Ring vs. Field: Comparing Two Common Sports Venues

[roni1]

What are the differences between ring and field?
It's for an article that I write.

 

[I like Serena]

What are the differences between ring and field?
It's for an article that I write.

Hi roni,

A field has more conditions/axioms than a ring.
In particular all elements in a field, except the zero-element ($0$), must have a multiplicative inverse.
And multiplication in a field must be commutative.
Depending on the text you have, a ring may need to have a multiplicative identity ($1$) or not. Either way, a field must have it, and it must be distinct from the zero-element ($0$).

So for instance $\mathbb Z$ is a ring but not a field, since for instance $2$ does not have a multiplicative inverse.
And $\mathbb Q$ is a field, since every non-zero element in it does have a multiplicative inverse.
In this example of a ring ($\mathbb Z$), multiplication happens to be commutative, and $1$ exists as part of the ring. That's why this example is called a commutative ring with identity.

 

[roni1]

Thanks...
I have a question:
Why the multiplicative indentity (1) distincit from the zero-element (0)?

 

[I like Serena]

I have a question:
Why the multiplicative indentity (1) distincit from the zero-element (0)?

Formally it is because it says so in the definition of a Field:

Additive and multiplicative identity: there exist two different elements 0 and 1 in F such that a + 0 = a and a · 1 = a.
​

The rationale is that a field is supposed to be a group with respect to multiplication - excluding the zero-element.
It can't be if 1 and 0 are the same element. After all, we already know that 0 does not have an inverse, which follows from the distributive properties.

The definition of a Ring does not have this restriction - assuming it does have a multiplicative identity. But then a ring is not a group with respect to multiplication to begin with.

 

[HOI]

If one drops the requirement that "\(\displaystyle 1\ne 0\)" what happens?

By the definition of 0, we have that x+ 0= x for all x. Since this a field we have the distributive law: y(x+ z)= yx+ yz. Taking z= 0, y(x+ 0)= yx+ y0. But x+ 0 = x so we have yx= yx+ y0. Since "cancelation" works in a field, y0= 0 for all y. But if 0= 1, the multiplicative identity, then y1= y. Together, those give y= 0 for all y. That is, if we allow the additive identity and multiplicative identities in a field to be the same element of the field, the field would have only a single member.

Some texts replace the condition that "\(\displaystyle 1\ne 0\)" in the definition of a field with "the field contains more that one element".