Set theory. What does this mean?

[omoplata]

If \mathbb{Z} is the set of integers, what does \mathbb{Z/2Z} mean?

 

[henry_m]

It is the integers modulo 2.

\mathbb{Z} is the set of integers, and 2\mathbb{Z} is the set of even integers, and \mathbb{Z}/\mathbb{2Z} is the set of integers modulo the ideal of even integers.

The notation A/B crops up all over the place in many different contexts, so here's a general explanation of roughly what it means:

A is some structure (in this case a ring, but it could be a group, a Lie algebra, a topological space...) and B is some special substructure of A (or in some contexts sometimes an equivalence relation on A). A/B is what you get when you regard the members of B as being equivalent. The exact definition depends somewhat on context. So in this case, we regard all the even integers as equivalent and by extension all the odd integers as equivalent, leaving just two elements in \mathbb{Z}/(2\mathbb{Z}): the set of odd integers and the set of even integers (the equivalence classes).

More generally, we could take the set \mathbb{Z}/(n\mathbb{Z})=\{[0],[1],\ldots,[n-1]\}, where the square brackets mean 'equivalence class of', so [k] = \{\ldots, k-2n, k-n,k,k+n,k+2n,\ldots\} as the integers modulo n.

Hope that's useful; it's hard to know what level to pitch it at. If you want more details or less jargon I'll be happy to rephrase.

 

[micromass]

Hi omoplata!

That is actually a notation from abstract algebra. You probably know that 2\mathbb{Z} is all the multiples of 2 (thus all even numbers).

Now, on \mathbb{Z}, we can put the following equivalence relation:

x\sim y~\Leftrightarrow~x-y\in 2\mathbb{Z}

Then \mathbb{Z}/2\mathbb{Z} is simply the quotient of this equivalence relation (thus the set of all equivalence classes).

It is easy to see that

\mathbb{Z}/2\mathbb{Z}=\{2\mathbb{Z},2\mathbb{Z}+1\}

thus the set contains two elements. We can easily put an addition and multiplication on this set and this gives us a field with two elements.

I hope that is clear!

 

[omoplata]

I get it now. Thanks a lot.

 

[HallsofIvy]

Note that although henry_m and Micromass are using different words, they are saying the same thing!

 

[omoplata]

Note that although henry_m and Micromass are using different words, they are saying the same thing!

Well, yeah. I understood that.