Not every finite ring is a field. Finite rings with zero divisors are not fields.

Every finie field is a subring of an infinite field: if F is a fileld, it is a subfield of F(x), the field of rational functions over F. It is also a subfield of its own algebraic closure.

As integral domains can be embedded in a field (field of quotients), I was wondering whether one could generate a finite field of quotients from an infinite ring. For instance, consider an infinite integral domain R with 1 and a maximal ideal M such that R/M is a finite field. The question is the existence of such maximal ideal M in R satisfying the finiteness of R/M. Clearly, one such example is the ring of integers Z (infinite) with maximal ideal (p) , p being a prime element so that Z/(p) is a finite field. Are there other (both commutative and non-commutative) rings satisfying this ? But, yeah, this is not embedding. I believe I have heedlessly used this word. 

On a different note, in some infinite fields, one can find a finite subring (which is also a field), say Z/pZ  in the polynomial ring Z/pZ (x), which is an infinite field or finite Galois Fields in their algebraic closure. Does there exist infinite field where the subring is not a field? What are the necessary conditions for the existence of a finite subring and it being also a field in an infinite field?

In a local ring (say R), the set of non-units forms an ideal, say J. Also, it has a unique maximal ideal which is also J and, R/J is a division ring. So, if R is commutative, then R/J is a field (commutative division ring). Is this the improvisation? If so, then how do the non-units still exist in a field?

Hello Sanjeet,

 in a  field, everything but zero is invertible so is a unit.

In an integral domain (commutative ring without any zero-divisor), some elements are not units. They become unit in the fraction ring  (or the  quotient ring, not to be confused with the quotient by an ideal).

Still, we wonder what you mean by "improvisation" in this context.

Best Regards,

Gabriel

Dear Maisnam, I will answer your last question in your second answer:

When you take the quotient ring R/J, with J maximal, there are no non-units. you can view this by two ways: 

First, R/J is a field ( it is a theorem, also, it is easy to reprove it ).

Second, you have already take all non-units in J, so no non-unit will "descend" to the quotient.

Also, I have a claim: First you have to prove that J formed by all non-units is a maximal ideal in R, and the only one if you want your ring to be locally.

Dear Dr. Breuer, 

With due respect, I would like to state that the first paragraph did not mean "yeah, oops. Sorry". In the second paragraph, I was asking if there was any necessary condition for the existence of a finite subring in an infinite field. Thereafter, I added that if, in case, such subring existed, was there any condition for it being a field too? Of course, the reals have integers as subrings which are not a field. Besides, this subring is not finite. So, how do you ascertain the existence of a finite subring in an infinite field? Anyway, thank you for your answer.

And, by the way, for the definition and properties of local rings, you can refer to "David S. Dummit, Richard M. Foote - Abstract Algebra" instead of Wikipedia.

Dear Dr. Thomas,

I was referring to the conditions one could insert in such rings to enhance their structures.

Dear Dr. Zekraoui,

Many thanks for your answer.

Consider Z there is a exact sequence 0------pZ-------Z-------Z/pZ~Z_p

1. A subring (with 1) of a field is an integral domain.

2. A finite integral domain is a field.

3. Therefore a finite subring of a field is a finite field.

Proof:

1 and 3 are self evident.

2 is true because the nonzero elements of an integral domain form a monoid with cancellation. A finite monoid with cancellation is a group, because cancellation means that the left and right multiplications are injective, and finiteness implies they are also surjective, hence invertible.

The same reasoning applies without the assumption of commutativity: a finite subring of a division ring is a finite ring without zero divisors, hence a finite division ring, and by Wedderburn's theorem it is a field.

Dear Maisnam, 

I totally agree with you in this remark:  " how do you ascertain the existence of a finite subring in an infinite field? "

Every finite commutative ring is a field, so it has non- zero characteristic . It can not be a subring of an infinite field.

Not every finite ring is a field. Finite rings with zero divisors are not fields.

Every finie field is a subring of an infinite field: if F is a fileld, it is a subfield of F(x), the field of rational functions over F. It is also a subfield of its own algebraic closure.

Yes, you have right, I forgot to say without zero-divisors (i.e. integral domain)

Also, yes you have right for the field of fractions of F[X], even F is a finite field, F[X] is not and so as, F(X) its fraction field.

1. Consider an infinite ring. Can it be embedded into a finite field? Can it have a finite quotient field?

No infinite set can be embedded in a finite set. But all infinite rings with 1 have quotient fields, because they have maximal ideals. Sometimes those fields are finite, as in case Z/pZ.

2. Consider, on the other hand, an infinite field. Does it always contain infinite subrings? 

First at all, every infinite field is itself an infinite ring. But if you want proper subsets, then consider the following cases: If the field has characteristic 0, then Z is always an infinite subring. If it has characteristic p and contains at least one element x which is transcendental over Fp, then Fp[x] is such a subring. If it has characteristic p and contains only Elements which are algebraic over Fp, then it does not contain subrings which are not fields. However, it contains infinitely many proper infinite subfields, which are automatically subrings.

3. Can we improvise a local ring to generate a field(finite or infinite)?

A local ring R is a ring with a unique maximal ideal m. The quotient ring R/m is always a field.

If you want the ring to be embedded in the field, the answer is no. Consider the ring Z/4Z. It has a unique maximal ideal consisting of the even elements {0, 2}. But it cannot be embedded in any field because 2 x 2 = 0 (it has 0-divisors, so is not a domain). 

Consider infinite ring of integer Z and finite field  Zp, where p is prime number. Then  homomorphism f: Z --------->   Zp  defined by f(x) = \bar{x}= x mod (p) is a surjective ring homomorphism.