What are the axioms of a ring and field??

I have to following axioms:

A1)

a+b \epsilon R, \forall a,b \epsilon R

A2)

a+b = b+a, \forall a,b \epsilon R

A3)

(a+b)+c = a+(b+c), \forall a,b,c \epsilon R

A4)

\exists 0 \epsilon R s.t, a+0 = a = 0+a, \forall a \epsilon R

A5)

\exists b \epsilon R s.t, a+b = 0 = b+a, \forall a \epsilon R

M1)

a \cdot b \epsilon R, \forall a,b \epsilon R

M2)

a \cdot b = b \cdot a, \forall a,b \epsilon R

M3)

(a \cdot b) \cdot c = a \cdot (b \cdot c), \forall a,b,c \epsilon R

M4)

\exists 1_{R} \epsilon R s.t, a \cdot 1_{R} = a = 1_{R} \cdot a, \forall a \epsilon R

M5)

\exists b \epsilon R s.t, a \cdot b = 1_{R} = b \cdot a, \forall a \epsilon R

D1)

a \cdot (b+c) = a \cdot b + a \cdot c, \forall a,b,c \epsilon R

D2)

(a+b) \cdot c = a \cdot c + b \cdot c, \forall a,b,c \epsilon R

If, a \cdot b = 0, then, a = 0, b = 0, or, a = b = 0, \forall a,b \epsilon R

I want to know which ones you need to CHECK to make sure something is a field of ring.

Wiki says you need A1) - A5), M1), M3), M4), D1) and D2) for a ring and for a field you need the previous axioms along with M2) and M5).

My problem is my lecture notes do not mention closure (which I think is incorrect) also neither does my textbook (Rings, Fields and Groups: An Introduction to Abstract Algebra - I think this again is wrong, but makes me doubt myself more) also in the textbook, it says that you need Z for a field. Is this true or does this just become true due to the presence of the other axioms??

Please remember I want the axioms, and therefore the things you must check to see if a set is a Ring/Field. If for example Z is true, but is just induced due to the presence of the other axioms, please state that.

Thanks for any help, rep will be given!

(P.S. Please excuse my LaTeX, I didn't put any spaces in!)

Reply 1

sonofdot

Z is the condition for a ring to be an integral domain. It can be deduced from the existence of a multiplicative inverse in fields (M5) - if ab=0 and

a \not= 0

, what is

a^{-1}ab

?

Obviously you do need closure for a ring, but I think having well-defined operations implies closure, so you don't always need to write it down.

Reply 2

adie_raz

Original post by sonofdot

Z is the condition for a ring to be an integral domain. It can be deduced from the existence of a multiplicative inverse in fields (M5) - if ab=0 and

a \not= 0

, what is

a^{-1}ab

?

Obviously you do need closure for a ring, but I think having well-defined operations implies closure, so you don't always need to write it down.

if ab=0 and

a \not= 0

, what is

a^{-1}ab

? 0

Thank you for your answer, but just to be clear are you saying that wiki is correct??

The link for a ring is:

http://en.wikipedia.org/wiki/Ring_(mathematics)

and for a field is:

http://en.wikipedia.org/wiki/Field_(mathematics)

Thanks :smile:

Reply 3

sonofdot

Original post by adie_raz

if ab=0 and

a \not= 0

, what is

a^{-1}ab

I think the article is right.

a^{-1} a b = b

so b=0 and Z can be deduced to be true.

Reply 4

adie_raz

Original post by sonofdot

I think the article is right.

a^{-1} a b = b

so b=0 and Z can be deduced to be true.

Thanks

Reply 5

SsEe

Original post by adie_raz

My problem is my lecture notes do not mention closure (which I think is incorrect) also neither does my textbook (Rings, Fields and Groups: An Introduction to Abstract Algebra - I think this again is wrong, but makes me doubt myself more) also in the textbook, it says that you need Z for a field. Is this true or does this just become true due to the presence of the other axioms?

For closure, you probably have the operations defined as functions RxR -> R. That ensures closure so it's not given as a numbered axiom. But you still need to check it. Eg, if you're showing that some set of matrices is a ring, you need to multiply two of them together to check the result still has the same form and hence show that multiplication really is a function RxR -> R.

Reply 6

adie_raz

Original post by SsEe

Thanks very much, indeed the operations are defined as RxR -> R :smile:

Reply 7

adie_raz

Another question on the same lines:

F* = F \ {0} correct??

This is supposed to be true, but due to axiom M5), surely this F* = F as all elements of F are already invertible...