Re: Please help, Thanks!

quasi <quasi@xxxxxxxx> wrote:
On Mon, 17 Mar 2008 18:40:58 EDT, steven <chasedtime@xxxxxxxxx> wrote:

define a ring R={0,1,a,b} by the multiplication table below
* | 0 1 a b
-------
0 | 0 0 0 0
1 | 0 1 a b
a | 0 a b 1
b | 0 b 1 a

To define a ring, you also have to define how elements _add_.

No, the above multiplication table suffices.

prove that as rings, R is not Isomorphic to Z_2 x Z_2.

Hint: Count the elements of order 2 (i.e. such that x^2 = 1).

That doesn't work. Instead, it is necessary and sufficient to
consider any single product of elements neither 0 nor 1, i.e. any
entry in the lower-right quarter of the multiplication table(s),
where one finds non-trivial idempotents, units, zero-divisors.

Alternate hint: Note that R is commutative, and that all nonzero
elements have inverses -- hence R is a field. But, assuming you have,
as a known theorem, the fact that the multiplicative group of a finite
field is cyclic, it follows that a,b have order 3. Now look at Z_2 x Z_2.

That's quite a sledgehammer for such a little chestnut.
A smaller more apropos hammer is to simply observe that
in a finite ring every element divides 0 or 1, i.e. every
non-unit is a zero-divisor. This classic pigeonhole proof
is closely connected to Euclid's proof that there are
infinitely many primes - see my prior post [1].

--Bill Dubuque

[1] http://google.com/groups?selm=y8zbr8kv5vl.fsf_-_%40nestle.csail.mit.edu
.

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