Question

What are rings? Examples?

Answer 1

Since I am not that good at Math,

Answer 2

best to google for a precise definition
examples would include the ring of integers, or the integers modulo some number
also rings of matrices

Answer 3

*

Answer 4

^Haven't seen that from TuringTest in a while!

Answer 5

I go to Openstudy when Google fails me.

Answer 6

* = Don't knw....)

Answer 7

do you know what a field is?

Answer 8

An answer to a question about why the the euler phi function is multiplicative: In general, if R and S are rings, then R×S is a ring. The units of R×S are the elements (r,s) with r a unit of R and s a unit of S.

Now if gcd(A,B)=1, then
Z/⟨AB⟩≅Z/⟨A⟩×Z/⟨B⟩
This is essentially the Chinese remainder theorem.

But the number of units in the ring Z/⟨n⟩ is ϕ(n). So the number of units in Z/⟨AB⟩ is ϕ(AB) and the number of units in Z/⟨A⟩×Z/⟨B⟩ is ϕ(A)ϕ(B)

Where the squares were 'expectation value' signs.

Answer 9

I really need to know Math, because this is the only field I know of:

Answer 10

You could turn Pi radians and see a completely different field, though!

Answer 11

what are the squares in the above?

Answer 12

\[\left \langle x\right \rangle\]

Answer 13

I don't really want a thorough definition- I won't understand the usefulness of it now- just what it means in this problem I need to know.

Answer 14

oh i see
ok what this means is that in \(\mathbb{Z}/n=\{0,1,...,n-1\}\) the number of units is \(\phi(n)\)
"unit" means an element that has a multiplicative inverse, and in \(\mathbb{Z}/n\) if \(x\) is a unit, then it is relatively prime to \(n\)

Answer 15

Z/x means 'integers up until, but not including, x', or 'coprime integers relative to x up until, but not including, x'?

Answer 16

first one
so \(\mathbb{Z}/5=\{0,1,2,3,4\}\)

Answer 17

with addition and multiplication performed mod 5

Answer 18

this structure makes it a ring
yes probably you can ignore that, although it provides a snap way to prove this, because it is true that \(\mathbb{Z}/n\times \mathbb{Z}/n\equiv\mathbb{Z}/mn\) if (m,n)=1

Answer 19

typo there sorry

Answer 20

\[\mathbb{Z}/m\times \mathbb{Z}/n\equiv\mathbb{Z}/mn\]

Answer 21

and the isomorphism is more or less trivial but an example will explain it
lets take \(m=5,n=4\)

Answer 22

is it clear what \(\mathbb{Z}_5\times \mathbb{Z}_4\) is ?

Answer 23

all ordered pairs where the first coordinate is an element of \(\mathbb{Z}_5\) and the second coordinate is an element of \(\mathbb{Z}_4\)

Answer 24

you add and multiply component wise

Answer 25

http://upload.wikimedia.org/math/2/1/4/21467e485cbafd1a308e388211257ad6.png

Answer 26

where of course the addition and multiplication are performed in their respective "rings" i.e mod 5 for the first coordinate and mod 4 for the second

Answer 27

oh no it is much simpler

Answer 28

\((1,2)+(2,3)=(3,1)\) in my example

Answer 29

you are multiplying or adding two ordered pairs, you end up with an ordered pair.

Answer 30

When you write Z/n, you use mod n for the 1st element?

Answer 31

yes that is the standard notation. \(\mathbb{Z}/n\) is usually written as \(\mathbb{Z}_n\)

Answer 32

and what this says is that if (m,n)=1 there is an isomorphism between the two rings

Answer 33

lets see how it works in my example
take an element of \(\mathbb{Z}_{20}\) say 13

Answer 34

what does it look like in \(\mathbb{Z}_5\times \mathbb{Z}_4\) in other words, what is the corresponding ordered pair?

Answer 35

the remainder when you divide 13 by 5 is 3, so first coordinate is 3
the remainder when you divide 13 by 4 is 1, so second coordinate is 1
therefore \(\phi^{-1}(13)=(3,1)\) and \(\phi((3,1))=13\)

Answer 36

this gives an isomorphism between the two rings,
which is another way of saying that they are essentially identical, they just look different

Answer 37

chinese remainder theorem says since (m,n)=1 we can always solve this, that is you can solve
\(x\equiv 3(\text{ mod }5 )\) and
\(x\equiv 1 (\text{ mod }4 )\) which means we know how to map \((3,1)\) to \(\mathbb{Z}_{20}\) by solving the above

Answer 38

although it is easer to go the other way

Answer 39

now the fact that \(\phi\) is multiplicative pops right out, because \(\phi(n)\) is the number of "units" (numbers relatively prime to) \(n\) in \(\mathbb{Z_n}\)
and the number of units is \(\mathbb{Z}_m\times \mathbb{Z_n}\) is \(\phi(m)\times \phi(n)\)
since if \((a,b)\) is a unit in \(\mathbb{Z}_m\times \mathbb{Z_n}\) then \(a\) is a unit in \(\mathbb{Z}_m\) and \(b\) is a unit in \(\mathbb{Z_n}\)

since \(\mathbb{Z}_m\times \mathbb{Z_n}\cong\mathbb{Z}_{mn}\)
you get taht \(\phi(m)\times \phi(n)=\phi(mn)\)

Answer 40

i think you can prove this without resorting to the word "ring" but it is going to require the chinese remainder theorem no matter what