Math 614, Fall 2008
Supplementary Problem Set #3: Solutions
1. Let M denote the ideal generated by all the monomials of R = R of positive degree.
Well show that M is not nitely generated. Suppose (mi , ni ), 1 i h are such that the
i = xmi y ni are genera
Math 614, Fall 2003
Problem Set #1: Solutions
1. (a) This is clear if R = K. If not, choose f R K of degree d > 0, and replace f
by a scalar multiple so that it has leading coecient 1: say f = xd + cd1 xd1 + + c0 .
Then xd + cd1 xd1 + + c0 f = 0 shows tha
Math 615, Winter 2007
Problem Set #3 Solutions
1. When one localizes at () P Supp (M/IM ), a regular sequence in I on M maps
to a regular sequence in IRP on MP by atness: one has MP /IMP = 0 by (). So the
depth cannot decrease. Pick a maximal regular sequ
Math 615, Winter 2007
Problem Set #4 Solutions
1. From 6. of Problem Set #1 with n = 3 and xij = xi+j1 the i are a Grbner basis for
o
hlex: and generating relations come from the checks in Buchbergers criterion: the ones
coming from cases (1) and (2) of t
Math 614, Fall 2010
Problem Set #3: Solutions
1. (a) The union of a chain of non-nitely generated submodules is a submodule: any two
elements belong to one of the modules in the chain, and, hence the union has the closure
properties needed to be a submodu
Math 614, Fall 2010
Problem Set #2: Solutions
1. (a) If X is irreducible and U = is open and not dense, let V be its closure. Then X
is the union of V U and Z = X U , which are proper and closed. For the converse,
suppose that every nonempty open set is d
Math 614, Fall 2010
Problem Set #4: Solutions
1. (a) Clearly, (v, u) is in the kernel. If r1 u+r2 v = 0 then u|(r2 v). Since GCD(u, v) = 1,
this implies u|r2 , say r2 = ur. Then r1 u+ruv = 0, u = 0, and r1 +rv = 0, so that r1 = rv.
But then (r1 , r2 ) = r
Math 614, Fall 2010
Problem Set #5: Solutions
1. By class results, Supp(M R N ) = Supp(M ) Supp(N ) = V (I) V (J) = V (I + J).
If r I (resp., J), the map M M (resp., N N ) induced by multiplication by r is 0,
and it follows that the map induced by multipl
Math 614, Fall 2008
Supplementary Problem Set #5: Solutions
1. Compare each of two sets of generators with their union. It thus suces to do the case
where one, u1 , . . . , un , is contained in the other, u1 , . . . , un+k . By induction on k, we may
n
as
Math 614, Fall 2008
Supplementary Problem Set #4: Solutions
1. If u frac(T ) then u frac(Si ) for all i. If u is integral over T , it is integral over every
Si , and therefore in every Si . But then u T .
2. This ring is not Noetherian. Consider the decre
Math 614, Fall 2010
Problem Set #1: Solutions
1. (a) Use induction on deg(g) where g is to be expressed in this way. If deg(g) d 1,
then we have g = a0 . If not, the form of the expression shows that a0 is the remainder when
g is divided by f in the divis