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Fix numbering
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aisejohan committed May 1, 2021
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10 changes: 5 additions & 5 deletions more-morphisms.tex
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Expand Up @@ -8122,7 +8122,7 @@ \section{Bertini theorems}
\end{lemma}

\begin{proof}
In order for assumption (2) to hold, the elements $v_1, v_2, v_3$
In order for assumption (3) to hold, the elements $v_1, v_2, v_3$
must be $k$-linearly independent in $V$ (small detail omitted).
Thus we may choose a basis $v_1, \ldots, v_r$ of $V$ incorporating
these elements as the first $3$. Recall that
Expand All @@ -8146,7 +8146,7 @@ \section{Bertini theorems}
$$
By our choice of basis we see that $f_1$ cannot be zero because this
would mean $v_1 = 0$ and hence $H_{v_1} = X$ which contradicts
assumption (2). Hence $\sum x_if_i$ is a nonzerodivisor in
assumption (3). Hence $\sum x_if_i$ is a nonzerodivisor in
$A[x_1, \ldots, x_r]$.
It follows that every irreducible component of $H_U$ has dimension
$d + r - 1$ where $d = \dim(X) = \dim(A)$. If $U' = U \cap D(f_1)$
Expand All @@ -8169,7 +8169,7 @@ \section{Bertini theorems}
dimension at most $d + r - 2$.
On the other hand, $(H_U \setminus H_{U'}) \times_U V(f_2, \ldots, f_r)$
is an $r$ dimensional affine space over $\Spec(A/(f_1, \ldots, f_r))$
and hence assumption (3) tells us this has dimension at most
and hence assumption (4) tells us this has dimension at most
$d + r - 2$. We conclude that $H_U$ is irreducible for every $U$ as above.
It follows that $H_{univ}$ is irreducible.

Expand All @@ -8180,8 +8180,8 @@ \section{Bertini theorems}
\ref{varieties-lemma-geometrically-irreducible-function-field}.
Choose a nonempty affine open $U = \Spec(A)$ of $X$
contained in $X \setminus H_{v_1}$ which meets the irreducible
component $Z$ of $H_{v_2} \cap H_{v_3}$ whose existence is averted in
assumption (2). With notation as above we have to prove that
component $Z$ of $H_{v_2} \cap H_{v_3}$ whose existence is asserted in
assumption (3). With notation as above we have to prove that
the field extension
$$
\text{Frac}(A[x_1, \ldots, x_r]/(x_1f_1 + \ldots + x_rf_r))/
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