# Ring/Field help.

• November 2nd 2009, 03:38 PM
Ring/Field help.
From what I understand, fields, as far as numbers Zeta are concerned, are zeta in some prime number. I can't seem to figure out why.

Also, an inverse of the product for any number in the ring is another number or itself that can be multiplied with it and yield 1. In a field, all the numbers have an inverse, but in zeta 6 (for example), which is not a field the only numbers that have an inverse are 1 and 5, in this case being themselves. Not sure I got everything down. What are these numbers called, btw? I am studying abroad in Spain, and in Spanish they are "unidades" but I don't know what they are in English.

Thanks.
• November 2nd 2009, 07:57 PM
tonio
Quote:

From what I understand, fields, as far as numbers Zeta are concerned, are zeta in some prime number. I can't seem to figure out why.

Also, an inverse of the product for any number in the ring is another number or itself that can be multiplied with it and yield 1. In a field, all the numbers have an inverse, but in zeta 6 (for example), which is not a field the only numbers that have an inverse are 1 and 5, in this case being themselves. Not sure I got everything down. What are these numbers called, btw? I am studying abroad in Spain, and in Spanish they are "unidades" but I don't know what they are in English.

Thanks.

Unidades, en castellano = units, in english. And indeed: the only two units in $\mathbb{Z}_6$ are the residue classes 1 and 5.

In a field every NON-ZERO elements has an inverse.

Of all the rings $\mathbb{Z}_n$ of residues classes modulo n, the only ones that are fields are those with n a prime number, since then for any non-zero element m of it, which has an integer representative $0\leq m , as (m,p) = 1, we can write by the Euclidean algorithm $am+bp=1 \,,\,\,a\,,\,b\in \mathbb{Z}\Longrightarrow am=1-bp$, and thus a (mod p) is the mult. inverse of p.

You better get a good algebra book and try to read this there since otherwise it'll be pretty hard for you to grab all this stuff.

Tonio
• November 2nd 2009, 11:37 PM
It's not required of me to understand why the only fields are of prime numbers in zeta though, so I think I am alright. But am I right about units of a non-field being multiplicate inverses?

Thanks.
• November 3rd 2009, 07:44 AM
tonio
Quote:

It's not required of me to understand why the only fields are of prime numbers in zeta though, so I think I am alright. But am I right about units of a non-field being multiplicate inverses?

Thanks.

You mean "...units of a non field HAVING multiplicative inverses"? Yes, this is right: a unit in a ring is an element u s.t. uv=1 for some element v in the ring.

Tonio