Does (R × Z,+, ·) Form a Ring Without a Multiplicative Identity?

[hsetennis]

Homework Statement

Let (R,+, ·) be an algebraic object that satisfies all the axioms for a ring
except for the multiplicative identity. Define addition and multiplication in
R × Z by
(a, n) + (b,m) = (a + b, n + m) and
(a, n) · (b,m) = (ab + ma + nb, nm).
Show that (R × Z,+, ·) is a ring that contains a subset in one-to-one
correspondence with R that has all the properties of the algebraic object
(R,+, ·).

Homework Equations

7 conditions for a ring: commute+, associate+, 1, 0, inverse+, associate×, distribute

The Attempt at a Solution

Trying to prove associativity, and I think I'm making some silly computational error, but I've been at it for hours and I can't catch what I'm missing:

$[(a,n)*(b,m)]*(c,p) = (ab+ma+nb,nm)*(c,) = (abc+mac+nbc+pab+pma+pnb+nmc,cp)$
while
$(a,n)*[(b,m)*(c,p)] = (a,n)*(bc+pb+mc,cp) = (abc+apb+amc+cpa+nbc+npb+nmc,bccp+pbcp+mccp)$

 

[Fredrik]

(a, n) · (b,m) = (ab + ma + nb, nm).

$[(a,n)*(b,m)]*(c,p) = (ab+ma+nb,nm)*(c,) = (abc+mac+nbc+pab+pma+pnb+nmc,cp)$
while
$(a,n)*[(b,m)*(c,p)] = (a,n)*(bc+pb+mc,cp) = (abc+apb+amc+cpa+nbc+npb+nmc,bccp+pbcp+mccp)$

Not sure how you managed to get a different number of factors in the terms. They should all have three factors. In the first line, (c,) should of course be (c,p) (undoubtedly just a typo). At the end of the first line, you should have nmp, not cp. Similarly, you should have ncp at the end of the second line, instead of bccp+pbcp+mccp.

Edit: I meant, assuming that the thing in the middle of the second line is correct, the second component of the thing on the right should be ncp. But the thing in the middle is wrong too. Its second component should be mp, not cp. So the second component of the thing on the right will be nmp, not ncp.