- #1
mathmari
Gold Member
MHB
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Hey! ![Eek! :eek: :eek:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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:
Let $a\in I$ and $r\in R$.
We suppose that $ar$ is invertible, then $$1=(ar)(ar)^{-1}\Rightarrow 1=a(r(ar)^{-1})$$ That means that $a$ has a right inverse.
Now it is left show that $a$ has also a left inverse. That would mean that $a$ is invertible, a contradiction.
Is this correct? (Wondering)
But how exactly can we show that $a$ has also a left inverse? (Wondering)
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:
Let $a\in I$ and $r\in R$.
We suppose that $ar$ is invertible, then $$1=(ar)(ar)^{-1}\Rightarrow 1=a(r(ar)^{-1})$$ That means that $a$ has a right inverse.
Now it is left show that $a$ has also a left inverse. That would mean that $a$ is invertible, a contradiction.
Is this correct? (Wondering)
But how exactly can we show that $a$ has also a left inverse? (Wondering)