Question

Give an example, with justification, to show that: 1.the union of integral domains need not be an integral domain; 2.all finite fields are not isomorphic.?

Answer 1

See explanation...

Explanation:

Example of two integral domains whose union is not one

A rich source of examples of such things is matrices.

For example, note that:

`((0, 1), (2, 0))^2 = ((0, 1), (2, 0)) ((0, 1), (2, 0)) = ((2, 0), (0, 2))`

That is `((0, 1), (2, 0))` is a square root of `2`.

We can identify an integral domain of matrices of the form:

`m((1, 0), (0, 1)) + n((0, 2), (1, 0)) = ((m, n), (2n, m))`

with `m, n in ZZ` which is isomorphic to the ring of numbers of the form `m+n sqrt(2)`.

Note that this has no zero divisors since `sqrt(2)` is irrational.

Similarly, we find:

`((0, 2), (1, 0))^2 = ((2, 0), (0, 2))`

So `((0, 2), (1, 0))` is also a square root of `2` and we can identify an integral domain of matrices of the form

`((m, 2n), (n, m))`

which is also isomorphic to the ring of numbers of the form `m+n sqrt(2)`

The union of these two sets of matrices is not closed under addition and therefore not a ring, let alone an integral domain.

For example:

`((0, 1), (2, 0)) + ((0, 2), (1, 0)) = ((0, 3), (3, 0))`

is not in either of the two integral domains.

Furthermore, if we add all elements required to make the set of matrices closed under addition and additive inverse, then we get a ring with zero divisors. For example:

`(((0, 3), (3, 0)) - ((3, 0), (0, 3)))(((0, 3), (3, 0)) + ((3, 0), (0, 3)))= ((-3, 3), (3, -3))((3, 3), (3, 3))= ((0, 0), (0, 0))`

Finite fields of different orders or characteristics cannot be isomorphic

For example `GF(2)` and `GF(3)` have `2` and `3` elements respectively, so there is not even a bijection between them as sets, let alone an isomorphism of fields.

All finite fields have `p^n` elements for some prime `p` and positive integer `n`. All finite fields of the same order are isomorphic. That is, a finite field is essentially determined by the values of `p` and `n`.