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\begin{flashcard}[intdom-incl-alg-to-integral-lemma]
\begin{lemma}
  \refsteplabel[Lemma 1]{hn_lemma1}
  Let $A \le B$ be an inclusion of integral
  domains, and assume the fraction field $K$ of
  $A$ is contained in $B$. If $b \in B$ is
  algebraic over $K$, then there exists $p \in A$
  non-zero such that $pb$ is \gls{intel} over $A$.
\end{lemma}

\begin{proof}
  \cloze{Suppose $g \in K[X]$ with $g(b) = 0$, $g \neq
  0$. By clearing denominators, we can assume $g
  \in A[X]$. Write
  \[ g(X) = a_N X^n + \cdots + a_0, \qquad a_n
  \neq 0, a_i \in A .\]
  Note
  \[ a_N^{N - 1} g = (a_N X)^n + a_{N - 1}(A_N
  X)^{N - 1} + a_{N - 2} a_N \cdot (a_N X)^{N - 2}
  + \cdots + a_0 a_N^{N - 1} .\]
  This is a monic polynomial in $a_N X$. Thus
  taking $X = b$, we thus have a monic polynomial
  killing $a_N b$. So $a_N b$ is \gls{intel} over
  $A$ and we take $p = a_N$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[UFD-intel-ff-is-integer]
\begin{lemma}
  \refsteplabel[Lemma 2]{hn_lemma2}
  Let $A$ be a UFD with fraction field $K$. Then
  if $\alpha \in K$ is \gls{intel} over $A$, we
  have $\alpha \in A$.
\end{lemma}

\begin{proof}
  \cloze{If $\alpha \in K$ is \gls{intel} over $A$, write
  $\alpha = \frac{a}{b}$, with $a, b$ having no
  commmon factor. We have $g \left( \frac{a}{b}
  \right) = 0$ for some monic polynomial $g$, say
  \[ g(X) = X^n + a_1 X^{n - 1} + \cdots + a_n .\]
  We have
  \[ \frac{a^n}{b^n} + a_1 \frac{a^{n - 1}}{b^{n -
  1}} + \cdots + a_n = 0 \]
  in $K$. So
  \[ a^n + a_1 ba^{n - 1} + \cdots + a_n b^n = 0 \]
  in $A$. So $b \mid a$, so $b$ must be a unit in
  $A$. Thus $\frac{a}{b} \in A$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[intel-subring-lemma]
\begin{lemma}
  \refsteplabel[Lemma 3]{hn_lemma3}
  Let $A \le B$ be integral domains, and $S
  \subseteq B$ the set of all elements in $B$
  \gls{intel} over $A$. Then $S$ is a subring of
  $B$.
\end{lemma}

\begin{proof}
  \cloze{If $b_1, b_2 \in S$, then $A[b_1]$ is a finitely
  generated $A$-module. Also, $b_2$ is
  \gls{intel} over $A$, and hence is \gls{intel}
  over $A[b_1]$. Thus $A[b_1][b_2] = A[b_1, b_2]$
  is a finitely generated $A[b_1]$-module. Thus
  $A[b_1, b_2]$ -s a finitely generated
  $A$-module. Since $A[b_1 \pm b_2], A[b_1 \cdot
  b_2] \subseteq A[b_1, b_2]$, we have $b_1 \pm
  b_2, b_1 \cdot b_2 \in S$ by the proposition.

  We also have $0, 1 \in S$ since $A \subset S$.}
\end{proof}
\end{flashcard}

\begin{flashcard}[hn-vzero-lemma]
\begin{lemma}[Hilbert's Nullstellensatz, Version 0]
  \refsteplabel[Lemma 4]{hn_lemma4}
  \cloze{Let $\KK$ be an algebraically closed field, and
  $F / \KK$ a field extension which is finitely
  generated as a $\KK$-algebra (i.e. $\exists$ a
  surjective $\KK$-algebra homomorphism $\KK[X_1,
  \ldots, X_d] \to F$). Then $F = \KK$.}
\end{lemma}

\begin{proof}
  \cloze{Suppose $\alpha \in F$ is algebraic over $\KK$,
  say with irreducible polynomial $f(X) \in
  \KK[X]$. Then $f$ is linear since $\KK$ is
  algebraically closed, hence of the form $c(X -
  \alpha)$. So $\alpha \in \KK$.

  Suppose we are given a surjective map
  \begin{align*}
    \KK[X_1, \ldots, X_d]
    &\to F \\
    x_i
    &\mapsto z_i \in F
  \end{align*}
  Then $z_1, \ldots, z_d$ generate $F$ as a field
  extension of $\KK$. Assume $z_1, \ldots, z_e$
  form a \gls{transc_basis} for $F / \KK$. Note
  that if $F \neq \KK$, we must have $e \ge 1$.
  Let $R = \KK[z_1, \ldots, z_e] \le F$ (note
  that this really is a polynomial ring since
  $z_1, \ldots, z_e$ are \gls{algindep}). Then
  $w_1 = z_{e + 1}, \ldots, w_{d - e} = z_d$ are
  algebraic over $L = \KK(z_1, \ldots, z_e)$. Let
  $S \le F$ be the set of elements of $F$
  \gls{intel} over $R$. $S$ is a subring of $F$ by
  \nameref{hn_lemma3}. By \nameref{hn_lemma1},
  there exists $p_1, \ldots, d_{d - e} \in R$
  with $t_i \defeq p_i w_i$ \gls{intel} over $R$.
  In particular, $t_1 \in S$. Choose $\frac{f}{g}
  \in \KK(z_1, \ldots, z_e) = L$, $f, g \in R$,
  with $f, g$ relatively prime. Then $g$ is
  relatively prime to $p_1, \ldots, p_{d - e}$.
  Here, we assume $e \ge 1$. Thus
  \[ p_1^{n_1} \cdots p_{d - e}^{n_{d - e}}
  \frac{f}{g} \notin \KK[z_1, \ldots, z_e] \]
  for any $n_1, \ldots, n_{e - d} \ge 0$. Since
  $z_1, \ldots, z_d$ generate $F$ as a
  $\KK$-algebra there exists $q \in \KK[X_1,
  \ldots, X_d]$ such that
  \[ \frac{f}{g} = q(z_1, \ldots, z_d) = q\left(z_1,
  \ldots, z_e, \ub{\frac{t_1}{p_1}}_{z_{e + 1}}, \ldots,
  \ub{\frac{t_{d - e}}{p_{d - e}}}_{z_d} \right)
  \tag{$*$} \label{lec7_l132} \]
  Let $n_j$ be the highest power of $X_{e + j}$
  appearing in $q$. Multiplying by $\prod_j
  p_j^{n_j}$ clears denominators on RHS of
  \eqref{lec7_l132}. So we have
  \[ p_1^{n_1} \cdots p_{d - e}^{n_{d - e}}
  \frac{f}{g} = q'(z_1, \ldots, z_e, t_1, \ldots,
  t_{d - e}) \tag{$**$} \label{lec7_l139} \]
  The RHS of \eqref{lec7_l139} lies in $S$ as
  $z_1, \ldots, z_e \in S$, $t_1, \ldots, t_{d -
  e} \in S$. Thus LHS lies in $S$. But LHS lies in
  $\KK(z_1, \ldots, z_e)$ and hence by
  \nameref{hn_lemma2}, lies in $\KK[z_1, \ldots,
  z_e]$, a contradiction. Thus $e = 0$, and $F$ is
  algebraic over $\KK$, so $F = \KK$ since $\KK$ is
  algebraically closed.}
\end{proof}
\end{flashcard}