How can we show that a has a left inverse?

[mathmari]

Hey!

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)

 

[Euge]

It looks good so far, and yes, if $a$ has both a left and right inverse, it is invertible (proof: if $ax = 1$ and $ya = 1$ for some $x,y\in R$, then $x = (ya)x = y(ax) = y$; therefore, $a$ is invertible). To see that $a$ has a right inverse, let $c = (ar)^{-1}$. Since $a\in I$ and $1\notin I$, then $1 - a\notin I$; otherwise, by closure of $I$ under addition, $1 = (1 - a) + a\in I$, a contradiction. So using the equation $arc = 1$, we have $1 - a = arc - a = a(rc - 1)$; thus $a(rc - 1)\notin I$, i.e., $a(rc - 1)$ is invertible. There exists $b\in R$ such that $a(rc-1)b = 1$. This shows that $a$ is right-invertible with $(rc-1)b$ as a right inverse.

 

[mathmari]

It looks good so far, and yes, if $a$ has both a left and right inverse, it is invertible (proof: if $ax = 1$ and $ya = 1$ for some $x,y\in R$, then $x = (ya)x = y(ax) = y$; therefore, $a$ is invertible). To see that $a$ has a right inverse, let $c = (ar)^{-1}$. Since $a\in I$ and $1\notin I$, then $1 - a\notin I$; otherwise, by closure of $I$ under addition, $1 = (1 - a) + a\in I$, a contradiction. So using the equation $arc = 1$, we have $1 - a = arc - a = a(rc - 1)$; thus $a(rc - 1)\notin I$, i.e., $a(rc - 1)$ is invertible. There exists $b\in R$ such that $a(rc-1)b = 1$. This shows that $a$ is right-invertible with $(rc-1)b$ as a right inverse.

I see... (Smile)

But how could we show that $a$ has also a left inverse? (Wondering)
Which $c$ do we take? (Wondering)

 

[Euge]

So sorry, I definitely need more rest. : ) I was thinking of left inverse bit instead did the right inverse in a roundabout way. I'll come back later.

Edit: So, let's try this again. We suppose $a\in I$ and $r\in R$ such that $ar\notin I$ and obtain a contradiction. Keep in mind the following fact: If $x\in I$ and $y\notin I$, then $x + y\notin I$. This is true, for otherwise $x + y\in I$ and closure under subtraction in $I$ would give $y = (x + y) - x\in I$, contrary to assumption.

If $r\notin I$, then $r^{-1}$ exists and $a = (ar)r^{-1}\notin I$, contrary to the assumption that $a\in I$. If $r\in I$, then since $ar\notin I$, $r + ar \notin I$, i.e., $(1 + a)r\in I$; also, $1 + a\notin I$ since $1\notin I$ and $a\in I$. So $(1 + a)^{-1}$ exists and $r = (1+a)^{-1}[(1 + a)r]\notin I$, a contradiction. Hence, $ar\in I$.

 

[mathmari]

If $r\in I$, then since $ar\notin I$, $r + ar \notin I$, i.e., $r(1 + a)\in I$; also, $1 + a\notin I$ since $1\notin I$ and $a\in I$. So $(1 + a)^{-1}$ exists and $r = [r(1 + a)](1 + a)^{-1}\notin I$, a contradiction.

When $r + ar \notin I$ doesn't it follow that $(1 + a)r\notin I$ ? (Wondering)

Then we have that $r =(1 + a)^{-1} [(1 + a)r]\notin I$, right? (Wondering)

We get that $(1 + a)^{-1} [(1 + a)r]\notin I$, because both of the terms $(1 + a)^{-1}$ and $(1 + a)r$ don't belong to $I$, right? (Wondering) So, we have that $ar\in I$. It is left to show also that $ra\in I$, or not? (Wondering)

We do that as follows:

We suppose that $ra\notin I$.

If $r\notin I$, then $r^{-1}$ exists and $a = r^{-1}(ra)\notin I$, a contradiction, since $a\in I$.

If $r\in I$, then since $ra\notin I$, $r + ar \notin I \Rightarrow r(1 + a)\notin I$.
Since $1\notin I$ and $a\in I$ we have that $1 + a\notin I$.
So $(1 + a)^{-1}$ exists and $r = [r(1 + a)](1 + a)^{-1}\notin I$, a contradiction, since $r\in I$.

Therefore, $ra\in I$. So, we conclude that $I$ is an ideal. Is $I$ the unique maximal right ideal of $R$ ? (Wondering)

 

[Euge]

To answer your first three questions - yes, yes, and yes. I made the correction. But in your argument that followed, the part $r + ar \notin I$ should be replaced by $r + ra\notin I$. Otherwise your proof is correct.

To answer your last question, the answer is yes. Note that you have just proved that $I$ is a two-sided ideal, not just a right ideal.

