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\title{Algebraic Geometry, PS 1}
\author{Bryden Cais}
\begin{document}
\maketitle

\begin{exer}
    Recall that for any field $k$, two systems of algebraic equations $S,S^{'}$ define the same
    algebraic variety if and only if $(S)=(S^{'}).$  Let $(S)$ be the $k[T_1,\ldots, T_n]:=k[T]$ ideal
    generated by the elementary symmetric polynomials $S_i$ of degree at most $n$ in the $T_i$, and let
    $(S^{'})$ be the ideal generated by the set of polynomials
    $$S^{'}=\left\{X_i:=\sum_{j=1}^n T_j^i:1\leq i\leq n\right\}.$$
    It is an elementary result from algebra that $(S)$ consists of \textit{all} symmetric polynomials in the $T_i$,
    (see, for example \cite[p. 550]{artin}) and therefore that $(S^{'})\subset(S)$.  The formula
    $$j!S_j=X_1^j-X_j+\sum_{i=1}^{\floor{\frac{j}{2}}}\binom{j}{i}\left(X_j-X_iX_{j-i}\right)$$
    may be proved by induction and shows that $j!S_j\in (S^{'})$ for each $1\leq j\leq n$.
    Now if $\cha(k)\not| n!$ then $j!$ is invertible in $k$ for each $1\leq j\leq n$ and therefore
    $S_j\in (S^{'})$.  Moreover, the representation of $S_j$ given above is unique so we see that
    $(S)=(S^{'})$ if and only if $\cha(k)>n.$
\end{exer}

\begin{exer}
    Consider the systems
    \begin{align*}
        S&=\left\{T_1^2+T_2,T_1\right\}\\
        S^{'}&=\left\{T_2^2T_1^2+T_2^3+T_2+T_1T_2,T_2T_1^2+T_2^2+T_1\right\}
    \end{align*}
    of algebraic equations over $\Q$.
    The system $S^{'}$ is equal to
    $$\left\{T_2(T_2T_1^2+T_2^2+T_1)+T_2,T_2T_1^2+T_2^2+T_1\right\},$$
    which is equivalent to the system
    $$\left\{T_2,T_2T_1^2+T_2^2+T_1\right\},$$
    which is in turn equivalent to
    $$\left\{T_2,T_1\right\}.$$
    On the other hand, $S$ is equivalent to
    $$\left\{T_2,T_1\right\},$$
    since if $T_1$ and $T_1^2+T_2$ vanish on any variety $V$ then so does $T_2$ and conversely (because we are working over a field).
    Thus, $S$ and $S'$ define the same $\Q$ varieties.
\end{exer}

\begin{exer}
    Let $X\subset\A_k^n,Y\subset\A_k^m$ be two affine algebraic $k$
    varieties.  Let $X,Y$ be generated by the systems $S=\{f_1,\ldots,f_r\}\subset k[T_1,\ldots,T_n]$ and
    $S=\{g_1,\ldots,g_s\}\subset K[U_1,\ldots,U_m]$.  Let
    $$I=\{f_1,\ldots,f_r,g_1,\ldots g_s\}\subset k[T_1,\ldots,T_n,U_1,\ldots,U_m].$$  We then define the cartesian product of $X,Y$ as
    $$X\times Y(K):=\sol(I;K).$$
    Clearly, $X\times Y(K)=X(K)\times Y(K)$ for any $k$ algebra $K$ (we just concatenate solutions).
\end{exer}

\begin{exer}
    Let us show that an arbitrary intersection of algebraic varieties is again an algebraic variety and that
    any finite union of algebraic varieties is an algebraic variety.
    \begin{enumerate}
\item Let $S$ be a finite index set and let $V_s$ for $s\in S$ be algebraic $k$ sets and let $I_s=I(V_s)$.  Define $I=\prod_{s\in S} I_s$.  Then
    we claim that $$V(I)=\bigcup_{s\in S} V_s.$$  To see this, observe that if $x\in V_s$ for some $s$ then
    every $f\in I_s$ vanishes at $x$, so any finite sum of products of elements of all the $I_r$ vanishes at $x$ (since one of the factors
     \textit{must} be in $I_s$); i.e. $x\in V(I)$.
    Conversely, let $x\in V(I)$ and suppose that $x\not\in V_s$ for all $s$.  Then for each $s$ there exists $f_s$ such that $f_s(x)\neq 0$,
    whence $f=\prod_{s\in S}f_s(x)\neq 0$, but $f\in I,$ a contradiction.

