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

Click For Summary
SUMMARY

The discussion centers on proving that the ideal M = {(pa,b) | a, b are integers and p is prime} is a maximal ideal in the ring Z x Z. Participants highlight two methods for proving maximality: demonstrating that the quotient ring R/M is a field or showing that any ideal J containing M must equal M or R. The consensus is that constructing the quotient ring R/M is the more straightforward approach, with specific references to isomorphisms involving Z/pZ and Z x Z / Z x {0}.

PREREQUISITES
  • Understanding of ideals in ring theory
  • Familiarity with quotient rings and isomorphisms
  • Knowledge of prime numbers and their properties
  • Basic concepts of bijective functions in algebra
NEXT STEPS
  • Study the construction of quotient rings in abstract algebra
  • Learn about maximal ideals and their properties in ring theory
  • Explore isomorphisms involving Z/pZ and their implications
  • Investigate the relationship between ideals and fields in algebraic structures
USEFUL FOR

Students of abstract algebra, mathematicians focusing on ring theory, and anyone interested in understanding the properties of maximal ideals and quotient rings.

Rfields
Messages
3
Reaction score
0

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.
 
Physics news on Phys.org
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}.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
931
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
986
  • · Replies 31 ·
2
Replies
31
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K