 

[I like Serena]

Suppose we pick $R = \{ \mathbb R^\infty \to \mathbb R^\infty \}$ with regular addition and function composition.
Then I think that is a ring. (Thinking)

Let $I$ be the non-invertible elements of $R$ and suppose that $I$ is an ideal.
Let $a: (x_1, x_2, ...) \mapsto (x_2, ...)$ and $b: (x_1, x_2, ...) \mapsto (0, x_1, x_2, ...)$.
Then neither is both left and right invertible, so $a, b \in I$.
But $ab=1$, so we found for $r=b$ that $ar \notin I$, which is a contradiction.
(For the record, $a$ is right invertible but not left invertible. (Nerd))

Therefore $I$ is not an ideal. (Worried)

 

[mathmari]

But in your argument that followed, the part $r + ar \notin I$ should be replaced by $r + ra\notin I$. Otherwise your proof is correct.

Ah ok... (Smile)

To answer your last question, the answer is yes. Note that you have just proved that $I$ is a two-sided ideal, not just a right ideal.

Is $I$ the unique maximal right ideal of $R$ because of the following:

Suppose that $J$ is a proper ideal.
It must hold that $1\notin J$, because otherwise $J=R$.
Therefore, $J$ must contain non-ivertible elements.
So, it must be $J\subset I$, since $I$ contains all the non-invertible elements.
Is this correct? (Wondering)

Does this imply that $I$ is the unique maximal right ideal of $R$ ? (Wondering)

 

[mathmari]

Let $a: (x_1, x_2, ...) \mapsto (x_2, ...)$ and $b: (x_1, x_2, ...) \mapsto (0, x_1, x_2, ...)$.
Then neither is both left and right invertible, so $a, b \in I$.
But $ab=1$, so we found for $r=b$ that $ar \notin I$, which is a contradiction.

Why do we have that $a, b \in I$ and why do we have that $ab=1$ ? (Wondering)

 

[Euge]

Suppose we pick $R = \{ \mathbb R^\infty \to \mathbb R^\infty \}$ with regular addition and function composition.
Then I think that is a ring. (Thinking)

Let $I$ be the non-invertible elements of $R$ and suppose that $I$ is an ideal.
Let $a: (x_1, x_2, ...) \mapsto (x_2, ...)$ and $b: (x_1, x_2, ...) \mapsto (0, x_1, x_2, ...)$.
Then neither is both left and right invertible, so $a, b \in I$.
But $ab=1$, so we found for $r=b$ that $ar \notin I$, which is a contradiction.
(For the record, $a$ is right invertible but not left invertible. (Nerd))

Therefore $I$ is not an ideal. (Worried)

Is this supposed to be a counterexample to the problem? It does not contradict the original statement because $(I,+)$ is not an additive subgroup of $(R,+)$. The identity mapping, which is invertible, is the sum of two noninvertible functions : $c : (x_1,x_2,\ldots)\mapsto (x_1,0,0,0,\ldots)$ and $d : (x_1,x_2,\ldots)\mapsto (0,x_2,x_3,\ldots)$.

 

[Euge]

Is $I$ the unique maximal right ideal of $R$ because of the following:

Suppose that $J$ is a proper ideal.
It must hold that $1\notin J$, because otherwise $J=R$.
Therefore, $J$ must contain non-ivertible elements.
So, it must be $J\subset I$, since $I$ contains all the non-invertible elements.
Is this correct? (Wondering)

Does this imply that $I$ is the unique maximal right ideal of $R$ ? (Wondering)

Yes, it looks good. Since no proper ideal of $R$ contains a unit, $I$ contains every proper ideal. In particular, every maximal right ideal is contained in $I$, so $I$ is the unique maximal right ideal.

 

[I like Serena]

Why do we have that $a, b \in I$ and why do we have that $ab=1$ ? (Wondering)

$a$ is not left invertible, because whichever function we apply next, we can never recover $x_1$.
So $a$ is not invertible and thus $a \in I$.

Similarly, whichever function we apply first before applying $b$, the result has a first entry that is always $0$ and can never be $x_1$ anymore. So $b$ is not right invertible and so $b\in I$.

$(a\circ b)(x_1, x_2, ...) = a(0, x_1, x_2, ...) = (x_1, x_2, ...)$
So $ab = 1$.

Is this supposed to be a counterexample to the problem? It does not contradict the original statement because $(I,+)$ is not an additive subgroup of $(R,+)$. The identity mapping, which is invertible, is the sum of two noninvertible functions : $c : (x_1,x_2,\ldots)\mapsto (x_1,0,0,0,\ldots)$ and $d : (x_1,x_2,\ldots)\mapsto (0,x_2,x_3,\ldots)$.

Yes. I came up with it before I read your edited response with a full-proof proof.
Anyway, that explains it - I forgot to check if $I$ is an additive subgroup. (Tmi)

 

[mathmari]

I see... (Smile)

Thank you very much! (Yes)