$ or $T_1=\pm i\sqrt{3} T_0$.  It is easy to see that these are precisely the 9 points listed
    above.  Of course, a more intuitive way to think about this problem is that the points of order $3$ are those points $P$ such that the tangent line
    at $P$ has third point of intersection $P$ (this is just the requirement that $2P=-P$).  But this is equivalent to requiring that
    the tangent line at $P$ intersect $X$ with multiplicity $3$ (the maximum multiplicity, according to Bezout's theorem),
    i.e. that $P$ be an inflection point of $X$.

    Now it can be seen that any line containing two of the eight points of order $3$ contains a third, and hence that this is
    an example of a configuration of $8$ points not satisfying the assumption of Lemma 1.

 \end{exer}

  \begin{exer}
    Let $X$ be the curve given by $T_1^2T_0-T_2^3$.  For any field $K$, define the map $f:X(K)\setminus (1,0,0)\rightarrow K$ by
    $$(T_0,T_1,T_2)\mapsto \frac{T_2}{T_1}.$$  Observe that this is a well defined map since $T_1=0$ iff $T_2=0$ and we are
    excluding the point $(1,0,0)$; in addition, the value of $f$ is independent of the homogenous coordinates chosen.  We now show
    that $f$ induces an isomorphism of the group structure on $X(K)$ with the additive group $K^{+}.$  Recall that $X$ is {\em rational},
    that is, we have the parameterization $(1,t^3,t^2)$ (excluding the point at infinity).  Thus, let $P=(1,\a^3,\a^2),Q=(1,\b^3,\b^2)$ be
    any two points of $X(K)$.
    Let us determine $P+Q$.  Passing to affine coordinates by setting $y=T_1/T_0,x=T_2/T_0$ and writing $y=mx+b$ for the line
    (in affine coordinates) passing through $P,Q$, we see that $\a^2,\b^2$ are
    both roots of $(mx+b)^2=x^3$ and that the third root, $\c$ is the $x$ coordinate of the sum $P+Q$.  Then we have $\a^2+\b^2+\c=m^2$.
    By now this process is standard, and we find that the $x$ coordinate of $P+Q$ is given by:
    \begin{align*}
        \left(\frac{\a^3-\b^3}{\a^2-\b^2}\right)^2-(\a^2+\b^2)&=\frac{\a^6+\b^6-2\a^3\b^3-(\a^2-\b^2)(\a^4-\b^4)}{(\a^2-\b^2)^2}\\
        &=\frac{\a^2\b^2(\a-\b)^2}{(\a^2-\b^2)^2}\\
        &=\left(\frac{\a\b}{\a+\b}\right)^2.
    \end{align*}
    It then follows that the $y$ coordinate of $P+Q$ is given by
    $$\left(\frac{\a\b}{\a+\b}\right)^3.$$
    Therefore, in homogenous coordinates, we have
    $$P+Q=\left(1,\left(\frac{\a\b}{\a+\b}\right)^3,\left(\frac{\a\b}{\a+\b}\right)^2\right)$$ if $\a\neq -\b$ (that is, if $P\neq -Q$)
    and $P+Q=(0,1,0)$ if $P=-Q$.  Now we have
    \begin{align*}
        f(0,1,0)&=0\\
        \intertext{and}
        f(P+Q)&=f\left(1,\left(\frac{\a\b}{\a+\b}\right)^3,\left(\frac{\a\b}{\a+\b}\right)^2\right)\\
        &=\frac{\left(\frac{\a\b}{\a+\b}\right)^2}{\left(\frac{\a\b}{\a+\b}\right)^3}\\
        &=\frac{\a+\b}{\a\b}\\
        &=f(P)+f(Q),
    \end{align*}
    so that $f$ is indeed a group homomorphism.  We have already seen that $f$ is bijective, so we are done.
 \end{exer}

  \begin{exer}
    This is similar to the above.  Let $X$ be given in projective coordinates by $T_1^2T_0-T_2^2(T_2+T_0)$.  For any field $K$,
    define the map $f:X(K)\setminus(1,0,0)\rightarrow K$ by
    $$(T_0,T_1,T_2)\mapsto \frac{T_1-T_2}{T_1+T_2}.$$  As before, it is not difficult to
    see that this is a well defined map and that $f(0,1,0)=1\in K^*$. We claim
    that $f$ induces an isomorphism of the group structure on $X(K)$ with the multiplicative group $K^*.$  As above, $X$ is
    parameterized by $(1,t(t^2-1),t^2-1)$ (excluding the point at infinity).  Thus, let $P=(1,\a(\a^2-1),\a^2-1),Q=(1,\b(\b^2-1),\b^2-1)$ be
    any two points of $X(K)$.  Observe first that
    \begin{align*}
        f(P)&=\frac{\a-1}{\a+1}\\
        f(Q)&=\frac{\b-1}{\b+1}.
    \end{align*}
    Let us determine $P+Q$.  Proceeding as above, set $y=T_1/T_0,x=T_2/T_0$ and write $y=mx+b$ for the line
    (in affine coordinates) passing through $P,Q$.  Then we see that $\a^2-1,\b^2-1$ are
    both roots of $(mx+b)^2=x^3+x^2$ and that the third root, $\c$ is the $x$ coordinate of the sum $P+Q$.  Then we have $\a^2+\b^2+\c-2=m^2-1$.
    By now this process is standard, and we find that the $x$ coordinate of $P+Q$ is given by:
    \begin{align*}
        \left(\frac{\a(\a^2-1)-\b(\b^2-1)}{\a^2-\b^2}\right)^2-(\a^2+\b^2-1)&=\left(\frac{\a^2+\b^2+\a\b-1}{\a+\b}\right)^2-(\a^2+\b^2-1)\\
        &=\frac{\a^2\b^2-\a^2-\b^2+1}{(\a+\b)^2}\\
        &=\left(\frac{\a\b+1}{\a+\b}\right)^2-1.
    \end{align*}
    We therefore see that
    $$P+Q=\left(1,\frac{\a\b+1}{\a+\b}\left(\left(\frac{\a\b+1}{\a+\b}\right)^2-1\right),\left(\frac{\a\b+1}{\a+\b}\right)^2-1\right),$$
    so that
    $$f(P+Q)=\frac{\a\b-\a-\b+1}{\a\b+\a+\b+1}=\frac{(\a-1)(\b-1)}{(\a+1)(\b+1)}=f(P)f(Q).$$
    Thus $f$ is a homomorphism.  The fact that $f$ is bijective follows from the rationality of our curve and the definition of $f$.

 \end{exer}

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