Related to fractional ideals - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:54:59Z http://mathoverflow.net/feeds/question/52341 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/52341/related-to-fractional-ideals Related to fractional ideals ashpool 2011-01-17T19:03:02Z 2011-01-25T21:55:12Z <p>$K$ a field, $A\subset K$ a subring, $M\subset K$ an $A$-submodule. Define $$(A:_{K}M):= \lbrace s\in K|sM\subset A\rbrace$$ Then it is easy to see that </p> <p>$$M\subset A\Longleftrightarrow A\subset (A:_{K}M),$$</p> <p>$$A\subset M\Longrightarrow (A:_{K}M)\subset A$$</p> <p>But I couldn't show the reverse implication of the second. It is true if $M$ is invertible, and I'm guessing that it is true only if $M$ is invertible. Any ideas?</p> http://mathoverflow.net/questions/52341/related-to-fractional-ideals/52373#52373 Answer by Joël for Related to fractional ideals Joël 2011-01-18T02:52:58Z 2011-01-19T15:51:58Z <p>This is not true, you are guessing right. Here is a counter-example: take $k$ a field, $A=k[X,Y]$, $K=k(X,Y)$. Let $M=XA+YA$. Then using the fact that $A$ is a UFD one see easily that $(A:_K M)=A$. (indeed, let $P/Q \in (A:_K M)$ with $P,Q \in A$, $Q \neq 0$. We have $XP/Q \in A$ so $Q$ divides $XP$. Similarly $Q$ divides $YP$. Since $X$ and $Y$ are two non-equivalent irreducible elements of $A$, this implies that $Q|P$, hence $P/Q \in A$).</p> <p>Yet $A \not \subset M$.</p> http://mathoverflow.net/questions/52341/related-to-fractional-ideals/53287#53287 Answer by gp for Related to fractional ideals gp 2011-01-25T21:55:12Z 2011-01-25T21:55:12Z <p>Let me add just something to this.</p> <p>Let $K$ be the quotient field of $A$. If $M$ is a fractional ideal of $A$, then the fractional ideal $M_v = (A\colon_K(A\colon_KM))$ is called "the divisorial closure of $M$". In this case (when $K$ is the quotient field of $A$) one can show that $(A\colon_KM) \subset A \Longleftrightarrow A \subseteq M_v$.</p> <p>Indeed, $(A\colon_KM) \subset A$ implies $M_v = (A:_K(A\colon_KM)) \supset (A\colon_KA) = A$. Conversely, $A \subset M_v$ implies $A = (A\colon_KA) \supset (A\colon_KM_v)$. Now it is easily seen that $(A\colon_KM_v) = (A\colon_KM)$. Thus $A \supset (A\colon_KM)$.</p> <p>So, if $M_v = M$ (i.e., if $M$ is "divisorial"), then the converse of your second statement holds. Now, invertible ideals are divisorial but there are plenty of divisorial ideals that are not invertible (see for example Bill Heinzer's paper "Integral domains in which each non-zero ideal is divisorial", Mathematika (1968), 15: 164-170), so it is not true that the reverse implication of your second statement is true "only" if $M$ is invertible. </p> <p>More on the divisorial closure (and more generally on star operations) is on Robert Gilmer's "Multiplicative ideal theory". </p>