Download Math 614, Fall 2015 Problem Set #1: Solutions 1. (a) Since every

Math 614, Fall 2015
Problem Set #1: Solutions
1. (a) Since every polynomial has unique remainder in Zx + Z upon division by X 2 − 3,
the quotient by (X 2 − 3) has the form Z + Zx where x is the image of X and x2 = 3.
The ideal generated by 2x in this ring contains all even multiples of x and all multiples of
2x2 = 6. The quotient is isomorphic as an abelian group with Z/6Z⊕Z/2Z with respective
generators that are the images of 1 and x. It has 12 elements.
(b) Since Z ⊆ A, A = Z[x2 , 2x] which is spanned over Z by the even powers of x, (x2 )k ,
and the even multiples of the odd powers of x, 2x(x2 )k . These are clearly in A, and sums
of these are closed under multiplication. The quotient is a direct sum of copies of Z/2Z
with the images of the odd powers of x as generators.
2. For all i ∈ N, mi = Ker (R → Ki ) is maximal (here, r to the i th element in its
sequence). We shall show Spec (R) = {mi : i ∈ N}. For i ≥ 1, let ei ∈ R be 0 on every
integer except i, and have value 1 on i. Since ei (1 − ei ) = 0, every prime ideal contains
ei or fi = 1 − ei . Every element g of the ring that vanishes at i for i ≥ 1 is a multiple
of fi : in fact, g = fi g. Hence fi generates the maximal ideal of elements that vanish at
i, which is mi . Thus, any prime P of R that is different from all the mi for i ≥ 1 must
contain all the ei for i ≥ 1. We claim that the ei generate m0 , and so P = m0 . Suppose
that g ∈ P0 . Since it vanishes at 0, it vanishes except for finitely many positive integers
i1 , . . . , it . The g = g(ei1 + · · · + eit ). Thus, the mi are all the primes of R, and all prime
ideals of R are maximal. Note that D(ei ) is {mi } and is open, so that every subset of
{mi : i ≥ 1} is open. The topology is therefore determined if we know which open sets
contain m0 . If D(g) contains m0 , g does not vanish at the origin, so the complement of
D(g) is finite. Thus the open sets are all subsets of {mi : i ≥ 1} and complements of finite
subsets of {mi : i ≥ 1}. (From this one sees, if Z+ is the set of positive integers, the map
between Spec (R) and {0} ∪ {1/i : iZ+ } ⊆ R such that m0 maps to 0 and mi to 1/i is a
homeomorphism.
3. (a) If f (x) 6= 0 f is nonzero on an open neighborhood of x, and has an inverse on that
neighborhood. Hence, every proper ideal consists of germs of functions that vanish at x
(vanishing at x is well-defined on equivalence classes). Thus, the functions that vanish at
x give the unique maximal ideal m. Evaluation at x is well-defined on equivalence classes,
and gives a surjection T R whose kernel is m. Hence, T /m ≡ R.
t
(b) By translation, we may assume the point is the origin. Let ft (x1 , . . . , xn ) = |x1 |1/2 .
2
Then ft = ft+1
, so (ft ) ⊆ (ft+1 ). Since ft+1 /ft → ∞ as x1 → 0, this is strict. 4. Call the family of closed sets {V (Iλ ) : λ ∈ Λ} and the open sets {D(fµ ) : µ ∈ M }.
The condition for V (Iλ1 ), . . . , V (Iλs ) and D(fµ1 ), . . . , D(fµt ) to all intersect is that there
is
Pa prime containing all the Iλj and not containing any of the D(fk ). This implies that
I is disjoint from the multiplicative system generated by the fµk . But then the ideal
j=1
Pj
I = λ∈Λ Iλ is disjoint from the multiplicative system W generated by all the fµ : each
element of I is in a sum of finitely many Iλ , and each element of W is a product of powers
of finitely many fµ . By a class theorem, there is a prime containing I and disjoint from
W , and this prime is in the intersection of all the sets. 5. Let A = R − P and B = R − Q. The multiplicative system AB = {ab : a ∈ A, b ∈ B} is
generated by A and B. A prime is contained in P and Q iff it is disjoint from both A and
B, i.e., iff both A and B are in its complement, which is equivalent to the condition that
the prime be disjoint from AB. Hence, there exists a prime contained both P and Q unless
0 ∈ AB. But if a ∈ A and b ∈ B are such that ab = 0, the D(a)∩D(b) = D(ab) = D(0) = ∅,
and P ∈ D(a) while Q ∈ D(b), so that P and Q have disjoint open neighborhoods.
