## 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.
1. 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.
2. 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 [cx]m=[cy]m: Assume that [x]b=[y]b. Thus b divides x-y. Therefore m=bc divides c(x-y)=cx - cy, hence [cx]m=[cy]m.
• 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([1]b) = [1]m, but in fact f([1]b) = [c]m, and [1]m is not equal to [c]m since m does not divide c-1 (since c > 1 but c < m).
3. 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) = (u/(3u2-v2))s-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.
4. 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 of homomorphism f([1]10) must be the multiplicative identity in Z/2Z x Z/5Z, which is ([1]2, [1]5). If f is a homomorphism, then it must take sums to sums; i.e., f([2]10)= f([1]10+[1]10)=f([1]10)+f([1]10) =2(f([1]10)) = 2([1]2, [1]5) =([2]2, [2]5). Similarly, f([i]10)=([i]2, [i]5) 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) =f([m+n]10)=([m+n]2, [m+n]5) =([m]2+[n]2, [m]5+[n]5) =([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([1]10)=([1]2, [1]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.
5. 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([3]9)=([3]3, [3]3)=([0]3, [0]3)=f([0]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.]