OTOH, addition in RxC is defined by
(a,z) + (a',z') = (a+a',z+z') and multiplication by
(a,z) * (a',z') = (a*a',z*z') (all operators have their usual meaning)

This encourages us to map (1,0,0,0) to (-1,i) and the other
units to powers thereof. Mopr precisely, define the following map
phi: P4 -> R x C
   (a,b,c,d) |-> (-a+b-c+d, a*i -b -c*i +d)

First we need to show that this is well-defined - we have not yet
mentioned the cancellation identity (x,x,x,x)=0.
But indeed  phi(x,x,x,x) = (-x+x+x+x, x*i-x-x*i+x) = (0,0)
and in general (because phi is R-linear when viewed as a map
R^4 -> R x C)  phi(a+x,b+x,c+x,d+x) = phi(a,b,c,d).
Thus phi is indeed *well-defined*.

We now have a map of sets phi: P4 -> RxC.
In fact phi(a,b,c,d) = (0,0) implies -a+b-c+d = 0 and a-c=0 and b-d=0.
From this it follows that a=b=c=d, i.e. (a,b,c,d)=0.
Hence phi is an *injective* map.

We observe that phi is onto, for if (a,z) in R x C is arbitrary,
then let  u = max{Re(z),0}, v = max{-Re(z),0}, w = max{Im(z),0}, x =
max{-Im(z),0}.
Then u,v,w,x are non-negative numbers and z = (u-v)+i*(w-x).
Hence
  phi(w,v,x,u) = (-w+v-x+u, z).
Let b = a-(-w+v-x+u).
If b>=0, then
  phi(w,v+b/2,x,u+b/2) = (a,z)
and if b<0, then
  phi(w-b/2,v,x-b/2,u) = (a,z)
hence phi is *surjective*.

Next note that
phi((a,b,c,d) + (e,f,g,h)) = phi(a+e, b+f, c+g, d+h)
  = (-(a+e)+(b+f)-(c+g)+(d+h), (a+e)*i - (b+f) - (c+g)*i + (d+h))
  = (-a+b-c+d, a*i -b -c*i + d) + (-e+f-g+h, e*i-f-g*i+h)
  = phi(a,b,c,d) + phi(e,f,g,h).
Hence phi is *compatible* *with* *addition*.

Next
phi((a,b,c,d) * (e,f,g,h))
  = phi(ah+bg+cf+de, ae+bh+cg+df, af+be+ch+dg, ag+bf+ce+dh)
  = (-(ah+bg+cf+de)+(ae+bh+cg+df)-(af+be+ch+dg)+(ag+bf+ce+dh),
    (ah+bg+cf+de)*i-(ae+bh+cg+df)-(af+be+ch+dg)*i+(ag+bf+ce+dh))
  = ( (-a+b-c+d)*(-e+f-g+h), (a*i-b-c*i+d)*(e*i-f-g*i+h) )
  = (-a+b-c+d, a*i-b-c*i+d) * (-e+f-g+h,e*i-f-g*i+h)
  = phi(a,b,c,d) * phi(e,f,g,h)
Hence phi is *compatible* *with* *multiplication*.

Until now we have thus shown that phi is a ring isomorphism.

The norm of an element (a,b,c,d) of P4 is defined as
sqrt( aa + bb + cc + dd - (2/3)(ab + bc + cd + de + ac + bd)).
(This definition is based on a "belief" uttered in a usenet
discussion).
To avoid square roots, consider the expression
   ||(a,b,c,d)||^2 = aa + bb + cc + dd - (2/3)(ab + bc + cd + de + ac
+ bd).
Likewise the first definition of a norm on R x C that comes to mind
is defined by
   ||(x,z)||^2 = r^2 + z*\bar{z} = r^2 + Re(z)^2 + Im(z)^2.
However, this norm is not so natural as it seems.
Any choice of positive factors alpha, beta allows us to
define
   ||(x,z)||^2 = alpha * r^2 + beta *(Re(z)^2 + Im(z)^2)
and (since RxC has zero divisors and ||.|| cannot be multplicative
anyway)
there's no specific reason to prefer one choice over the others --
after all
they all define the same topology on RxC.

Let's calculate this for phi(a,b,c,d):
||phi(a,b,c,d)||^2 = ||(-a+b-c+d, a*i-b-c*i+d)||^2
   = alpha*(-a+b-c+d)^2 + beta*((a-c)^2+(d-b)^2)
   = alpha*(a^2+b^2+c^2+d^2+2(ac+bd-ab-ad-bc-cd)) + beta*(a^2+c^2-2ac
+d^2+b^2-2bd)
   = (alpha+beta)*(a^2+b^2+c^2+d^2) -2(alpha*(ab+ad+bc+cd) + (beta-
alpha)*(ac+bd))
It thus turns out that with the specific choice alpha=1/3, beta=2/3,
we have
||phi(a,b,c,d)|| = ||(a,b,c,d)||

We have thus proved the

THEOREM.
  P4 is isomorphic as a metric R-algebra to the RxC with norm
  ||(r,z)|| = sqrt( 1/3 * r^2 + 2/3 * (Re(z)^2+Im(z)^2) ).

hagman