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@@ -37,5 +37,6 @@ Christian Merten (\href{mailto:cmerten@mathi.uni-heidelberg.de}{cmerten@mathi.un
 \input{rav21.tex}
 \input{rav19.tex}
 \input{rav20.tex}
+\input{rav22.tex}
 
 \end{document}
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+\documentclass{lecture}
+
+\begin{document}
+
+\section{The real Nullstellensatz}
+
+When $k$ is algebraically closed, Hilbert's Nullstellensatz implies
+$\mathcal{I}(\mathcal{V}_{k^{n}}(I)) = \sqrt{I}$ for all ideal
+$I \subseteq k[T_1, \ldots, T_n]$. In this section we try to compute
+$\mathcal{I}(\mathcal{V}_{k^{n}}(I))$ when $k$ is a real-closed field.
+
+\begin{definition}[]
+    Let $(k, \le)$ be an ordered field and let $A$ be a commutative $k$-algebra with unit.
+    An ideal $I \subseteq A$ is called a \emph{real ideal} if it satisfies the following condition: If
+    $\lambda_1, \ldots, \lambda_r > 0$ in $k$ and $a_1, \ldots, a_r \in A$ satisfy
+    \[
+    \sum_{j=1}^{r} \lambda_j a_j^2 \in I
+    ,\] then $a_j \in I$ for all $j$.
+    $A$ is a \emph{real algebra} if the zero ideal in $A$ is
+    a real ideal.
+\end{definition}
+
+\begin{satz}
+    Let $(k, \le)$ be an ordered field and let $Z \subseteq k^{n}$ be a subset. Then the ideal
+    $\mathcal{I}(Z)$ is a real ideal.
+\end{satz}
+
+\begin{proof}
+    If $Z = \emptyset$, then $\mathcal{I}(Z) = \mathcal{I}(\emptyset) = k[T_1, \ldots, T_n]$ is
+    a real ideal. Now assume $Z \neq \emptyset$. In this case, if $P_1, \ldots, P_r \in k[T_1, \ldots, T_n]$
+    and $\lambda_1, \ldots, \lambda_r > 0$ in $k$ are such that
+    $\sum_{j=1}^{r} \lambda_j P_j^2 \in \mathcal{I}(Z)$, then
+    for all $x \in Z$, $\sum_{j=1}^{r} \lambda_j P_j^2(x) = 0$ in $k$. Since
+    $k$ is an ordered field and $\lambda_j > 0$ for all $j$, this implies
+    that for all $j$, $P_j(x) = 0$, i.e. $P_j \in \mathcal{I}(Z)$.
+\end{proof}
+
+Recall that if $k$ is an arbitrary field and $I \subsetneq k[T_1, \ldots, T_n]$ is a proper ideal,
+then finding a common zero $x \in L^{n}$ to all polynomials $P \in I$ for some extension $L$ of $k$
+is equivalent to finding a homomorphism of $k$-algebras
+\[
+\varphi\colon k[T_1, \ldots, T_n]/I \longrightarrow L
+.\] Indeed, the correspondence is obtained by sending such a $\varphi$
+to $x = (x_1, \ldots, x_n)$ where $x_i = \varphi(T_i \text{ mod } I)$. The basic
+result should be about giving sufficient conditions for such homomorphisms to exist.
+
+\begin{theorem}[Real Nullstellensatz I]
+    Let $(k, \le )$ be an ordered field and let $k^{(r)}$ be the real closure of $k$. Let
+    $I \subseteq k[T_1, \ldots, T_n]$ be a real ideal. Then there exists a homomorphism
+    of $k$-algebras
+    \[
+    k[T_1, \ldots, T_n] / I \longrightarrow k^{(r)}
+    .\] In particular, if $I \subsetneq k[T_1, \ldots, T_n]$ is a proper real ideal, then
+    $\mathcal{V}_{k^{r}}(I) \neq \emptyset$.
+    \label{thm:real-nullstellensatz}
+\end{theorem}
+
+Let $(k, \le)$ be an ordered field. For the proof of \ref{thm:real-nullstellensatz}, we need
+two lemmata:
+
+\begin{lemma}
+    Let $I \subseteq k[T_1, \ldots, T_n]$ be a real ideal. Then $\sqrt{I} = I$. Moreover,
+    if $\mathfrak{p} \supset I$ is a minimal prime ideal containing $I$, then
+    $\mathfrak{p}$ is real.
+\end{lemma}
+
+\begin{lemma}
+    Let $\mathfrak{p} \subseteq k[T_1, \ldots, T_n]$ be a prime ideal. Then the fraction field
+    \[
+    K \coloneqq \operatorname{Frac}\left( k[T_1, \ldots, T_n]/\mathfrak{p} \right) 
+    \] is a real field if and only if the prime ideal $\mathfrak{p}$ is real. In that case
+    $K$ can be ordered in a way that extends the order of $k$.
+\end{lemma}
+
+\end{document}