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

In summary: Similarly, Zx{0} = {(a,b) | a is an integer and b = 0}. This should help you see what ZxZ/M is.In summary, to prove that M is a maximal ideal in Z x Z, one can either show that whenever there is an ideal J containing M, then either M equals J or J equals the entire ring R, or one can show that the quotient ring R/M is a field. Using the given information, it is easier to show that R/M is a field. To construct R/M, one can use the set of all cosets of M (congruence classes modulo M). Since p is a fixed prime integer, Z/M is isomorphic to a
  • #1
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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

Homework Equations





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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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
  • #5
To clarify what I mean, consider that {0}xZ = {(a,b) | a = 0 and b is an integer}.
 

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

1. What is abstract algebra?

Abstract algebra is a branch of mathematics that studies algebraic structures such as groups, rings, and fields. It focuses on the inherent structure and properties of these structures rather than specific numbers or calculations.

2. What is an ideal in abstract algebra?

In abstract algebra, an ideal is a subset of a ring that is closed under addition, subtraction, and multiplication by elements of the ring. It is a generalization of the concept of a multiple in integer arithmetic.

3. How is an ideal proven to be maximal?

To prove that an ideal is maximal, one must show that it is not properly contained in any other ideal of the ring. This can be done by assuming the existence of a larger ideal and arriving at a contradiction, thus showing that the original ideal is indeed maximal.

4. How is a quotient ring constructed?

A quotient ring is constructed by taking a ring and an ideal of that ring, and then forming a new ring whose elements are cosets of the ideal. This new ring inherits its operations and properties from the original ring, but with some elements identified as equivalent.

5. What is the significance of proving an ideal to be maximal?

Proving an ideal to be maximal is important because it allows for the construction of a quotient ring, which is a useful tool in abstract algebra. Additionally, maximal ideals have important applications in other areas of mathematics, such as algebraic geometry and commutative algebra.

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