# left invertible elements in a ring with unit.

My question is:

Let $R$ be a ring with the unit $e$ and $a \in R$. If $a^{\circ}\triangleq\{x\in R \;| \;ax=0\}=\{0\}$, as the following conter-example given by Matthias Klupsch, we know that it is not left invertible in general. Then what additional conditions on $R$ or on $a$ will make that $a$ is left invertible?

If we have known $a$ is right invertilbe, then what will happen?

${\bf Notes:}$ Let $A$ and $B$ be two sets, and $f : A \to B$ be a mapping, then it is well-known that:

(1) $f$ is injective iff $f$ is left invertible.

(2) $f$ is subjective iff $f$ is righted invertible.

Thanks!

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If you take $R = \mathbb{Z}$, you have $2^\circ = \{0\}$, but $2$ is not left-invertible. Generalizing this example, if your statement would be correct, then every integral domain would be a field. –  Matthias Klupsch Oct 27 '12 at 8:05
Maybe you meant the converse? –  wj32 Oct 27 '12 at 8:08
Thanks very much! But if we already have $a^{\circ}\triangleq\{x\in R \;| \;ax=0\}=\{0\}$, then what additional conditions will make that $a$ is left invertible? –  Spring Xiao Oct 27 '12 at 9:21
For commutative rings, an element is invertible if and only if it is not contained in any maximal ideal. –  Arthur Oct 27 '12 at 10:00
Why the tag [banach-algebras]? –  Davide Giraudo Oct 27 '12 at 14:07

This question can also be considered in categorical setting: a ring $R$ itself is a (pre-)additive category with one object, meaning that it has $+,-$ operations (and of course $0_{AB}$ arrow) on the hom-sets (now we have only one hom-set for the one object).

In preadditive category, $ax=ay \implies a(x-y)=0 \overset{\text{cond.}}\implies x-y=0 \implies x=y$, hence your condition means exactly that $a$ is left cancellable.

So, basically, you are asking whether left cancellable elements ('epimorphic' if the composition is read from left to right) are left invertibles as well. And the answer is: no in general. (Matthias and others wrote nice simple counterexamples in the comments.)

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Thanks very much for your answer! But if we consider $R$ as a Banach algebra instead of a ring, or other ring special property, then what can we say about may question? –  Spring Xiao Oct 27 '12 at 14:02
Well.. if you want left cancellable elements to make left invertible, I think, you can do it, by perhaps extending your structure with formal left inverses, I guess, even for Banach algebras. Why do you need the left inverses? –  Berci Oct 27 '12 at 14:52
Thanks! I wang to prove an element to be invertible. I have known it is right invertible, I need to prove it is left invertible. –  Spring Xiao Oct 28 '12 at 1:46
Thanks very much for all helps! –  Spring Xiao Oct 31 '12 at 7:16

If you look at the polynomial ring $R[X]$ over an integral domain $R$, for any $p \in R[X] \setminus R$, we have $p^{\circ} = \{ 0_{R[X]} \}$. However, $p$ is not invertible. This serves as a counterexample.

With regards to the latest edit, let $\mathbb{R}^{\mathbb{N}}$ denote the $\mathbb{R}$-vector space of all sequences of real numbers. Consider the $\mathbb{R}$-linear map $F_{\text{Right}}$ on $\mathbb{R}^{\mathbb{N}}$ that shifts a sequence to the right by a single position. For example, $${F_{\text{Right}}}(a_{1},a_{2},a_{3},a_{4},\ldots) = (0,a_{1},a_{2},a_{3},\ldots).$$ Clearly, $F_{\text{Right}}$ has a left-inverse, which is the $\mathbb{R}$-linear map $F_{\text{Left}}$ on $\mathbb{R}^{\mathbb{N}}$ that shifts a sequence to the left by a single position. However, there is no $\mathbb{R}$-linear map that acts as a right-inverse of $F_{\text{Right}}$, because all sequences in the range of $F_{\text{Right}}$ have a $0$ in the first position. This implies that in the endomorphism ring of $\mathbb{R}^{\mathbb{N}}$, where ring multiplication corresponds to composition of $\mathbb{R}$-linear maps, there exist elements that are left-invertible but not right-invertible.

Without much pain, one can just as easily produce a ring $R$ that has right-invertible-but-not-left-invertible elements.

Conclusion In general, left-invertibility does not imply right-invertibility, and right-invertibility does not imply left-invertibility. However, in square-matrix rings over a field, the two imply each other.

Addendum Given a ring $R$, if $a \in R$ satisfies (i) $a^{\circ} = \{ 0_{R} \}$ and (ii) $a$ is right-invertible, then $a$ is automatically left-invertible. Indeed, suppose that $ab = 1_{R}$ for some $b \in R$. Then $$a(1_{R} - ba) = a - a(ba) = a - (ab)a = a - 1_{R} \cdot a = a - a = 0_{R}.$$ As $a^{\circ} = \{ 0_{R} \}$, it follows that $1_{R} - ba = 0_{R}$, or equivalently, $ba = 1_{R}$. Therefore, $a$ is left-invertible.

The foregoing argument shows why right-invertibles in a square-matrix ring over a field must also be left-invertibles. Let $A \in {\text{M}_{n}}(\mathbb{F})$ be right-invertible, i.e., $AB = \mathbf{I}_{n}$ for some $B \in {\text{M}_{n}}(\mathbb{F})$. Viewing $A$ and $B$ respectively as linear transformations $T_{A}$ and $T_{B}$ on $\mathbb{F}^{n}$, we have $T_{A} \circ T_{B} = \text{id}_{\mathbb{F}^{n}}$. Hence, $T_{B}$ is injective and $T_{A}$ is surjective. However, the Dimension Theorem from linear algebra tells us that surjective linear operators on finite-dimensional vector spaces are also injective, so \begin{align} A^{\circ} &= \{ X \in {\text{M}_{n}}(\mathbb{F}) \,|\, AX = 0_{{\text{M}_{n}}(\mathbb{F})} \} \\ &= \{ X \in {\text{M}_{n}}(\mathbb{F}) \,|\, T_{A} \circ T_{X} = 0_{\mathcal{L}(\mathbb{F}^{n},\mathbb{F}^{n})} \} \\ &= \{ X \in {\text{M}_{n}}(\mathbb{F}) \,|\, T_{X} = 0_{\mathcal{L}(\mathbb{F}^{n},\mathbb{F}^{n})} \} \\ &= \{ 0_{{\text{M}_{n}}(\mathbb{F})} \}. \end{align} Therefore, $BA = \mathbf{I}_{n}$, i.e., $A$ is left-invertible.

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You probably mean $p \in R [X] \setminus R$. –  Zhen Lin Oct 27 '12 at 11:28
Thanks very much! –  Spring Xiao Oct 27 '12 at 14:03
@ZhenLin: Thanks! I've made the correction. You're right. There might be invertible elements in $R$. –  Haskell Curry Oct 28 '12 at 2:13
One positive result is that if $a^\circ = \{0\}$ and if $a$ is right-invertible, then $a$ is invertible. Indeed, if $ab = 1$, then $a(1-ba) = a - (ab)a = a - a = 0$. So $1-ba \in a^{\circ}$, which means that $ba = 1 = ab$. Hence $a$ is invertible with $a^{-1} = b$.