Commutative algebra, Homework 2
(1) (a) Let y1 = y(2,0) , y2 = y(1,1) and y3 = y(0,2) . Clearly y1 y3 − y22 is in ker(ϕ), hence
(y1 y3 − y22 ) ⊆ ker(ϕ). Now let f (y1 , y2 , y3 ) be in ker(ϕ). We simplify f by subtracting
elements in (y1 y3 − y22 ) until f is 0, which will show ker(ϕ) = (y1 y3 − y22 ). Note that
all relations with integer coefficients among v1 = (2, 0), v2 = (1, 1) and v3 = (0, 2) in
N2 are obtained by multiplying v1 + v3 = 2v2 by an integer.
Namely, in order for f 6= 0 to be in ker(ϕ), there have to be two distinct monomials
y1s1 y2s2 y3s3 and y1t1 y2t2 y3t3 occurring in f that have the same image under ϕ. This means
that s1 v1 + s2 v2 + s3 v3 = t1 v1 + t2 v2 + t3 v3 . Cancelling terms until max(si , ti ) = 0
for i = 1, 2, 3 we see that the result comes from the relation kv1 + kv3 = 2kv2 for
some k > 0. As cancelling terms is the same as cancelling factors yi , we see that
y1s1 y2s2 y3s3 − y1t1 y2t2 y3t3 contains some factor y1k y3k − y22k , which contains a factor y1 y3 − y22 ,
hence is in (y1 y3 − y22 ). Therefore we can reduce the number of monomials occurring
in f by subtracting elements of this ideal, as claimed.
P
P
(b) If A is linearly dependent over Q, then we have some relation α∈I nα α = β∈J nβ β
in Nn , where all nα and nβ are positive integers, and I, J ⊆ A are disjoint and nonQ
Q
n
empty. Then α∈I yαnα − β∈J yβ β is in ker(ϕ). Letting zγ for γ in A be a variable,
Q
Q
n
we see that α∈I zαnα − β∈J zβ β is a non-zero polynomial in K[{zα }α ] because I and
J are disjoint and non-empty, and all nα , nβ are positive (so at least one zγ occurs
in one term but not in the other). Replacing zγ with ϕ(yγ ) then gives a non-trivial
polynomial relation among the ϕ(yγ ) with γ in I ∪ J.
(c) Assume that β1 , . . . , βm in A for m ≥ 0 are linearly independent in Qn . Suppose
that the ϕ(yβi ) are algebraically dependent, so let f (z1 , . . . , zm ) in K[z1 , . . . , zm ] be
non-zero such that f (ϕ(yβ1 ), . . . , ϕ(yβm )) = 0. Then certainly two distinct monomials
sm and z t1 . . . z tm that occur in f must have the same image in K[x , . . . , x ]
z1s1 . . . zm
1
n
m
1
Pm
P
when we replace zγ by ϕ(yγ ) = xγ . Therefore m
i=1 ti βi . Since for some
i=1 si βi =
j, si 6= tj , this is a contradiction.
By (b) and Proposition 5.10, it follows that the Q-dimension of the Q-span of A in Qn
equals the transcendence degree of im(ϕ). By Theorem 5.9 this also equals the Krull
dimension of im(ϕ).
(2) For A ⊆ I, let ψA be the function with ψA (i) = 0 if i is in A and 1 otherwise. Note that
the association f 7→ Zf gives a bijection between R∗ -orbits in R and subsets of I, the orbit
corresponding to A ⊆ I being R∗ ψA . So if we show that a union of such R∗ -orbits is an ideal
of R not equal to R precisely when the collection of corresponding As has the properties
in (a) below, we automatically get a bijection between such ideals and such collections of
subsets of I.
(a) I is the zero locus of ψI , the 0-function, which is in Q. ∅ is the zero locus of the
R∗ -orbit of ψ∅ , which is the constant function 1; this is not in Q as Q 6= R. If A is in
ZQ then ψA is in Q, so if B ⊇ A then ψB = ψB ψA is in Q and B = ZψB is in ZQ . If
A and B are in ZQ then Q contains ψA and ψB , hence also ψA + ψB − ψA ψB = ψA∩B ,
so A ∩ B is in ZQ .
(b) Let Q = ∪A∈Z R∗ ψA . Note that, for r in R and A in Z, rψA = uψB for B = A ∪ Zr ,
which is in Z, and u in R∗ . So it is clear that Q contains the 0-function (because
Z contains at least one element, namely I) and that it is closed under multiplication
by R. Now take uψA and vψB in Q where u and v are in R∗ , A and B are in Z. Then
uψA − vψB vanishes on A ∩ B, so is of the form rψA∩B for some r in R, hence lies in Q.
