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- Non-invertible elements form an ideal 2 answers
Let $R$ be a ring and $I$ the set of non-invertible elements of $R$.
If $(I,+)$ is an additive subgroup of $(R,+)$, then show that $I$ is an ideal of $R$ and so $R$ is local.
I have done the following:
Since $(I,+)$ is an additive subgroup of $(R,+)$, we have that $\forall a,b \in I$ : $ab\in I$.
But how can we show that it holds that $ax\in I, \forall a\in I, \forall x\in R$ ?