Don't use a PID+UFD for 5G

tex/H113/ring-flavors.tex

@@ -847,10 +847,10 @@ \section{\problemhead}
 \end{problem}
 
 \begin{dproblem}
-	How many prime ideals of $\ZZ[\sqrt 2]$ are \emph{not} maximal ideals?
+	How many prime ideals of $\ZZ[\sqrt{2017}]$ are \emph{not} maximal ideals?
 	\label{prob:dedekind_sample}
 	\begin{hint}
-		Show that the quotient $\ZZ[\sqrt2]/I$ has finitely many elements
+		Show that the quotient $\ZZ[\sqrt{2017}]/I$ has finitely many elements
 		for any nonzero prime ideal $I$.
 		Therefore, the quotient is an integral domain, it is also a field,
 		and thus $I$ was a maximal ideal.
@@ -860,13 +860,13 @@ \section{\problemhead}
 		We contend every other prime ideal is maximal.
 
 		Indeed, let $I$ be any ideal (not necessarily prime),
-		and let $a + b \sqrt 2$ be a nonzero element of it.
-		Then $I$ also contains $(a^2-2b^2)$.
+		and let $a + b \sqrt{2017}$ be a nonzero element of it.
+		Then $I$ also contains $(a^2-2017b^2)$.
 		That means when taking modulo $I$ we may take modulo the integer
-		$n \coloneqq |a^2-2b^2| \neq 0$.
+		$n \coloneqq |a^2-2017b^2| \neq 0$.
 
 		So every element in $R$ is equivalent modulo $I$
-		to an element of the form $x + y \sqrt 2$,
+		to an element of the form $x + y \sqrt{2017}$,
 		where $x,y \in \{0, 1, \dots, n-1\}$.
 		In other words, the quotient $R/I$ has at most finitely many elements.