Re: Existence of (co)product

In article <dsq63r$fc3$1@xxxxxxxxxxxxxxxxx>,
kj <socyl@xxxxxxxxxxxxxxxxx> wrote:
In <4573q9F5baftU1@xxxxxxxxxxxxxx> Marc Olschok <invalid@xxxxxxxxxxx> writes:

Not quite true, for example Q x Q = Q in Field.

I assume this means that Q is equal to its categorical product with
itself, but I can't figure out why. Would someone be kind enough
to help me out here? Does it have to do with Q being terminal (or
initial) in Field?

Each prime field is initial in Field if you allow the
zero map: if F is a prime field (i.e., either Q or the field of p
elements, p a prime), and K is any field, then if char(K)=char(F)
there is a unique embedding map from F to K; and if char(K) is
different from char(F), then there is a unique map from F to K, namely
the 0 map.

If you do not allow the zero map (by requiring that 1 be mapped to 1),
then no field is initial (or terminal), since you cannot map a field
of one characteristic into a map of a distinct characteristic.

Now, since the poster's statement only holds if you do NOT allow the
zero map, I will assume we do not allow the zero map.

Why is Q the categorical product with itself in Field? Remember the
universal property of the product: if A and B are objects in the
category, then P is a categorical product of A and B if and only if
you have arrows p_A:P->A and p_B:P->B, such that for any object O and
any pair of arrows a:O->A, b:O->B, there exists a unique arrow c:O->P
such that a=p_a c and b = p_b c.

Here, the claim is that P=Q. Clearly, both projection maps will be the
identity. Let F be any field, and f,g:F->Q be field morphisms. Since
any ring homomorphism from a field must be either injective or the
zero map, and we are not allowing the zero map, it follows that both f
and g must be injections. The only subfield of Q is Q itself, and F
must be characteristic zero, hence f and g are really the identity,
and therefore, the identity map is the unique map from F into P that
will make the corresponding diagram commute.

For pretty much the same reason, you will find that if F is a prime
field, then F x F = F in the category of fields with nonzero morphisms.

Note that if we allow the zero map, then the assertion would be false:
just take F=Q, and let f = id and g = 0. Then any map from F to Q
would have to be either 0 (in which case the triangle with f does not
work) or the identity (in which case the triangle with g does not
work), so Q would not be the product of Q with itself.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

.

Relevant Pages

  • Re: Data compression in roguelikes
    ... Store the random number seed for map generation. ... Compress the mostly zero map with zlib (do not roll your own ... I'm not sure it's worth compressing for ...
    (rec.games.roguelike.development)
  • Re: F-Isomorphism proof
    ... assuming that q is not the zero map. ... and the only ideals of a field are the trivial ones). ...
    (sci.math)
  • Re: F-Isomorphism proof
    ... >> map that somehow takes into account the fact that we are considering ... field and q is not the zero map (because F is not the zero ring and q ... Using the term isomorphism ...
    (sci.math)
Quantcast