- 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**/b**Z**->**Z**/m**Z**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**/b**Z**and**Z**/m**Z**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).

- Show that this function is well-defined; i.e., if
[x]
- 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)/(3u^{2}-v^{2}) = (u/(3u^{2}-v^{2}))s-v/(3u^{2}-v^{2}). Since (u/(3u^{2}-v^{2})) and v/(3u^{2}-v^{2}) are rational, 1/r is in**Q**[s]. Thus**Q**[s] is a field.

- Is
- Are
**Z**/10**Z**and**Z**/2**Z**x**Z**/5**Z**isomorphic as rings? Why or why not? Yes, they are isomorphic as rings. In fact there is a unique isomorphism. If f:**Z**/10**Z**and**Z**/2**Z**x**Z**/5**Z**is an isomorphism, then by definition of homomorphism f([1]_{10}) must be the multiplicative identity in**Z**/2**Z**x**Z**/5**Z**, 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**/10**Z**and**Z**/2**Z**x**Z**/5**Z**both have 10 elements, f must be surjective and hence bijective. Thus f is an isomorphism. - Are
**Z**/9**Z**and**Z**/3**Z**x**Z**/3**Z**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**/9**Z**->**Z**/3**Z**x**Z**/3**Z**, 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.]