• Dec 4th 2008, 05:44 PM
Luck of the Irish
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 :)
• Dec 4th 2008, 11:11 PM
Gamma
easier way to look at 1
Consider the fact that \$\displaystyle \mathbb{Z}_{60}\cong\mathbb{Z}_{5}\oplus\mathbb{Z} _{12}\$ and \$\displaystyle \mathbb{Z}_{12}\cong\mathbb{Z}_3\oplus\mathbb{Z}_4\$.

Applying these relations we have the following:
\$\displaystyle \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 \$\displaystyle \mathbb{Z}_3\$ or \$\displaystyle \mathbb{Z}_5\$ respectively (Lagrange). Furthermore, it is obvious that 1 and 3 have order 4 in \$\displaystyle \mathbb{Z}_4\$. Thus any combination of these elements in the 3 copies of \$\displaystyle \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 \$\displaystyle 3*3*3-1\$ elements, you oughta be able to prove why there are no more and no less.
• Dec 5th 2008, 12:30 AM
Gamma
Problem 2
Certainly \$\displaystyle \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 \$\displaystyle \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 \$\displaystyle \mathbb{Z}\$ in the second coordinate, and would have to contain (3) in the first coordinate, but since (3) is maximal in \$\displaystyle \mathbb{Z}\$ this means any ideal satisfying this is necessarily all of \$\displaystyle \mathbb{Z}\$ or else (3) again, and so if B were an ideal containing A then either \$\displaystyle B=(3)\oplus\mathbb{Z}\$ or \$\displaystyle 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.