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Mon, 27 Feb 2006 22:59:35 -0600
> $R[x]$ is a subring of $Q(R)[x]$ where $Q(R)$ is the quotient
> field of $R$.
I find your notation a little confusing. In Axiom notation I think
you mean, for example:
$R[x]$ = POLY INT
$Q(R)[x]$ = POLY FRAC INT
Is that right? If so, then I agree that:
> '1/x' is in neither of these polynomial rings,
But I do not agree with your continuation of the statement:
> (((No matter what $R$ is!))). This is a FACT of mathematics,
> the meaning of what an indeterminate x over a ring or
> field is. 1/x ONLY lives in $Q(R[x]) = Q(Q(R)[x]$, where
> $(1/x)\times x = 1$.
By $Q(R[x])$ I suppose you mean, for example:
$Q(R[x])$ = FRAC POLY INT
I don't understand why you write $Q(Q(R)[x]$ meaning for example:
$Q(Q(R)[x])$ = FRAC FRAC POLY INT
but certainly '1/x' does "live" in these domains, if by "live" you
mean that a suitable function '/' can be found by a selection
from these domains.
However '1/x' is also lives in other polynomial rings such as:
$Q(R[x])[x]$ = DMP(![x], FRAC POLY INT)
where $(1/x)\times x \ne 1$. $Q(R[x])[x]$ satisfies all of the
axioms of a ring and the 'x' in $Q(...)[x]$ remains transcendent
over the 'x' in $R[x]$. You can see that a valid selection for
the function '/' exists in this domain by doing::
)sh DMP([x], FRAC POLY INT)
It is convenient that you choose to write $\times$ above because
we need to distinquish two different multiplications: the $\times$
in DMP and the '*' in 'POLY INT'.
It is only when we coerce $Q(R[x])[x]$ into $Q(R[x])$, which
means that we map the $\times$ and '/' of 'DMP(![x], FRAC POLY INT)'
into the $*$ and '/' of 'FRAC POLY INT' that we can equate these
two uses of $x$.
In making this claim I have not violated any FACT of mathematics.
forwarded from http://wiki.axiom-developer.org/address@hidden