Spivak: basic question on the existence of multiplicative identity The properties explained up to this point are:

$$
\begin{matrix}
\text{(P1)} & \text{(Associative law for addition)} & a+(b+c)=(a+b)+c. \\
\text{(P2)} & \text{(Existence of an additive identity)} & a+0=0+a=a. \\
\text{(P3)} & \text{(Existence of additive inverses)} & a+(-a)=(-a)+a=0. \\
\text{(P4)} & \text{(Commutative law for addition)} & a+b=b+a. \\
\text{(P5)} & \text{(Associative law for multiplication)} & a\cdot(b\cdot c)=(a\cdot b)\cdot c. \\
\end{matrix}
$$
(P6) If $a$ is any number, then
$$a\cdot1=1\cdot a=a.$$
Moreover, $1\neq0$.
(The assertion that $1\neq0$ may seem a strange fact to list, but we have to list it, because there is no way it could possibly be proved on the basis of the other properties listed—these properties would all hold if there were only one number, namely, $0$.)

I don’t understand why is it important to make the assertion that $1 ≠ 0$. Could someone elaborate on that and maybe give some examples?
 A: Take $ R = \{0\}\subset \Bbb Z.$ With the same operations as in $\Bbb Z$, the set $R$ satisfies your axioms: $$\begin{align} \text{P1)}& 0+(0+0) = (0+0)+0 \\ \text{P2) }&\text{$0$ is in the set $R$, and for any element $a$ of $R$ satisfies } a + 0 = 0 + a \\ \text{P3)} &\text{ True because } 0 = -0 \\ \text{P4) }& 0 +0 = 0 + 0 \\ \text{P5) }& 0\cdot(0\cdot0) = (0\cdot0)\cdot0\end{align}$$
P6 is the one that's strange in this case, but $R$ satisfies it nonetheless. We require only that there be an element of (I'll call it $e$ instead of "$1$") $e\in R$ such that, for any other element $a\in R$ we have $$a\cdot e = e\cdot a = a$$
There's not much left to do other than check if this holds true for some element of $R$, but $R$ only has one of those, so the only thing to check is whether or not $$0\cdot 0 = 0\cdot 0 = 0$$
Obviously this is true, so $0$ is the multiplicative identity in $R$, or put otherwise, $0 = e$ or, recovering the old symbol, $0 = 1$.
It's then a matter of choice whether or not we allow such trivial rings. In the case we do allow them you then have to consider the case $0=1$ in theorems, proofs, etc. Many authors do this, while others (such as Spivak I guess) choose to avoid this "nuisance" by including $0 \neq 1 $ in the axioms.