\item Now let $S$ be an arbitrary index set and let $V_s,I_s$ be as above.  Set $I=\sum_{s\in S} I_s$.  Now we claim
that $$\bigcap_{s\in S}V_s=V(I).$$  Let $x\in V_s$ for all $s$.  Then $I_S$ vanishes at $x$ for each $s$, so that $I$
also vanishes at $x$; that is, $x\in V(I).$  Conversely, since $I_s\subset \sum_{s\in S} I_s$, we see that
$V(I)\subset V(I_s)$ (since the $V$ operator is inclusion reversing) and hence that $V(I)\subset \bigcap V_s$.
\end{enumerate}
\end{exer}

\begin{exer}
    Let $$S,S^{'}\subset k[T_1,\ldots,T_n]$$ be two systems of linear equations and suppose that $\sol((S);k)=\sol((S^{'});k)$.
    Since $S$ is linear, we can identify $S$ with a subset of $k^n$
    via
    $$\sum_{i=1}^n a_iT_i\mapsto (a_1,\ldots,a_n).$$
    Indeed, we can in fact view $S$ as a subspace of $k^n$ (as a vector space)
    since if $f(c_1,\ldots,c_n)=0$ for some $f\in S$ and $c=(c_1,\ldots,c_n)\in k^n$ then $f(\lambda c_1,\ldots,\lambda c_n)=0$
    also since $f$ is linear.  It is not difficult to see that under this identification, $\sol((S);k)$ is just the orthogonal
    complement of $S$ in $k^n$ with respect to the usual inner product.  Now observe that any space is determined
    uniquely by its orthogonal complement, so that the \textit{vector spaces} $S,S^{'}$ must be equal.  This implies that
    the \textit{ideals} $(S),(S^{'})$ are equal.  The converse of this result is trivial.
\end{exer}

\begin{exer}
    We show that if a ring $A$ is Noetherian then so is the polynomial ring $A[X]$.  For suppose that $A[X]$
    is not Noetherian.  Then there exists a nonstabilizing chain of ideals $I_1\subset I_2\subset\ldots\subset A[X]$.
    Let $f_1\in I$ be a polynomial of minimal degree in $X$.  Define $f_i$ for $i\geq 2$ inductively to be a polynomial
    of minimal degree in $I\setminus(f_1,f_2,\ldots,f_{i-1})$.
    Let $f_i$ have degree $d(i)$ in $X$ and leading coefficient $a_i$.  Then
    we claim that $(a_1)\subset (a_1,a_2)\subset\ldots\subset A$ is a nonstabilizing chain of ideals.  For suppose that
    it stabilizes: then there exists some $n$ such that $(a_1,\ldots a_n)=(a_1,\ldots,a_m)$ for all $m\geq n$, so that we can
    write
    $$a_{n+1}=\sum_{i=1}^n b_ia_i$$ where $b_i\in A$.  Then
    the polynomial $f_{n+1}-\sum_{i=1}^n b_iX^{d(n)-d(i)}f_i$ is in $I\setminus (f_1,\ldots,f_n)$
    (since $f_{n+1}\not\in (f_1,\ldots, f_n)$), but is of lower degree than $f_{n+1}$.  This is a contradiction and our
    chain of ideals in $A$ does not terminate.  The contrapositive then shows that if $A$ is Noetherian, so is $A[X]$, and by induction,
    this implies that $A[T_1,\ldots,T_n]$ is Noetherian.
\end{exer}

\begin{exer}
    Let $A=k[T_1,T_2]/(T_1^2-T_2^3)$.  Let $K$ denote the field of fractions of $A$.  Then $T_1/T_2\in K\setminus A$ and
    $$\left(\frac{T_1}{T_2}\right)^2=\frac{T_1^2}{T_2^2}=\frac{T_2^3}{T_2^2}=T_2.$$  Therefore,
    $T_1/T_2$ is a root of the polynomial $X^2-T_2$.  It is therefore integral over $A$.
\end{exer}

\begin{exer}
    Let $X,Y$ be two affine algebraic sets in $K^n$.  Suppose that $f\in I(X)\cap I(Y)$.  Then $f$ vanishes on $X$ and on $Y$
    so that $f$ vanishes on $X\cup Y$ and hence $f\in I(X\cup Y).$  Conversely, if $f\in I(X\cup Y)$ then $f$ vanishes on $X$ and
    on $Y$ and therefore $f\in I(X)$ and $f\in I(Y)$.  In short, this shows that
    $$I(X\cup Y)=I(X)\cap I(Y).$$  We do not always have $I(X)\cap I(Y)=I(X)I(Y)$.  For example, let $Y=X=\sol(T_1;K).$
    Then $I(X)=I(Y)=(T_1)$ (since this is a prime ideal it is already radical) and hence $I(X)\cap I(Y)=(T_1).$  On the other
    hand, $I(X)I(Y)=(T_1^2)$.
\end{exer}

