diff --git a/kaplansky_direct_finite_char2_v2.txt b/kaplansky_direct_finite_char2_v2.txt
index d120aac..90b18f1 100644
--- a/kaplansky_direct_finite_char2_v2.txt
+++ b/kaplansky_direct_finite_char2_v2.txt
@@ -99,7 +99,7 @@ is the smallest such positive integer. Examples:
 
   - Q, R, C: characteristic 0.
   - F_p = Z/pZ: characteristic p.
-  - F_{p^r}: finite field with p^r elements, also characteritic p.
+  - F_{p^r}: finite field with p^r elements, also characteristic p.
   - Specifically F_2 = {0, 1}: characteristic two, and 1 + 1 = 0.
 
 In this paper the field K has characteristic two, so 2 = 0 as a
@@ -359,8 +359,8 @@ order l and finite sums a_out, b_out in K[G] such that
 
       a_out b_out = 1            and            b_out a_out != 1.
 
-The elements a_out, b_out are given by explicit formulas (Section
-5 below in this document; Section 02-criterion in the original).
+The elements a_out, b_out are given by explicit formulas (PARTS III
+and IX below; Section 02-criterion of the original).
 
 What each hypothesis does:
   (a) exact incidence gives the "subgroup average" identity;
@@ -553,8 +553,10 @@ cross-terms inside any one slot vanish, so J_0 C = e P. Put
       a_mat = U J_0 + (1 - e) I_m,
       b_mat = C T + (1 - e) I_m.
 
-Then a_mat b_mat = I_m, because U (eP) T = U T (using UP=U, PT=T)
-and the (1-e) parts do not mix. So far, a failure of STABLE
+Then a_mat b_mat = I_m. Indeed UJ_0 C T = U (eP) T = e (U P T) =
+e (U T) = e I_m, using UP=U, PT=T, and UT=I_m from Step 5; the
+(1-e) terms give (1-e)^2 = 1-e and are killed by the e-factors, so
+the total is e I_m + (1-e) I_m = I_m. So far, a failure of STABLE
 finiteness in Mat_m(K[L]).
 
 Step 8. Detect the reverse defect. Set Delta = b_mat a_mat - I_m.