(c) Suppose Q is a prime ideal. If A ⊆ I then ψA ψI\A is the zero function, which lies in
Q, hence either ψA or ψI\A is in Q. Therefore A or I \ A must lie in ZQ .
Conversely, suppose that Q is not a prime ideal. Multiplying by elements in R∗ , there
are subsets A and B of I such that ψA ψB = ψA∪B is in Q but neither ψA nor ψB are
in Q. Then I \ A cannot be in ZQ : if it were then (A ∪ B) ∩ (I \ A) = B \ A ∩ B would
be in ZQ , and as it is contained in B, B would be in ZQ , contradiction.
(d) Suppose that we have ideals J1 ⊆ J2 ( R. By (b) and (a) this gives ZJ1 ⊆ ZJ2 (the
inclusion is clear as we can only get more functions). If J1 is prime but there exists A
in ZJ2 \ ZJ1 , then I \ A must be in ZJ1 by (c), hence ZJ2 must contain A ∩ (I \ A) = ∅,
which is not possible. Therefore any prime ideal of R is maximal, and R has Krull
dimension 0.
(3) (a) We have morphisms of F -algebras F [x, y] → F [x] given by h(x, y) 7→ h(x, 0), and
F [x, y] → F [y] given by h(x, y) 7→ h(0, y). Therefore we have a morphism of F -algebras
F [x, y] → F [x] × F [y] given by h(x, y) 7→ (h(x, 0), h(0, y)). Clearly the image is contained in A. The image equals A because it maps c + xh1 (x) + yh2 (y) + (xy) to
(c + xh1 (x), c + yh2 (y)), and every element of A is of this form for suitable h1 (x) in
F [x], h2 (y) in F [y]. As for the kernel, it is clear that xy is in it, hence the kernel contains (xy). If h(x, y) is in the kernel, then h(0, y) = h(x, 0) = 0. Using division with
remainder by x, h(x, y) = q(x, y)x + r(y), so r(y) = h(0, y) = 0 and h(x, y) = q(x, y)x.
Using division with remainder on q(x, y) we write q(x, y) = q1 (x, y)y + r1 (x), so that
0 = h(0, y) = r1 (x)x = 0. It follows that r1 (x) = 0, and h(x, y) is in (xy). By the first
isomorphism theorem, we obtain an F -algebra isomorphism R → A, given by mapping
h(x, y) + (xy) to (h(x, 0), h(0, y)).
We compute the zero divisors in A. Suppose (f1 (x), g1 (y)) is a zero divisor. Then
(f1 (x), g1 (y))(f2 (x), g2 (y)) = (0, 0) for some non-zero (f2 (x), g2 (y)). If f2 (x) 6= 0 then
f1 (x) = 0, and if g2 (y) 6= 0 then g1 (x) = 0. Conversely, if at least one of f1 (x) and
g1 (y) is zero then clearly (f1 (x), g1 (y)) is a zero divisor. So the set of zero divisors of
A is
{(xh1 (x), 0) with h1 (x) in F [x]} ∪ {(0, yh2 (y)) with h2 (y) in F [y]} ,
and the set of zero divisors of R is what corresponds to this under the isomorphism.
Explicitly, {xh1 (x) + (xy) with h1 (x) in F [x]} ∪ {yh2 (y) + (xy) with h2 (y) in F [y]}.
(b) The isomorphism in (a) induces an isomorphism Ra → Az as z = (x, 0) the image of a
in A. The projection A → F [x] to the first coordinate is an F -algebra homomorphism,
and it induces a map Az → F [x]x as it maps z to x. This map is an isomorphism.
(x),0)
(x)
Namely, it is surjective as for n ≥ 0, (xfz n+1
is in Az and maps to xf
= fx(x)
n . It is
xn+1
injective as well: (f (x),g(y))
maps to fx(x)
n , so this is zero if and only if f (x) is 0. But
zn
(f (x),g(y))
(xf (x),0·g(y))
(0,0)
= 1 .
then
=
zn
z n+1
(c) By (a), the set of non-zero divisors in A is U = {(f (x), g(y)) in A with f (x), g(y) 6= 0}.
Projection onto the first factor gives an F -algebra morphism A → F [x], hence by
combining with the inclusion F [x] ⊂ F (x), such a morphism A → F (x). As this maps
(x),g(y))
U to units in F (x), we obtain such a morphism U −1 A → F (x) mapping (f(f1 (x),g
to
1 (y))
f (x)
f1 (x) .
