If $R$ is a simple ring and $n \in \Bbb Z$, show that either $nR = 0$ or $nr=0$ implies that $r=0$

Question

If $R$ is a simple ring , $n \in \Bbb Z$, show that either $nR = 0$, or $nr=0$ implies that $r=0$ for $r \in R$.

Since $nR$ is an ideal of $R$, then either $nR=0$, or $nR = R$. In this second case, I am not sure why it is the case that if $nr = 0$, that necessarily $r = 0$.

Side note, I know a theorem that states that a commutative ring $R$ is simple iff it is a field, but in this case I don't have commutativity. I also know that if $\text{char}(R)= k$, then $kR=0$, but I don't know that the characteristic is non-zero, and not sure about the case where $n$ is not a multiple of $k$.

Answer 1

Consider the set of elements $r \in R$ with $nr = 0$. This is a two-sided ideal, so it must be either $0$ or all of $R$.

Answer 2

Recall that any ring $R$ is an $R$-module (i.e. every ring is a module over itself).

By the definition of simplicity of $R$, assuming $R \neq 0$, there exists two possible submodules of $R$, namely: $0$ and $R$ itself.

Recall: (Schur's Lemma) Let $U,V$ be simple $R$-modules. Then, for an arbitrary R-homomorphism $f:U \to V$, we have the following: $$f = 0, \\\text{or}\\f \text{ is an isomorphism}$$ Proof: c.f. 9.2.1a

Now, define an $R$-module $R$ with a simple left-sided ideal $I$ and suppose that $IR = {\{\sum_{i}r_ix_i \space | \space r_i \in I, \space x_i \in R\}}$.

By definition of the simplicity of $R$, in the case of $IR \neq 0$, we have that $IR = R$; thus, for any $x \in R$, we have $Ix = R$.

Define an $R$-homomorphism $\phi : I \to R$ such that $$\phi (r) = rx.$$ Since we know that $\phi(x) \neq 0$ for any $x \in R$, it is obvious that $\text{ker}\phi \neq I$, and since, by definition of simplicity, the function is either the zero map or an isomorphism.

It immediately follows that $f$ is indeed an isomorphism (i.e. $I \cong R$)

Therefore, $\text{ker}\phi = \{0\}$, which proves that, as in the original question, when $nr = 0$, $r = 0$.