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Problem is this:

Suppose for any element $r$ in a ring $R$ with unity $1$, $r$ or $1-r$ is invertible. Then show that non-invertible elements form an ideal of $R$.

It is crystal clear that $-r$ is non-invertible if $r$ is. So I tried to show the sum of two non-invertible elements is non-invertible, and the product of non-invertible elements with any ring element is non-invertible but I'm stuck.

Any help will be appreciated.

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  • $\begingroup$ Just to be clear, are you assuming that exactly one of $r$ and $1-r$ is invertible? $\endgroup$ – Ben Sheller Sep 30 '15 at 5:57
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Assuming the statement of the problem is that exactly one of $r$ or $1-r$ is invertible for any $r\in R$:

Let $J$ denote the set of noninvertible elements in $R$. Suppose $r,s\in J$, and that $r+s$ is invertible. Observe that $r$ noninvertible implies that $-r$ is noninvertible, so $1-(-r)=1+r$ is invertible. Then we have that:

$s=(r+s)-r\implies 1-s=1-(r+s)+r\implies (1-s)(r+s)^{-1}=(1+r)(r+s)^{-1}-1$.

But note that $(1-s)(r+s)^{-1}$ is invertible, so it must be that $(1+r)(r+s)^{-1}-1$ is invertible, but $(1+r)(r+s)^{-1}$ is invertible, so it must be that $(1+r)(r+s)^{-1}-1=-(1-(1+r)(r+s)^{-1}$ is noninvertible. Therefore, we have a contradiction, so $r+s$ is noninvertible.

(**)Hopefully, it is clear that the product of two invertible elements is again invertible. Now, let $r\in J$ and $q\in R$, and suppose $qr$ is invertible. We split the proof into two cases: First, suppose that $q$ is invertible. Note that $qr$ invertible implies that there exists $t\in R$ such that $tqr=1=qrt$. But then $t$ is invertible, so $qr=t^{-1}qrt$ and using $q$ invertible implies $r=q^{-1}t^{-1}qrt$, but then $r$ is the product of invertible elements and hence is invertible, a contradiction.

For the other case, suppose $q$ is noninvertible. Then $1-q$ must be invertible, so by the argument of the previous case, $(1-q)r$ must be noninvertible, and hence $1-(r-qr)$ is invertible. But then rearranging yields $-r+(1+qr)$ is invertible, but $-r$ is noninvertible, and $-qr$ is invertible so $1+qr$ is noninvertible, so $1-(r-qr)$ is the sum of noninvertible elements, and by what we have above, must be noninvertible, a contradiction.

Therefore, $J$ is a (left) ideal in $R$. To show it is also a right ideal, the proof is similar.

EDIT: At the request of @Junkhook9000 in the comments below, here is a proof that $J$ is an ideal which only requires the weaker property that at least one of $r$ and $1-r$ is invertible (as opposed to having exactly one of them invertible). First, prove (as above, starred) that the product of an invertible element with a noninvertible element is noninvertible. Now, suppose that $r,s\in J$ but that $r+s$ is invertible. Then there exists some element $q\in R$ with $q(r+s)=1$ and rearranging yields $r=q^{-1}(1-qs)$. But $1-qs$ is invertible since $qs$ is the product of an invertible element with a noninverible element and hence $qs$ is noninvertible. Therefore $r$ is the product of two invertible elements and is therefore invertible, a contradiction. So $J$ is closed under addition. Now suppose $rs$ is invertible. Then $(rs)r$ is noninvertible since $r\in J$, so $1-rsr$ is invertible. However, $1-rsr=(rs)^{-1}(rs-r)=-(rs)^{-1}(1-s)r$, which is noninvertible since $rs$ is invertible and since $1-s$ is invertible because $s$ is not, but the product of an invertible element (in this case, $-(rs)^{-1}(1-s)$) times a noninvertible element ($r$) is noninvertible. Therefore $rs$ cannot be invertible, so $rs\in J$.

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  • $\begingroup$ it says 'either r or 1-r.'...not so clear i guess. tqr=1 means tq is the just left inverse of r...can it lead to a contradiction? $\endgroup$ – Mathcho Sep 30 '15 at 6:51
  • $\begingroup$ @Mathcho Yes, that would still be a contradiction. $\endgroup$ – Ben Sheller Sep 30 '15 at 16:01
  • $\begingroup$ could you tell me why? i don't see why it must be two sided inverse as you said in your posting. $\endgroup$ – Mathcho Oct 1 '15 at 9:04
  • $\begingroup$ @Mathcho I've added something about why it is enough. Hopefully that helps. $\endgroup$ – Ben Sheller Oct 2 '15 at 16:10
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    $\begingroup$ Thanks for the reply! I went through it and the first part looks good. I think the second part can be shown (like in the comments above) by saying: for $r \in J$ and $s \in R$ if $rs$ invertible then there exists $x \in R$ such that $(rs)x = 1$ so $r(sx) = 1$ and $r$ invertible, a contradiction (likewise for left invertible). Please correct me if I am wrong. $\endgroup$ – Junkhook9000 May 9 '18 at 3:09
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While Ben's solution looks fine, I'd like to put my solution here since it is a bit simpler and covers the weaker case (at least one of $r$ or $1-r$ invertible) pretty nicely.

If $a$ not invertible, then $ar$ not invertible for any $r \in R$; otherwise there exists some $s \in R$ such that $(ar)s = 1 \Leftrightarrow a(rs) = 1 \Leftrightarrow a$ invertible, a contradiction.

Take $a, b \in R$ not invertible and consider $a-b$. If $a-b$ invertible then there is some $r \in R$ such that $(a-b)r = 1 \Leftrightarrow ar - br = 1 \Leftrightarrow -br = 1-ar$. By above, $ar$ not invertible, so $1-ar$ must be. Also by above, $br$ is not invertible, a contradiction.

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  • $\begingroup$ It seems to me you are assuming $R$ is commutative. How did you get $a(rs) = 1 \Leftrightarrow a$ is invertible? $\endgroup$ – No One Oct 16 '18 at 18:19
  • $\begingroup$ @no one Good point, but I think we can still get by without commutativity. This should read right invertible throughout the proof, then the left inverse case should be the same (show $a$ not left invertible implies $ra$ not left invertible, then consider $r(a-b)$ for $a, b$ not left invertible to get a contradiction). I'll try to fix up the proof when I get a chance. $\endgroup$ – Junkhook9000 Oct 17 '18 at 22:11

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