For a ring if we have: $\forall a\in R$ with $za=az=z$ does that mean $z=0$

Question

For a ring if we have: $\forall a\in R$ with $za=az=z$ does that mean $z=0$?

If we have $x+z=x$ for all $x\in R$ ($x+z=z+x$ as it is a ring) then we show z is unique and call it 0. For 0 it is true that $0a=a0=0$ (which is one of my favourite statements, as it shows the multiplicative identity and additive one are different (or the same for a ring of 1 element))

Is it true that an element $z\in R$ such that $az=za=z$ $\forall a\in R\implies z=0$

If the ring has unity then I can see this is true (in the sense that the multiplication table wont be able to have the intersecting-on-the-diagonal row of 0s and es it'd need to define, they'd "overlap") but that isn't formal.

I'd like to formally prove this.

Answer 1

Certainly if it holds for all $a \in R$, it holds for $a = 0$.

Then your statement becomes $0z = z0 = z$, but $0z = 0$ (by definition of $0$), so $z = 0$.

Answer 2

Of course trivially, if $a = 0$ then $0z = z$ so $z = 0$. But let's exclude this trivial case and suppose $az = za = z$ for all nonzero $a$. Then must $z = 0$?

There are two cases.

• First, suppose $e + e \ne 0$, where $e$ is the identity. Then choose $a = e + e$, and we have $z + z = z$ by distributivity, hence $z = 0$.

• Otherwise, $e + e = 0$. Then for any $x, y \ne 0$, $$ (x + y)z = xz + yz = z + z = (e + e)z = 0 $$ If $x + y \ne 0$, then $(x + y)z = z$ so this implies $z = 0$. Otherwise, we get that $x + y = 0$ for all nonzero $x, y \in R$. Adding $y$ to both sides, $x = y$, and the ring must consist of only two elements. We have $R = \mathbb{Z} / 2\mathbb{Z}$, and $z = 1$, and indeed $az = z$ for all nonzero $a$. This is the only exception.