Using the second coordinate we obtain an F -algebra morphism U −1 A → F (y)
(f (x),g(y))
g(y)
−1 A → F (x) ×
(f1 (x),g1 (y)) to g1 (y) . Together they give such a morphism U
(x),g(y))
g(y)
F (y), mapping (f(f1 (x),g
to ( ff1(x)
(x) , g1 (y) ). We claim it is an isomorphism. Namely, if
1 (y))
(f (x),g(y))
(f1 (x),g1 (y)) is in the kernel, then f (x) = 0 and g(y) = 0, so the map is injective. For
mapping
f (x) in F [x] and g(y) in F [y], (xf (x), yg(y)) is in A, and if f1 (x) 6= 0, g1 (y) 6= 0, then
(xf (x),yg(y))
−1 A and maps to ( f (x) , g(y) ). So it is surjective.
(xf1 (x),yg1 (y)) is in U
f1 (x) g1 (y)
The isomorphism R → A of F -algebras gives such an isomorphism of the corresponding
total rings of fractions. Combining it with the isomorphism U −1 A → F (x) × F (y) we
have just constructed, we get an isomorphism of F algebras from the total ring of
7→ ( hh(x,0)
, h(0,y)) ).
fractions of R to F (x) × F (y), given by hh(x,y)+(xy)
1 (x,y)+(xy)
1 (x,0) h1 (0,y)
(d) Under the isomorphism in (a), P maps to Q = {(f, g) in F [x] × F [y]|f (0) = g(0) = 0},
and we get an isomorphism RP ' AQ . Note that A \ Q is contained in the nonzero divisors, so by Example 6.2(2), we can view AQ as a subring of the total ring of
(x),g(y))
fractions, which by (b) is isomorphic with F (x) × F (y). AQ has as elements (f(f1 (x),g
1 (y))
with f (0) = g(0) and f1 (0) = g1 (0) 6= 0. Under the isomorphism in (c) this element is
g(y)
mapped to ( ff1(x)
(x) , g1 (y) ), which lies in F [x](x) × F [y](y) ⊂ F (x) × F (y). But the image
consists precisely of those ( hh(x)
, k(y) ) with hh(0)
= kk(0)
: clearly an element in the
1 (x) k1 (y)
1 (0)
1 (0)
image is of this form, and conversely, scaling k(y) by F ∗ , we may assume h(0) = k(0),
so that h1 (0) = k1 (0), and then (h(h(x),k(y))
is in AQ and maps to this element.
1 (x),k1 (y))
(e) As in the previous part, the powers of a are not zero divisors, so we can transport everything to A under the isomorphism. Then we want to compute Az with
z = (αx, βy). Mapping Az to F (x) × F (y) we obtain all elements of the form
g(y)
α−n f (x) β −n g(y)
( αfn(x)
, yn ) with n ≥ 0 and f (0) = g(0). Those lie in F [x]x ×
xn , β n y n ) = (
xn
g(y)
F [y]y and in fact we get everything in that this way: if s, t ≥ 0 then ( fx(x)
s , yt ) =
s+t+1 t+1
( α (αx)xs+t+1f (x) , β
in A.
s+t+1 y s+1 g(y)
(βy)s+t+1
) is in the image as (αs+t+1 xt+1 f (x), β s+t+1 y s+1 g(y)) is
(4) We shall use that, for an R-module M and prime ideal P of R, m
1 6= 0 in MP if and only if
Ann(m) ⊆ P , and that Ann(M ) = ∩m∈M Ann(m).
(a) Note that ep = (0, . . . , 0, 1, 0, . . . ) with 1 in the position for p has annihilator (p). Hence
e
Ann(M ) ⊆ ∩p (p) = (0) and Ann(M ) = (0). Also, (p) lies in Supp(M ) as 1p 6= 0 in
n!m
0
M(p) . But M(0) is trivial: if m is in M then m
1 = n! = 1 for n 0 as m has finite
order under addition. Hence Supp(M ) = Spec(Z) \ {(0)}.
(b) Since M ⊆ N is a Z-submodule, Ann(N ) ⊆ Ann(M ), so Ann(N ) = (0) as well.
Because localizing is exact, we find Supp(M ) ⊆ Supp(N ). But Ann(m) = (0) for
0
m = (1, 1, . . . ), so m
1 6= 1 in M(0) . Therefore Supp(M ) = Spec(Z).