[FOM] Transfinite Euclidean Algorithm
joeshipman at aol.com
Wed Nov 14 09:47:02 EST 2007
>from: hendrik at topoi.pooq.com
>> Commutative rings exist in which there is no Euclidean algorthm, but
>> there is a "division algorithm" in which the appropriate "norm" with
>> respect to which the remainder decreases takes values in a more
>> well-ordered set than the integers. Can anyone give a simple example
> such a ring?
>polynomials over the integers with ordinal exponents but only a
>finite number of terms in each polynomial?
How does that work? What happens when you divide x^omega + x + 1 by x^3?
If you say the quotient is x^(omega-3) and the remainder is (x+1), then
you have more than just ordinal exponents.
You can try to fix this by allowing exponents which are themselves
linear polynomials in an indeterminate, which we can call y. So an
element of the ring is a finite sum of terms of the form n * (x^p(y)),
where n is an integer and p(y) is a linear polynomial in y with integer
coefficients. But then x^(-1) is in the ring, so x becomes an
invertible element, and I'm not sure whether the resulting ring gives
what we want because units don't count for unique factorization.
There is probably a simple way to fix this, but it would be especially
nice if there were already an example in the literature.
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