Proving r*r=q in S, a Ring with Identity

[Achieve]

Let S={p,q,r} and S=(S,+,*) a ring with identity. Let p be the identity for + and q the identity for *. Use the equation
r*(r+q)=r*r+r*q to deduce that r*r=q.

Attempt of a solution
r*r=r*(r+q)- r*q
=r*r+r*q - r*q
But I'm not finding a clever way to deduce what is required.
Any type of help would be appreciated.

 

[Euge]

Since S is closed under $+$, $r+q$ is either $p$, $q$, or $r$. If $r+q = q$ or $r+q = r$, then we get $r = p$ or $q = p$, which are both false. So $r+q = p$. Substituting $p$ for $r+q$ in the equation

$r * (r + q) = r * r + r * q$

gives

$r * p = r * r + r * q$.

Since $p$ is the additive identity in S, $r * p = p$; since $q$ is the multiplicative identity, $r * q = r$. Therefore

$p = r * r + r$.

Adding $q$ to both sides results in $q = r * r + p$, or, $q = r * r$.

 

[Achieve]

"Since p is the additive identity in S, r∗p=p"
Can you please explain why. I understand if p is the identity then r+p=r but not that statement.

 

[Euge]

It's because $r * p = r * (p - p) = r * p - r * p = p$.

 

[Deveno]

A basic fact about rings, is that under addition, they form an abelian group.

In this case, the abelian group has order 3, so either of $q$ or $r$ is a generator.

Since the map $a \mapsto -a$ is a group isomorphism for any abelian group (in particular it maps a generator to a generator), it must be the case that either:

$-q = r$, or $-q = q$.

If $-q = q$, then $q + q = p$, in which case $q$ has order 2. But 2 does not divide 3, so this violates Lagrange.

Therefore, $-q = r$, that is $r + q = p$.

So from our given equation:

$r\ast(r + q) = r\ast r + r\ast q$ we get:

$r \ast p = r\ast r + r \ast q$ (from the above)

$r \ast p = r\ast r + r$ (since $q$ is the multiplicative identity).

Now, to see that $r \ast p = p$, note that:

$r \ast p = r\ast(p + p)$ (since $p + p = p$, since it is the additive group identity)

$= r \ast p + r\ast p$. Hence:

$p = -(r \ast p) + r\ast p = -(r\ast p) + r\ast p + r\ast p = p + r\ast p = r\ast p$.

Thus (continuing our first argument):

$p = r\ast r + r$, so that $r\ast r$ is the additive inverse of $r = -q$.

So $r\ast r = -(-q) = q$ (in a group, the inverse of an inverse is the original element).

****************************

The above is tedious, and complicated by notation. In point of fact, there is no reason not to denote $p$ by $0$, and $q$ by $1$. Thus:

$S = \{0,1,r\}$.

It is plain to see we must have $r = 1 + 1$. Now we can just calculate:

$(1+1)(1+1) = (1+1)+(1+1) = [(1+1)+1] + 1 = 0 + 1 = 1$

because $(1+1)+1 = 3\cdot 1 = 0$ (the "3rd power" of 1, written additively) by Lagrange.