The equivalent condition to an element being invertible in a ring
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Show that an element $u$ of a ring is invertible with $v=u^{-1}$ if and only if $uvu=u$ and $v$ is the only element satisfying this condition. This is part of an exercise from Jacobson.
abstract-algebra
asked Jan 25 at 3:00
Smart YaoSmart Yao
1369
$endgroup$
add a comment |
$begingroup$
Show that an element $u$ of a ring is invertible with $v=u^{-1}$ if and only if $uvu=u$ and $v$ is the only element satisfying this condition. This is part of an exercise from Jacobson.
abstract-algebra
asked Jan 25 at 3:00
Smart YaoSmart Yao
1369
$endgroup$
add a comment |
$begingroup$
Show that an element $u$ of a ring is invertible with $v=u^{-1}$ if and only if $uvu=u$ and $v$ is the only element satisfying this condition. This is part of an exercise from Jacobson.
abstract-algebra
asked Jan 25 at 3:00
Smart YaoSmart Yao
1369
$endgroup$
Show that an element $u$ of a ring is invertible with $v=u^{-1}$ if and only if $uvu=u$ and $v$ is the only element satisfying this condition. This is part of an exercise from Jacobson.
abstract-algebra
abstract-algebra
asked Jan 25 at 3:00
Smart YaoSmart Yao
1369
asked Jan 25 at 3:00
Smart YaoSmart Yao
1369
asked Jan 25 at 3:00
Smart YaoSmart Yao
1369
asked Jan 25 at 3:00
Smart YaoSmart Yao
1369
asked Jan 25 at 3:00
Smart YaoSmart Yao
1369
1369
add a comment |
add a comment |
1 Answer
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$begingroup$
$u(1-vu+v)u=uvu=u$ and by uniqueness $1-vu+v=v$. Similarly, $u(1-uv+v)u=uvu=u$ implies $1-uv+v=v$. The proof completes.
answered Jan 25 at 3:15
Smart YaoSmart Yao
1369
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$u(1-vu+v)u=uvu=u$ and by uniqueness $1-vu+v=v$. Similarly, $u(1-uv+v)u=uvu=u$ implies $1-uv+v=v$. The proof completes.
answered Jan 25 at 3:15
Smart YaoSmart Yao
1369
$endgroup$
add a comment |
$begingroup$
$u(1-vu+v)u=uvu=u$ and by uniqueness $1-vu+v=v$. Similarly, $u(1-uv+v)u=uvu=u$ implies $1-uv+v=v$. The proof completes.
answered Jan 25 at 3:15
Smart YaoSmart Yao
1369
$endgroup$
add a comment |
$begingroup$
$u(1-vu+v)u=uvu=u$ and by uniqueness $1-vu+v=v$. Similarly, $u(1-uv+v)u=uvu=u$ implies $1-uv+v=v$. The proof completes.
answered Jan 25 at 3:15
Smart YaoSmart Yao
1369
$endgroup$
$u(1-vu+v)u=uvu=u$ and by uniqueness $1-vu+v=v$. Similarly, $u(1-uv+v)u=uvu=u$ implies $1-uv+v=v$. The proof completes.
answered Jan 25 at 3:15
Smart YaoSmart Yao
1369
answered Jan 25 at 3:15
Smart YaoSmart Yao
1369
answered Jan 25 at 3:15
Smart YaoSmart Yao
1369
answered Jan 25 at 3:15
Smart YaoSmart Yao
1369
1369
add a comment |
add a comment |
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