Math 310: Problem set 7
Instructions: This problem set is due
Thursday, November 2, 2006.
Your goal is not only
to give correct answers but to communicate
your ideas well. Make sure you use good English.
- Let R be a commutative ring such that 1 and 0 are not equal.
If r is a unit in R, then show that r is not a 0-divisor:
Since r is a unit, there is an s in R such that rs=1.
But if r were a 0-divisor, then there would be a non-zero
t such that rt=0. Hence 0=s0=srt=1t=t, contrary to hypothesis.
- Let b and c be integers bigger than 1, and let m=bc.
Define the function f:Z/bZ -> Z/mZ
by f([x]b) = [cx]m.
- Show that this function is well-defined; i.e., if
[x]b=[y]b, show that
Assume that [x]b=[y]b. Thus
b divides x-y. Therefore m=bc divides c(x-y)=cx - cy,
- Show that this is a homomorphism of groups
if we regard Z/bZ and Z/mZ
as groups under addition: f([x]b+[x']b)
= f([x+x']b) = [c(x+x')]m
= [cx+cx']m = [cx]m + [cx']m
=f([x]b) + f[x']b), as we needed to show.
- Determine whether f is also a homomorphism of rings.
Justify why or why not: It is not a homomorphism of rings.
If it were, then f(b) = m,
but in fact f(b) = [c]m, and
m is not equal to [c]m
since m does not divide c-1 (since c > 1 but c < m).
- Let s denote the square root of 3, hence about 1.732...
Let Q[s] denote the set of all real numbers of the form
h(s), where h is a polynomial with rational coefficients (i.e., h is an element of Q[x]). Note that h(s) simplifies to a number of the form
us+v, where u and v are rational.
- Is Q[s] a ring? Justify: It is a ring.
As shown in class, since the set Q[s] is contained in a ring
(ther eals in this case), we only need to verify
that Q[s] is closed under addition and multiplication,
and that 1, 0 and -1 are all in Q[s] (which they are since
1=0s+1, -1=0s+(-1) and 0=0s+0).
If r and t are elements of Q[s], then r= us+v for some
rationals u and v, and t=u's+v' for some rationals u' and v'.
Thus r+t = (u+u')s+(t+t') which is in Q[s],
while rt = (us+v)(u's+v') = (uv'+vu')s + (3uu'+vv'),
which is also in Q[s]. Thus Q[s] is closed under addition
and multiplication. (It is also closed under taking additive inverses
since it is closed under multiplication: -r = (-1)r, for example.
Also, Q[s] has 1 and 0 in it and the other properties of rings hold,
such as associativity of addition and multiplication, and the distributive
law, since Q[s] is a subset of the reals, which is a ring.)
- Is Q[s] a field? Justify: We know Q[s]
is a subring of the reals, which is a field.
Thus every non-zero element of Q[s] has a multiplicative
inverse as a real number; we must show that the inverse is in Q[s].
But if r is in Q[s] and is not 0, then r = us+v
for some rationals u and v, where either u or v is not 0.
Now 1/r = 1/(us+v) = (us-v)/[(us+v)(us-v)] = (us-v)/(3u2-v2)
Since (u/(3u2-v2)) and v/(3u2-v2)
are rational, 1/r is in Q[s]. Thus Q[s] is a field.
- Are Z/10Z and Z/2Z x Z/5Z isomorphic as rings? Why or why not? Yes, they are isomorphic as rings.
In fact there is a unique isomorphism. If f:Z/10Z and Z/2Z x Z/5Z is an isomorphism, then by definition
f(10) must be the multiplicative identity in
Z/2Z x Z/5Z, which is
(2, 5). If f is a homomorphism, then
it must take sums to sums; i.e., f(10)=
=2(f(10)) = 2(2, 5)
=(2, 5). Similarly,
for all integers i. This means that the only possible homomorphism
f is forced on us: f([i]10)
has to be ([i]2, [i]5).
The only question is if this is a homomorphism, and if so, is it an isomorphism.
First, it is well-defined: if [i]10=[j]10,
then f([i]10)=f([j]10). [This is because
[i]10=[j]10 means that 10 divides i-j,
hence 2 and 5 both divide i-j, so [i]2=[j]2
and [i]5=[j]5, so
([i]2, [i]5)=([i]2, [i]5).]
Next it preserves addition: f([m]10+[n]10)
=([m]2, [m]5)+([n]2, [n]5)
=f([m]10)+f([n]10). Similarly, it preserves multiplication
(just use the same argument with multiplication in place of addition).
Finally, we have f(10)=(2, 5)
by the way we defined f. Thus f is a homomorphism of rings.
It is injective since if f([m]10)=f([n]10),
this means ([m]2, [m]5)=([n]2, [n]5),
hence [m]2=[n]2 (so 2 | m-n)
and [m]5=[n]5 (so 5 | m-n). Since
2 and 5 both divide m-n, so does 10 (since 2 and 5 are relatively prime,
by Corollary 3 on page 33). Thus [m]10=[n]10,
which means f is injective. But since f is injective
and since Z/10Z and Z/2Z x Z/5Z
both have 10 elements, f must be surjective and hence bijective.
Thus f is an isomorphism.
- Are Z/9Z and Z/3Z x Z/3Z isomorphic as rings? Why or why not? No, they are not isomorphic as rings.
If they were, there would be a unique homorphism, as in the preceding problem;
i.e., the isomorphism would have to be f:Z/9Z -> Z/3Z x Z/3Z, where f([i]9)=([i]3, [i]3)
for every integer i. But that means f(9)=(3, 3)=(3, 3)=f(9).
Thus f cannot be injective, so there is no isomorphism.
[In fact, they are not even isomorphic as additive groups!
But this wasn't part of the problem.]