Fractional Ideals) a Non-Zero Submodule M of K Is Called a Fraction Ideal If There Exist X R 0 Such That Xm R

LECTURE OUTLINE AND SOLUTION TO EXERCISE FOR MATH581A

WENHAN WANG

1. Dedekind domain 1. (Fractional Ideals) A non-zero submodule M of K is called a fraction ideal if there exist x R 0 such that xM R. If M is a fractional ideal, we 1 ∈ − { } ⊆ 1 deﬁne M − = x K xM R . Note that M − is also a fractional ideal. { ∈ | ⊆ } 1 1 1a. Suppose M N are two fractional ideals, then M − N − . ⊆ ⊇ 1 Proof. By deﬁnition, M − = x K xM R . Since M N, we have xM xN for all x K{. Th∈erefor| e ⊆x }K xM R ⊆ x ⊆ 1 ∈ { ∈ | ⊆ } ⊇ { ∈ K xN R = N − . | ⊆ } ! 1 1 1b. Show that M (M − )− . Give an example such that the inclusion is strict. ⊆

1 1 Proof. Let x M, then it is suffices to show that x (M − )− , i.e., 1 ∈ 1 ∈ xM − R. This holds because for each y M − , xy R, which follows ∈from the fact that yM R and x M∈. ∈ A counter example could be giv⊆ en in man∈ y ways (as mentioned in class): (i) Let R = k[x, y], where k is a field. Let M = (x, y) = xR + yR is an ideal in R, hence a fractional ideal of K = k(x, y). We 1 claim that M − = R. To see this, note that an element t K 1 ∈ satisfies t M − if and only if tM R. This means t(xR + yR) R, ∈which requires that txR ⊆R and tyR R. Hence ⊆ 1 1 ⊆ 1 ⊆ t (x− R) (y− R) = R. Therefore M − R and the inverse in∈clusion is ∩trivial. Hence our claim holds, wh⊆ich also yields that 1 1 (M − )− = R M is a strict inclusion. ⊃ (ii) Let R = k[x, y]/(y2 x3), where k is a field. Then R = k[t2, t3] − 2 3 2 3 ∼ 1 ⊂ k[t] as k-algebras. Let M = (t , t ) = t R + t R. Then M − = 1 1 2 3 t− (R + tR). So MM − = tR + t R + t R R. ⊂ (iii) Let R = k[x, y, z]/(xy z2) = k[s2, st, t2] as k-algebra. Let 2 2 − ∼ p = (s , st) = s R + stR, then p is a prime ideal because R/p ∼= k[t2] is an integral domain. p is not maximal because k[t2] is 1 1 not a field. First we calculate p− . Note that x p− if and only if xp R, which means that x(s2R + stR)∈ R. It is ⊆ 1 1 1 2 1 2⊆ 2 equivalent to that x s− (s− R t− R) = (s t)− (s tR+st R). 1 ∈2 1 2 ∩ 2 2 2 2 Therefore pp− = (s t)− (s R+stR)(s tR+st R) = (s , st, t ), the maximal ideal containing p.

! 1 2 WENHAN WANG

1 1 1 1 1c. Show that ((M − )− )− = M − . 1 1 1 1 1 Proof. Apply 1b to M − we get M − ((M − )− )− . Apply 1a to 1 1 ⊆ M (M − )− we get the inverse inclusion. ⊆ ! 1 1 1 1d. (M − )− = M if M is invertible, i.e., MM − = R. 1 1 1 Proof. It suﬃces to prove that (M − )− M. Since MM − = R, ⊆ then we can write 1R = i xiyi as a ﬁnite sum, with xi M and 1 1 1 ∈ yi M − . Then for any f (M − )− , we have f = f 1 = i fxiyi. ∈ 1 1 !∈ · Note that f (M − )− implies that fyi R for each i. Therefore f M since eac∈ h x M. ∈ ! ∈ i ∈ ! 1f. A principal fractional ideal is a fractional ideal of the form yR with y K. Show that all the principal fractional ideals forms a subgroup. ∈ Proof. First note that the identity element R = 1RR is a principal fractional ideal. Furthermore if M = xR and N = yR are two frac- 1 1 tional ideals, then MN − = (xR) (yR)− = (x/y)R is also a fractional ideal. Therefore all the principal ·fractional ideals is closed under mul- tiplication. ! 1g. Show that two invertible fractional ideals are isomorphic if and only if they give the same element in the class group. Proof. . Suppose M and N are two invertible fractional ideals and ⇒ 1 1 M = N as R-module. Then MN − = M R N − as R-modules. ∼ 1 1 ∼ 1 ⊗ Therefore MN − = N R N − = NN − = R as R-modules. This 1∼ ⊗ ∼ proves that MN − is a principal fractional ideal. 1 . Suppose M and N give the same element in Cl(R), then MN − is ⇐ 1 a principal ideal, say aR for some a K. Then M(aN)− = R. This ∈ gives that M = aN since M, N are invertible. Therefore M ∼= N by sending m m/a. ,→ ! E-mail address: [email protected]