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This question already has an answer here:

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

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marked as duplicate by Dietrich Burde, rschwieb abstract-algebra Apr 10 '16 at 22:39

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    $\begingroup$ If $ax$ were invertible, then $1=a(x(ax)^{-1})$, so that $a \notin I$. $\endgroup$ – Crostul Apr 10 '16 at 18:09
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    $\begingroup$ Wait, why would $ab\in I$ from the fact that $I$ is a subgroup ? (I mean, it's true since $I$ is actually an ideal, but it's not immediate.) $\endgroup$ – Captain Lama Apr 10 '16 at 18:09
  • $\begingroup$ @CaptainLama: Correct. In fact, it is not true in general that the non-invertible elements form an additive subgroup; in this problem, though, this is given as an assumption, but it seems that the OP has got a little confused about what she has to check. $\endgroup$ – Alex M. Apr 10 '16 at 18:16
  • $\begingroup$ If $ax$ is invertible there is a right and a left inverse. The right inverse is $(ax)(ax)^{-1}=1$ and the left inverse is $(ax)^{-1}(ax)=1$, or not? So, do we have to show also that $1=(\text{ something } ) a$, to conclude that $a\notin I$, or not? @Crostul $\endgroup$ – Mary Star Apr 10 '16 at 19:03
  • $\begingroup$ @rschwieb: The question that this is supposed to be a duplicate of is fairly different from this one: it has a very constraining assumption that is not to be found here, which makes that the proof found there not be applicable here. Could you please reconsider your close vote? $\endgroup$ – Alex M. Apr 11 '16 at 17:39
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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 $ax \in I$.

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  • $\begingroup$ Ah ok... Is $I$ also the unique maximal right ideal of $R$ ? $\endgroup$ – Mary Star Apr 11 '16 at 19:14
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    $\begingroup$ @MaryStar: Yes, it is: by assumption, $(I,+)$ is a subgroup of $(R,+)$. If $0 \ne 1$ (which is probably tacitly assumed), then $R$ is a local ring, so it will have a unique maximal ideal. Notice that $R/I$ is a field (this quotient essentially kills the non-invertible elements in $I$, leaving only the invertible ones). Since $R/I$ is a field, $I$ must be maximal. Since $R$ is local, $I$ must be unique then. $\endgroup$ – Alex M. Apr 11 '16 at 19:37
  • $\begingroup$ I see... Thanks a lot!! :-) $\endgroup$ – Mary Star Apr 12 '16 at 0:00
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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.

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