Math 614, Fall 2008 Problem Set #2: Solutions
1. In the category of sets, the empty set is an initial object and a set with one element
is a final object. In the category of commutative rings with identity,Zis an initial object
(the unique map Z→Rsends nto n·1R) and the 0 ring is a final object. In the category
of modules over a ring R, the 0 module is both an initial and a final object.
2. f=hbecause, by the universal mapping property of localization, the map R→T
given by gf =hf factors uniquely through S(since f(W) consists of units of S,gf(W)
consists of units of T).
Alternatively, every element of Shas the form r/w where r∈Rand w∈W. Then
(gf )(w)g(r/w) = g(w/1)g(r/w) = g(r/1) = (gf )(r)=(hf)(r) = h(r/1) = h(w/1)h(r/w) =
(hf)(w)h(r/w). Since (gf)(w)=(hf )(w), we can conclude that g(r/w) = h(r/w) provided
that (gf )(w) is invertible in T. But (gf)(w) = g(w/1) has inverse g(1/w), as required.
3. We have the localization homomorphism R→U−1Rthat sends r7→ r/1. Since
the image of Wconsists of units in U−1R, there is a unique R-algebra homomorphism
W−1R→U−1Rthat sends r/w to r/w (the two symbols have different meanings: in
the second instance, wis to be thought of as an element of U). Since the image of
e
V
consists of units in U−1Rwe have a unique homomorphism α:
e
V−1S∼
=U−1Rthat
sends (r/w)/(v/1) 7→ r/(vw). Similarly, since the image of U=V W consists of units
in
e
V−1S(the inverse of vw is (1/w)/v), we have a unique R-algebra homomorphism
β:U−1R→
e
V−1Sthat sends r/(vw)7→ (r/w)/(v/1). It is now obvious that α◦βis the
identity on U−1Rand that β◦αis the identity on
e
V−1S.
4. Let s=1 + √p
2. This element is integral over Z, since it is a root of x2−x−k= 0.
Clearly, sis in the fraction field of Z[√p], which is Q[√p]. Note that √p= 2s−1, so
Z[√p]⊆Z[s]. Since s2=s+k, the set Z+Zsis already closed under multiplication, and
so Z[s] = Z+Zs. Now suppose that a+b√pis integral over Z, where a, b ∈Q. We shall
show that a+b√p∈Z. If b= 0, this follows because Zis a PID, hence, a UFD, and is
normal. If b6= 0 the minimal monic polynomial of a+b√pover Qis (x−a)2−b2p= 0.
Since Zis normal, if a+b√5 is integral over Z, we must have that the coefficients are in
Z, i.e., that −2aand a2−b2pare in Z. Thus, we we may assume that a=n/2, where
n∈Z. By subtracting ns we obtain a new integral element of the form c√p, where cis
rational. To complete the proof, it suffices to show cmust be an integer. Then (c√p)2is
integral over Zand rational, and so must be an integer. Since pc2is an integer, we can
write c2=n/p. But when we write c=r/s is in lowest terms, r, s ∈Z,s > 0, r2/s2is also
in lowest terms, which implies that s2divides p. Thus, s= 1, as required.
5. (a) Let A=R−Qand B=R−Q0. Then a prime is contained in both Pand Q
if and only if it is disjoint from both Aand B, and this holds if and only if it is disjoint
from AB, since the complement of a prime is a multiplicative system. Thus, there exists
a prime contained in both Qand Q0if and only if AB does not contain the element 0,
i.e., if and only if there do not exist a∈R−Qand b∈R−Q0such that ab = 0. If
there is no such prime, then with a∈Aand b∈Bsuch that ab = 0, we have that
