On Monday 04 December 2006 11:16, William Stein wrote:
> On Mon, 04 Dec 2006 04:52:02 -0800, Joel B. Mohler
>
> <[EMAIL PROTECTED]> wrote:
> > On Sunday 03 December 2006 10:23, William Stein wrote:
> >> > If the predefined indeterminate belongs to a ring, which ring is it?
> >>
> >> Very likely it would just be ZZ[a,b,c,d...,z, A,B,C,...,Z].    
> >> Alternatively,
> >> they could be "formal indeterminates" that don't belong to some "formal
> >>  
> >> ring",
> >> but that's less easy to think about.
> >
> > What if you want to differentiate f(x)=x^x.  This is senseless in a
> > polynomial
> > ring.
>
> This is the same as saying that "1/2" is senseless if 1 and 2
> are integers.
>
> This is not a problem.  That the expression x^x does not define an element
> of the polynomial ring, is not a problem.
> Exponentiation of an element of the polynomial ring to the power
> of a non-integer would be a constructor for some other sort of object,
> namely a function in some general function space.

I'm not sure I entirely agree with this.  I see that it could be done this 
way, but it seems quite mathematically strange to me.  The fraction field of 
a ring is a well-known mathematical object, but the idea of putting a 
generator for a polynomial ring in an exponent is entirely undefined in any 
abstract sense (unless there's some mathematics I'm missing).

--
Joel

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