Abstract algebra: proving an ideal is maximal, Constructing quotient rings

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Homework Statement


M = {(pa,b) | a, b are integers and p is prime}
Prove that M is a maximal ideal in Z x Z

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The Attempt at a Solution



I know that there are two ways to prove an ideal is maximal:

You can show that, in the ring R, whenever J is an ideal such that M is contained by J, then M=J or J=R.

Or you can show that the quotient ring R/M is a field.

I think it will be much easier to show that R/M is a field, but I'm not familiar with how to construct it from the given information. My understanding is that it is the set of all cosets of M (congruence classes modulo M).

Can anyone point me in the right direction? Thanks.
 
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I assume that p is fixed in M?

Do you know what Z/pZ is isomorphic to, for a given prime p?

Do you know how to show that ZxZ / Zx{0} is isomorphic to Z?
 
P is fixed.

I know that Z/pZ is a field, for a given prime p. Is that what you mean?

To show that ZxZ /Zx{0} is isomorphic to Z, I'd need to create a bijective function from ZxZ / Zx{0} where f(a + b) = f(a) + f(b) and f(ab)= f(a)*f(b).

Are you hinting that since p is a prime integer, Z/M must be a field?
 
Take a look at what M is, when written in the form RxS for the sets R and S. You will see that both R and S are well known sets and that should help you figure out what ZxZ/M should look like.
 
To clarify what I mean, consider that {0}xZ = {(a,b) | a = 0 and b is an integer}.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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