\begin{exer}
    Let $I\subset k[T_1,T_2]$ be generated by $T_1^2 T_2,T_1T_2^3$ and suppose that $f\in \rad(I)$.  Then $f^N=aT_1^2 T_2+bT_1T_2^3$
    for some $N\in\N$ and $a,b\in k[T_1,T_2].$  This shows that $f^N$ is divisible by $T_1T_2$ and therefore that $f$ is divisible by $T_1T_2$
    (since $T_1$ and $T_2$ are both irreducible, clearly).  Hence $f\in (T_1T_2)$ and therefore that $\rad(I)\subset (T_1T_2).$  On the other
    hand, $(T_1T_2)^2=T_2(T_1^2T_2)\in I$ so that $(T_1T_2)\subset \rad(I)$ and therefore $\rad(I)=(T_1T_2)$.
\end{exer}

\begin{exer}
    We claim that the Zariski topology on $\A^n(K)$ is not Hausdorff.  For let $x,y\in\A^n(K)$ be distinct points.  Then
    if the Zariski topology were Hausdorff, we could find varieties $V,U$ such that $x\in\A^n(K)\setminus V$
    and $y\in\A^n(K)\setminus U$ and
    $$\A^n(K)\setminus V\cap\A^n(K)\setminus U=\A^n(K)\setminus (V\cup U)=\emptyset,$$
    or equivalently, $V\cup U=\A^n(K).$  But observe that $\A^n(K)$ is an irreducible variety
    and therefore we must have $V=\A^n(K)$ or $U=\A^n(K)$ and consequently we cannot separate $x,y$.
    The Zariski topology on $\A^n(K)$ is, however separable: Let $x,y$ be distinct points of $\A^n(K)$.
    Then the variety $V=\sol(T-x;K)$ contains $x$ and not $y$; consequently $O=\A^n(K)\setminus V$
    is an open set containing $y$ and not $x$.  The same is not true in general for the Zariski topology
    on $\A_k^n(K)$ is $k\neq K$.  As an example, let $k=\Q$ and $K=\Q(\sqrt{2}).$  Let $x=\sqrt{2}$ and
    $y=-\sqrt{2}$.  Then any algebraic $\Q$ set containing $x$ must consist of polynomials all divisible by $T_2-2$
    and therefore contains $y$ also.  I conjecture that if $x\not\in G_{K/k}y$ then $x,y$ can be separated in the
    Zariski topology, where $G_{K/k}$ denotes the galois group of $K/k$, which has a natural action
    on $K^n$ and hence on any variety.
\end{exer}

\begin{exer}
    Observe that any solution in $K^n$ to the system of algebraic equations
    $$T_1^3=T_2^3=T_1T_2(T_1+T_2)=0$$
    over any field $K$ must have $T_1=T_2=0$ (this is already implied by the first two equations over a field).
    Any polynomial $f\not\equiv 0$ vanishing at $T_1=T_2=0$ is certainly in the ideal $(T_1,T_2)$: if some term of $f$ is not
    divisible by either $T_1$ or $T_2$ then there exists some point of $K^n$ where $f$ does not vanish.  Conversely, any polynomial
    in this ideal vanishes at $T_1=T_2=0$.  Hence, $I(V)=(T_1,T_2)\subset k[T_1,\ldots T_n]$.
\end{exer}

\begin{exer}
    Observe that the set $S=\{(z_1,z_2)\in\C^2:|z_1|^2+|z_2|^2=1\}$ has a limit point in $\C^2$.  The closure of $S$ in the Zariski $\C^2$
    topology is contained in some closed set, that is, some variety $V$.  Now let $f\in I(V)$.  Then $f$ is a polynomial in $z_1,z_2$
    that vanishes on a subset of $\C^2$ having a limit point in $\C^2$.  From several complex variables we know that any holomorphic
    function that vanishes on a set having a limit point vanishes identically.  Therefore, $f\equiv 0$ and $I(V)=(0)$ so that
    $V=C^2$.  This shows that $S$ is dense in $\C^2$.
\end{exer}

\begin{thebibliography}{2}

    \bibitem{artin} Artin, M.  \textit{Algebra.} Prentice Hall, N.J. 1991.

\end{thebibliography}
\end{document}