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Can you help me to Show:

A unital ring $R$ contains an element which is right invertible but not left invertible if and only if $M_{2}(R)$ contains an invertible upper triangular matrix whose inverse is not upper triangular.

How can we use "If $ab = 1$, then $e := 1 − ba$ is an idempotent and $ae = eb = 0$" to prove the problem?

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The proof of the 'if' part will suggest a suitable matrix for the 'only if' part, so we'll proceed in that order.

Let $M =\left[ {\begin{array}{cc} x & y \\ 0 & z \\\end{array} } \right]$ and let its inverse be $N =\left[ {\begin{array}{cc} p & q \\ r & s \\\end{array} } \right]$, with $ r \neq 0$, so that $MN = NM = I $, where $I$ is the identity matrix in $M_2(R)$. Multiplying out $MN$ and $NM$ gives:

$MN =\left[ {\begin{array}{cc} xp + yr & xq + ys \\ zr & zs \\\end{array} } \right]$

$NM =\left[ {\begin{array}{cc} px & py + qz \\ rx & ry + sz \\\end{array} } \right]$

Equating both to $I$ gives 8 equations, one of which is: $zs = 1$. So if we can prove that $sz$ is not $1$, we will be done (because if an element has both left and right inverses, they should be equal). One of the remaining equations is $ry + sz = 1$, which means $ry$ shouldn't be $0$.

The proof of that is as follows: $zr = 0 \implies szr = 0 \implies (1 - ry)r = r(1 - yr) = 0 $. If either $ry$ or $yr$ is $0$, that would imply that $r$ is $0$ which contradicts the given information.

So $ 1 - sz = zs - sz = ry \neq 0 $ which means z is an element with a right inverse but not a left inverse (similarly, $x$ has a left inverse but no right inverse).

For the 'only if' part: Suppose that $ab = 1$ and $1 - ba = e \neq 0$. The previous paragraph suggests that we can take $z = p = a$ and $x = s = b$ so that $ry = e$. Let $y = 1 \implies r = e \neq 0 $, then $M$ and $N$ are:

$M =\left[ {\begin{array}{cc} b & 1 \\ 0 & a \\\end{array} } \right]$

$N =\left[ {\begin{array}{cc} a & -1 \\ e & b \\\end{array} } \right]$

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