Im studying for my final next week and I need a little help.

"+ denotes an external direct product"

1. Find the number of elements of order 4 in Z_60+Z_12+Z_12.
2. In the ring Z+Z show: A = {(3x,y)|x,y in Z} is a maximal ideal.

For number 1 the LCM(60,12)=60 then is that the number of elements of order 4?

For number 2, I need some serious help starting.

Thanks Guys

2. ## easier way to look at 1

Consider the fact that $\mathbb{Z}_{60}\cong\mathbb{Z}_{5}\oplus\mathbb{Z} _{12}$ and $\mathbb{Z}_{12}\cong\mathbb{Z}_3\oplus\mathbb{Z}_4$.

Applying these relations we have the following:
$\mathbb{Z}_{60}\oplus\mathbb{Z}_{12}\oplus\mathbb{ Z}_{12}\cong\mathbb{Z}_3\oplus\mathbb{Z}_3\oplus\m athbb{Z}_3\oplus\mathbb{Z}_4\oplus\mathbb{Z}_4\opl us\mathbb{Z}_4\oplus\mathbb{Z}_5$

It is clear that since 4 does not divide 3 or 5 there will be no elements of order 4 in $\mathbb{Z}_3$ or $\mathbb{Z}_5$ respectively (Lagrange). Furthermore, it is obvious that 1 and 3 have order 4 in $\mathbb{Z}_4$. Thus any combination of these elements in the 3 copies of $\mathbb{Z}_4$ will work so long as you choose 0 in the rest of the places. Note obviously not ALL of the componants can be 0 or else it is the identity and has order 1. I claim that there will be $3*3*3-1$ elements, you oughta be able to prove why there are no more and no less.

3. ## Problem 2

Certainly $\mathbb{Z}$ is a Euclidean Domain which can be shown to be a Principal Ideal Domain. In a PID every nonzero prime ideal is maximal. It is pretty clear that the second coordinate is all of $\mathbb{Z}$, so really this just amounts to showing that the first coordinate is maximal (I think anyway). The first coordinate is simply the principal ideal generated by (3) which is a prime ideal, so it is also maximal. Thus any ideal containing A would have to contain all of $\mathbb{Z}$ in the second coordinate, and would have to contain (3) in the first coordinate, but since (3) is maximal in $\mathbb{Z}$ this means any ideal satisfying this is necessarily all of $\mathbb{Z}$ or else (3) again, and so if B were an ideal containing A then either $B=(3)\oplus\mathbb{Z}$ or $B=\mathbb{Z}\oplus\mathbb{Z}$ and so A is maximal.

Not sure where exactly you got hung up on this one, but it is pretty easy to show both that E.D. implies PID, and then in a PID that prime ideals are maximal, just follow definitions, nothing too tricky about that, plus almost any algebra book would have proofs of these facts I should think.