(Alternative: apply 4. to all open sets that are neighborhoods of P or neighborhoods of
Q. If P and Q don’t have disjoint open neighborhoods, these have the FIP. A point in the
intersection of all corresponds to a prime P0 such that both P and Q are in the closure of
{P0 }, i.e. P0 ⊆ P ∩ Q.)
6. Call the K-subalgebra S. Then K ⊆ S ⊆ R. If S = K the result is clear. If not, let
f ∈ S with deg(f ) = d > 0, so that D = K[f ] ⊆ S ⊆ R. f is transcendental over K, i.e.,
K[f ] is isomorphic to a polynomial ring in one variable over K. Hence, D is a principal
ideal domain. Note that D ⊆ S ⊆ K[x]. The images of 1, x, . . . , xd−1 span K[x] over D.
Let T be the D-span of 1, x, . . . , xd−1 . To prove D = R, use induction on degree s of an
element g of K[x]. If s < d, then g is in the K-span of 1, x, . . . , xd−1 . If s ≥ d, we have
s = qf +ρ, where ρ is in the K-span of 1, x, . . . , xd−1 , and, by induction, since deg(q) < s,
q ∈ T . Multiplying by f shows that qf ∈ T , and so we are done. (Alternatively, f may
be chosen monic, and x satisfies the monic polynomial f (X) − f (x) = 0 with coefficients
in D, so that x is integral over D and R = D[x] is module-finite over D.) Since S is a
D-submodule of R, it is also finitely generated as a module. (In fact, a finitely generated
torsion-free module over a PID is free, and every submodule is free with at most the same
number of generators.) Then f together with the finite number of D-module generators of
S will generate S as a K-algebra. EC1 R is not Noetherian. If n ≥ 1, the product of any n consecutive
positive integers,
k+n−1
say k, k + 1, . . . , k + n − 1, is divisible by n! (the quotient is
). Hence, the same
n
is true for the product of any n consecutive negative integers. Thus, if n ≥ 1 and fn =
x(x − 1) · · · (x − n + 1)/n!,
S∞ fn maps Z to Z (if a factor in the numerator is 0). Let In =
(f1 , . . . , fn )R and I = n=1 In . If I is finitely generated, I = In for some n 0. Thus, it
Pp−1
suffices to show that if p is any odd prime, fp ∈
/ Ip−1 . Suppose (∗) fp = j=1 gj fj with
Pp−1
2p
all gj ∈ R. Evaluate both sides at 2p to get (∗∗) 2p
j=1 gj (p) j . p occurs as a
p =
factor twice in (2p)! (in 2p and in p) and twice in (p!)2 . Hence, p 6 | 2p
p . Every gj (p) ∈ Z,
since gj ∈ R. But p| 2p
for 1 ≤ p − 1, since it occurs as a factor in the numerator
j
(2p)(2p − 1) · · · (2p − j − 1) while it does not occur
as a factor in the denominator j!. This
2p
is a contradiction, since (∗∗) then implies p| p .
EC2 None of the rings described is finitely generated, and none is Noetherian: the ideal
mr generated by all of the xa y b for b/a < r is not finitely generated, and neither is the
ring Sr . If Sr were finitely generated, the generators could be expressed as polynomials
in finitely many of the monomials xai y bi , 1 ≤ i ≤ h, with bi /ai < r. Choose s < r larger
than the maximum of the bi /ai . Note that if b ≤ sa and b0 ≤ sa0 then (b + b0 ) < s(a + a0 ).
Thus, the ring generated by these monomials is contained in Ss , and is strictly smaller
than Sr . (This was not part of the problem, but if mr were finitely generated, one could
find finitely many generators that are monomials. One can the show by induction on the
degree that every element of Sr is in the ring generated by these monomials, which we
have already shown is impossible.)