Radical ideal of A ---> Jacobson radical of A

Thanks to Kestutis Cesnavicius
http://stacks.math.columbia.edu/tag/09XE#comment-457

more-algebra.tex

@@ -1603,7 +1603,7 @@ \section{Henselian pairs}
 \label{definition-henselian-pair}
 A {\it henselian pair} is a pair $(A, I)$ satisfying
 \begin{enumerate}
-\item $I$ is contained in the radical ideal of $A$, and
+\item $I$ is contained in the Jacobson radical of $A$, and
 \item for any monic polynomial $f \in A[T]$ and factorization
 $\overline{f} = g_0h_0$ with $g_0, h_0 \in A/I[T]$ monic
 generating the unit ideal in $A/I[T]$, there
@@ -1623,7 +1623,7 @@ \section{Henselian pairs}
 
 \begin{lemma}
 \label{lemma-idempotents-determined-modulo-radical}
-Let $(A, I)$ be a pair. If $I$ is contained in the radical ideal
+Let $(A, I)$ be a pair. If $I$ is contained in the Jacobson radical
 of $A$, then the map from idempotents of $A$ to idempotents of
 $A/I$ is injective.
 \end{lemma}
@@ -1739,10 +1739,10 @@ \section{Henselian pairs}
 \end{lemma}
 
 \begin{proof}
-Assume (2) holds. Then $I$ is contained in the radical of $A$, since
+Assume (2) holds. Then $I$ is contained in the Jacobson radical of $A$, since
 otherwise there would be a nonunit $f \in A$ not contained in $I$
 and the map $A \to A_f$ would contradict (2). Hence $IB \subset B$
-is contained in the radical of $B$ for $B$ integral over $A$
+is contained in the Jacobson radical of $B$ for $B$ integral over $A$
 because $\Spec(B) \to \Spec(A)$ is closed by
 Algebra, Lemmas \ref{algebra-lemma-going-up-closed} and
 \ref{algebra-lemma-integral-going-up}.
@@ -1761,8 +1761,8 @@ \section{Henselian pairs}
 finite map $A \to B = A/(I \cap \mathfrak m)$. The condition that
 $B \to B/IB$ induces a bijection on idempotents implies that
 $I \subset \mathfrak m$ (if not, then $B = A/I \times A/\mathfrak m$
-and $B/IB = A/I$). Thus we see that $I$ is contained in the
-radical ideal of $A$. Let $f \in A[T]$ be monic and suppose given a
+and $B/IB = A/I$). Thus we see that $I$ is contained in the Jacobson
+radical of $A$. Let $f \in A[T]$ be monic and suppose given a
 factorization $\overline{f} = g_0h_0$ with $g_0, h_0 \in A/I[T]$ monic.
 Set $B = A[T]/(f)$. Let $\overline{e}$ be the nontrivial idempotent
 of $B/IB$ corresponding to the decomposition
@@ -1800,18 +1800,18 @@ \section{Henselian pairs}
 reduction modulo $I$ factors as $\overline{f} = g_0 T^n$ where
 $T, g_0$ generate the unit ideal in $A/I[T]$. Thus by assumption
 we can factor $f$ as $f = g h$ where $g$ is a monic lift of $g_0$ and
-$h$ is a monic lift of $T^n$. Because $I$ is contained in the radical
+$h$ is a monic lift of $T^n$. Because $I$ is contained in the Jacobson radical
 of $A$, we find that $g$ and $h$ generate the unit ideal in $A[T]$
 (details omitted; hint: use that $A[T]/(g, h)$ is finite over $A$).
 Thus $A[T]/(f) = A[T]/(h) \times A[T]/(g)$ and we find a corresponding
 product decomposition $A'' = A''_1 \times A''_2$. By construction
 we have $A''_1/IA''_1 = A/I$ and $A''_2/IA''_2 = C'$. Since $A''_1$
-is integral over $A$ and $I$ is contained in the radical of $A$ we
+is integral over $A$ and $I$ is contained in the Jacobson radical of $A$ we
 see that $a$ maps to an invertible element of $A''_1$. Hence
 $A''_a = A''_1 \times (A''_2)_a$. It follows that $A \to A''_1$
 is integral as well as \'etale, hence finite locally free.
 However, $A''_1/IA''_1 = A/I$ thus $A''_1$ has rank $1$ as an
-$A$-module along $V(I)$. Since $I$ is contained in the radical
+$A$-module along $V(I)$. Since $I$ is contained in the Jacobson radical
 of $A$ we conclude that $A''_1$ has rank $1$ everywhere and
 it follows that $A \to A''_1$ is an isomorphism. Thus
 $A' \to A'_a \cong A''_a \to (A''_1)_a = A''_1 = A$ is the
@@ -1910,7 +1910,7 @@ \section{Henselian pairs}
 (Lemma \ref{lemma-characterize-henselian-pair})
 we get $B' = B'_1 \times B'_2$ with $C = B'_1/IB'_1$ and $C' = B'_2/IB'_2$.
 The image of $g$ in $B'_1$ is invertible and $(B'_2)_g = 0$
-because $IB'$ is contained in the radical of $B'$
+because $IB'$ is contained in the Jacobson radical of $B'$
 (for example because $(B', IB')$ is a henselian pair by
 Lemma \ref{lemma-integral-over-henselian-pair}).
 We conclude that $B'_1 = B_g$ is finite \'etale over $A$
@@ -1974,8 +1974,8 @@ \section{Henselian pairs}
 the section $\sigma_B$ is the unique $A'$-algebra map $B' \to A'$.
 We have $B' \otimes_{A'} B' = B' \times R$ for some ring $R$, see
 Algebra, Lemma \ref{algebra-lemma-diagonal-unramified}. In our case
-$R/I'R = 0$ as $B'/I'B' = A'/I'$. Since $I'$ is contained in the radical
-of $A'$ we see that $\Hom_{A'\text{-modules}}(R, A') = 0$ for example
+$R/I'R = 0$ as $B'/I'B' = A'/I'$. Since $I'$ is contained in the Jacobson
+radical of $A'$ we see that $\Hom_{A'\text{-modules}}(R, A') = 0$ for example
 by Algebra, Lemma \ref{algebra-lemma-characterize-zero-local}.
 Thus there is at most one $A'$-algebra map $B' \to A'$.
 Using the commutativity we obtain