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\title{Math 631, Lecture 1}
\author{Notes by Bryden Cais}
\date{Sept. 6, 2002}
\begin{document}
\maketitle

Fix a field $k$ and denote by $k[T]$ the polynomial ring in $n$ variables $k[T_1\ldots T_n].$  The ring $k[T]$ has the natural
structure of a $k$-algebra, since it is a ring, $k$ injects (via inclusion) into $k[T]$ and this injection realizes $k[T]$
as a $k$-vector space.

\begin{defn}
    A \textbf{system of algebraic equations} is any set $S\subset k[T]$.
\end{defn}

Now let $K$ be any $k$-algebra (in particular, this includes the cases when $K/k$ is a field extension), and let
$\alpha=(\alpha_1\ldots\alpha_n)\in K^n$.  Then we say that $\alpha$ is a $K$-\textbf{solution of} $S$ if
$F(\alpha)=0$ for all $F\in S$.  We denote the set of all $K$-solutions of $S$ $\sol(S;K)$.

\begin{exam}
    If $n=2$, $k=\Q$, and $S=\{F(T_1,T_2)\}$, then $\sol(S,k)=\{(T_1,T_2)\in\Q^2:F(T_1,T_2)=0\}.$  Similarly,
    we see that $\sol(S;\R)\subset\R^2$ is a real algebraic curve and that $\sol(S;\C)\subset\C^2$ has the structure
    or a Riemann surface.
\end{exam}
\begin{exam}
    We have $K^n=\sol(\{0\};K)$ for any $K$.
\end{exam}

Given any $k$-algebra homomorphism (that is, a ring homomorphism that is the identity on $k$)
$$\varphi :K\longrightarrow K^{'},$$ we have an induced homomorphism
$$\bigoplus^n\varphi:\sol(S;K)\longrightarrow\sol(S,K^{'})$$
given by the obvious homomorphism from $K^n$ to ${K^{'}}^n$ (since after all, $\sol(S;K)\subset K^n$).

\begin{prop}\label{Sideal}
    Let $(S)$ be the $k[T]$ ideal generated by $S$.  Then for any $K,S$, $\sol(S;K)=\sol((S);K)$.
\end{prop}
\begin{proof}
    Suppose $\alpha\in K^n$ is in $S$.  Then for every $F\in S$ we have $F(\alpha)=0,$ and therefore $\alpha\in \sol((S);K)$ since
    every $G\in(S)$ has the form
    $$G=\sum_{F_s\in S}A_sF_s,$$ where $A_s\in k[T]$.  The converse is even more obvious since $S\subset (S)$.
\end{proof}

\begin{thm}
    We have $\sol(S;K)=\sol(S^{'},K)$ for all $K$ if and only if $(S)=(S^{'})$.
\end{thm}
\begin{proof}
    The if direction is trivial.  For the reverse, observe that
    by Proposition \ref{Sideal} it suffices to show that $\sol((S);K)=\sol((S^{'}),K)$ for all $K$ implies $(S)=(S^{'})$.
    Let $\alpha=(\alpha_1\ldots\alpha_n)\in\sol(S;K)$ and define
    $$\varphi_{\alpha}:k[T]\longrightarrow K$$ by
    $\varphi(T_i)=\alpha_i$.  It is clear that $\varphi_{\alpha}$ is a homomorphism of $k$-algebras.  Moreover,
    we see that $\varphi_{\alpha}((S))=0$ so that $(S)\subset \ker\varphi_{\alpha}.$  We therefore obtain the commutative
    diagram
    \begin{eqnarray*}
        \begin{CD}
            k[T]             @=               k[T]\\
            @VV q V                  @V\varphi_{\alpha}VV \\
             k[T]/(S)  @>\bar{\varphi}_{\alpha}>>  K
        \end{CD}
        \end{eqnarray*}
    where $q$ is the natural quotient map.
    Observe that this diagram tells us two things:
    \begin{enumerate}
        \item Given any $\alpha\in\sol((S);K)$ we obtain $\bar{\varphi}_{\alpha}$, an element of $\hom{k[T]/S}{K}$.
        \item Given any element $\varphi\in \hom{k[T]/S}{K}$, we obtain the composite homomorphism
        $$\tilde{\varphi}=\varphi\circ q:k[T]\longrightarrow K$$
        satisfying $\tilde{\varphi}((S))=0$.  This then gives us $\alpha=(\tilde{\varphi}(T_1),\ldots \tilde{\varphi}(T_n))\in \sol((S);K)$.
    \end{enumerate}
    The upshot is that we have a bijection
    $$\hom{k[T]/(S)}{K}\leftrightarrow \sol(S;K).$$
    Therefore, if $\sol((S);K)=\sol((S^{'});K)$ for all $K$, we have a bijection
    $$\hom{k[T]/(S)}{K}\leftrightarrow \hom{k[T]/(S^{'})}{K}$$
    for all $K$.  Now let $K$ be the $k$-algebra
    $k[T]/(S)$.  Observe that the identity map is in
    $$\hom{k[T]/(S)}{k[T]/(S)},$$ so that we have some homomorphism
    $\varphi:k[T]/(S^{'})\longrightarrow k[T]/(S)$.  In particular, this tells us that if we lift $\varphi$ to a map
    $\tilde{\varphi}:k[T]\longrightarrow k[T]$ in the natural way, we have $\tilde{\varphi}(S)\subset (S^{'})$.  Interchanging
    ALMOST THERE!!!
    $S$ and $S^{'}$ then completes the proof.
\end{proof}

Observe that the hypothesis that $K$ be \textit{any} $k$-algebra is essential: If $S=x^2+1$, then $\sol(S;\R)=\sol(1,\R)$ but
we clearly do not have $(1)=(S)$.  Notice also that the above proof uses this variability of field in a crucial way.

\end{document}