How can we show that $I$ is an ideal?

Question

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.

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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 $ax\in I, \forall a\in I, \forall x\in R$ ?

Answer 1

I believe that somewhere in your textbook $R$ is assumed to be commutative. Otherwise, the product of a non-invertible element with an arbitrary element of the ring may turn out to be invertible (the examples given there are even stronger: they show that the product of two non-invertible elements might be invertible).

Let $a \in I$ and $x\in R$. Assume that $ax$ is invertible; there must exist, then, some $y \in R$ such that $(ax) y = 1$, which is equivalent to $a (xy) = 1$, which means that $xy$ is a right inverse for $a$, which will also be a left inverse because $R$ is commutative, so $a$ is invertible, which is a contradiction, therefore $ax$ is not invertible so $ax \in I$.

Answer 2

Since $(I,+)$ is an additive subgroup of $(R,+)$, we have that $∀a,b∈I > : ab∈I.$

This reasoning is wrong. Moreover, you don't need to prove this statement. The product of a non-invertible element and any other element is non-invertible. Therefore it is immediate that $IR\subset R$. The set $I$ is by assumption a subgroup w.r.t. addition, hence an ideal.