Mixing
Can every monoid be turned into a ring?
$begingroup$
It’s well-known that $mathbb{Q} / mathbb{Z}$ is an example of an abelian group which is not isomorphic to the additive group of any ring. But my question is, does there exist a monoid which is not isomorphic to the multiplicative monoid of any ring?
Or can you always find a binary operation $+$ which turns a monoid into a ring?
abstract-algebra ring-theory monoid
asked Jan 16 at 5:57
Keshav SrinivasanKeshav Srinivasan
2,35121444
$endgroup$
add a comment |
$begingroup$
It’s well-known that $mathbb{Q} / mathbb{Z}$ is an example of an abelian group which is not isomorphic to the additive group of any ring. But my question is, does there exist a monoid which is not isomorphic to the multiplicative monoid of any ring?
Or can you always find a binary operation $+$ which turns a monoid into a ring?
abstract-algebra ring-theory monoid
asked Jan 16 at 5:57
Keshav SrinivasanKeshav Srinivasan
2,35121444
$endgroup$
add a comment |
$begingroup$
It’s well-known that $mathbb{Q} / mathbb{Z}$ is an example of an abelian group which is not isomorphic to the additive group of any ring. But my question is, does there exist a monoid which is not isomorphic to the multiplicative monoid of any ring?
Or can you always find a binary operation $+$ which turns a monoid into a ring?
abstract-algebra ring-theory monoid
asked Jan 16 at 5:57
Keshav SrinivasanKeshav Srinivasan
2,35121444
$endgroup$
It’s well-known that $mathbb{Q} / mathbb{Z}$ is an example of an abelian group which is not isomorphic to the additive group of any ring. But my question is, does there exist a monoid which is not isomorphic to the multiplicative monoid of any ring?
Or can you always find a binary operation $+$ which turns a monoid into a ring?
abstract-algebra ring-theory monoid
abstract-algebra ring-theory monoid
asked Jan 16 at 5:57
Keshav SrinivasanKeshav Srinivasan
2,35121444
asked Jan 16 at 5:57
Keshav SrinivasanKeshav Srinivasan
2,35121444
asked Jan 16 at 5:57
Keshav SrinivasanKeshav Srinivasan
2,35121444
asked Jan 16 at 5:57
Keshav SrinivasanKeshav Srinivasan
2,35121444
asked Jan 16 at 5:57
Keshav SrinivasanKeshav Srinivasan
2,35121444
2,35121444
add a comment |
add a comment |
1 Answer
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$begingroup$
Well, an obvious necessary condition for a monoid $M$ to admit a ring structure is the existence of an element $0in M$ such that $0cdot x=xcdot 0=0$ for all $xin M$ (an absorbing element). This is not true of most monoids (for instance, it is not true of any nontrivial group).
Even this condition is not sufficient, though. For instance, consider the monoid $M={0,1,2}$ with operation $min$ (so $0$ is the absorbing element and $2$ is the identity element). This does not admit a ring structure, since the only ring with $3$ elements up to isomorphism is $mathbb{Z}/(3)$ and $M$ is not multiplicatively isomorphic to $mathbb{Z}/(3)$ (the non-absorbing non-identity element of $mathbb{Z}/(3)$ has an inverse, but the non-absorbing non-identity element of $M$ does not).
answered Jan 16 at 6:10
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Well, an obvious necessary condition for a monoid $M$ to admit a ring structure is the existence of an element $0in M$ such that $0cdot x=xcdot 0=0$ for all $xin M$ (an absorbing element). This is not true of most monoids (for instance, it is not true of any nontrivial group).
Even this condition is not sufficient, though. For instance, consider the monoid $M={0,1,2}$ with operation $min$ (so $0$ is the absorbing element and $2$ is the identity element). This does not admit a ring structure, since the only ring with $3$ elements up to isomorphism is $mathbb{Z}/(3)$ and $M$ is not multiplicatively isomorphic to $mathbb{Z}/(3)$ (the non-absorbing non-identity element of $mathbb{Z}/(3)$ has an inverse, but the non-absorbing non-identity element of $M$ does not).
answered Jan 16 at 6:10
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
Well, an obvious necessary condition for a monoid $M$ to admit a ring structure is the existence of an element $0in M$ such that $0cdot x=xcdot 0=0$ for all $xin M$ (an absorbing element). This is not true of most monoids (for instance, it is not true of any nontrivial group).
Even this condition is not sufficient, though. For instance, consider the monoid $M={0,1,2}$ with operation $min$ (so $0$ is the absorbing element and $2$ is the identity element). This does not admit a ring structure, since the only ring with $3$ elements up to isomorphism is $mathbb{Z}/(3)$ and $M$ is not multiplicatively isomorphic to $mathbb{Z}/(3)$ (the non-absorbing non-identity element of $mathbb{Z}/(3)$ has an inverse, but the non-absorbing non-identity element of $M$ does not).
answered Jan 16 at 6:10
Eric WofseyEric Wofsey
187k14215344
$endgroup$
add a comment |
$begingroup$
Well, an obvious necessary condition for a monoid $M$ to admit a ring structure is the existence of an element $0in M$ such that $0cdot x=xcdot 0=0$ for all $xin M$ (an absorbing element). This is not true of most monoids (for instance, it is not true of any nontrivial group).
Even this condition is not sufficient, though. For instance, consider the monoid $M={0,1,2}$ with operation $min$ (so $0$ is the absorbing element and $2$ is the identity element). This does not admit a ring structure, since the only ring with $3$ elements up to isomorphism is $mathbb{Z}/(3)$ and $M$ is not multiplicatively isomorphic to $mathbb{Z}/(3)$ (the non-absorbing non-identity element of $mathbb{Z}/(3)$ has an inverse, but the non-absorbing non-identity element of $M$ does not).
answered Jan 16 at 6:10
Eric WofseyEric Wofsey
187k14215344
$endgroup$
Well, an obvious necessary condition for a monoid $M$ to admit a ring structure is the existence of an element $0in M$ such that $0cdot x=xcdot 0=0$ for all $xin M$ (an absorbing element). This is not true of most monoids (for instance, it is not true of any nontrivial group).
Even this condition is not sufficient, though. For instance, consider the monoid $M={0,1,2}$ with operation $min$ (so $0$ is the absorbing element and $2$ is the identity element). This does not admit a ring structure, since the only ring with $3$ elements up to isomorphism is $mathbb{Z}/(3)$ and $M$ is not multiplicatively isomorphic to $mathbb{Z}/(3)$ (the non-absorbing non-identity element of $mathbb{Z}/(3)$ has an inverse, but the non-absorbing non-identity element of $M$ does not).
answered Jan 16 at 6:10
Eric WofseyEric Wofsey
187k14215344
answered Jan 16 at 6:10
Eric WofseyEric Wofsey
187k14215344
answered Jan 16 at 6:10
Eric WofseyEric Wofsey
187k14215344
answered Jan 16 at 6:10
Eric WofseyEric Wofsey
187k14215344
187k14215344
add a comment |
add a comment |
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