Mixing
Are simple commutative monoids monogeneous?
$begingroup$
Let $M$ be a simple commutative monoid.
Is there a surjective monoid morphism $mathbb Nto M$?
If non-monogeneous simple commutative monoids do exist, what's known about them?
Edit. The commutative monoid $(M,+)$ is simple if it satisfies the following equivalent conditions:
$bullet$ $M$ admits exactly two congruences,
$bullet$ up to isomorphism, there are exactly two surjective morphisms $Mto N$,
$bullet$ $Mne0$ and any nonzero surjective morphisms $Mto N$ is an isomorphism.
abstract-algebra monoid
asked Jan 7 at 13:39
Pierre-Yves GaillardPierre-Yves Gaillard
13.3k23181
$endgroup$
add a comment |
$begingroup$
Let $M$ be a simple commutative monoid.
Is there a surjective monoid morphism $mathbb Nto M$?
If non-monogeneous simple commutative monoids do exist, what's known about them?
Edit. The commutative monoid $(M,+)$ is simple if it satisfies the following equivalent conditions:
$bullet$ $M$ admits exactly two congruences,
$bullet$ up to isomorphism, there are exactly two surjective morphisms $Mto N$,
$bullet$ $Mne0$ and any nonzero surjective morphisms $Mto N$ is an isomorphism.
abstract-algebra monoid
asked Jan 7 at 13:39
Pierre-Yves GaillardPierre-Yves Gaillard
13.3k23181
$endgroup$
add a comment |
$begingroup$
Let $M$ be a simple commutative monoid.
Is there a surjective monoid morphism $mathbb Nto M$?
If non-monogeneous simple commutative monoids do exist, what's known about them?
Edit. The commutative monoid $(M,+)$ is simple if it satisfies the following equivalent conditions:
$bullet$ $M$ admits exactly two congruences,
$bullet$ up to isomorphism, there are exactly two surjective morphisms $Mto N$,
$bullet$ $Mne0$ and any nonzero surjective morphisms $Mto N$ is an isomorphism.
abstract-algebra monoid
asked Jan 7 at 13:39
Pierre-Yves GaillardPierre-Yves Gaillard
13.3k23181
$endgroup$
Let $M$ be a simple commutative monoid.
Is there a surjective monoid morphism $mathbb Nto M$?
If non-monogeneous simple commutative monoids do exist, what's known about them?
Edit. The commutative monoid $(M,+)$ is simple if it satisfies the following equivalent conditions:
$bullet$ $M$ admits exactly two congruences,
$bullet$ up to isomorphism, there are exactly two surjective morphisms $Mto N$,
$bullet$ $Mne0$ and any nonzero surjective morphisms $Mto N$ is an isomorphism.
abstract-algebra monoid
abstract-algebra monoid
asked Jan 7 at 13:39
Pierre-Yves GaillardPierre-Yves Gaillard
13.3k23181
asked Jan 7 at 13:39
Pierre-Yves GaillardPierre-Yves Gaillard
13.3k23181
edited Jan 7 at 20:54
Pierre-Yves Gaillard
asked Jan 7 at 13:39
Pierre-Yves GaillardPierre-Yves Gaillard
13.3k23181
asked Jan 7 at 13:39
Pierre-Yves GaillardPierre-Yves Gaillard
13.3k23181
asked Jan 7 at 13:39
Pierre-Yves GaillardPierre-Yves Gaillard
13.3k23181
13.3k23181
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Let $L={0,1}$, considered as a commutative monoid under multiplication. If $M$ is any commutative monoid, there is a homomorphism $f:Mto L$ which sends all invertible elements to $1$ and all non-invertible elements to $0$. If $M$ is simple, then either $f$ must be an isomorphism or $f$ must fail to be surjective.
If $f$ is not surjective, that means every element of $M$ is invertible so $M$ is a group. Then $M$ must also be simple as a abelian group, so it is cyclic of prime order.
So, the only simple commutative monoids (up to isomorphism) are $L$ and cyclic groups of prime order. In particular, they are all generated by a single element.
answered Jan 7 at 20:55
Eric WofseyEric Wofsey
183k13211338
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $L={0,1}$, considered as a commutative monoid under multiplication. If $M$ is any commutative monoid, there is a homomorphism $f:Mto L$ which sends all invertible elements to $1$ and all non-invertible elements to $0$. If $M$ is simple, then either $f$ must be an isomorphism or $f$ must fail to be surjective.
If $f$ is not surjective, that means every element of $M$ is invertible so $M$ is a group. Then $M$ must also be simple as a abelian group, so it is cyclic of prime order.
So, the only simple commutative monoids (up to isomorphism) are $L$ and cyclic groups of prime order. In particular, they are all generated by a single element.
answered Jan 7 at 20:55
Eric WofseyEric Wofsey
183k13211338
$endgroup$
add a comment |
$begingroup$
Let $L={0,1}$, considered as a commutative monoid under multiplication. If $M$ is any commutative monoid, there is a homomorphism $f:Mto L$ which sends all invertible elements to $1$ and all non-invertible elements to $0$. If $M$ is simple, then either $f$ must be an isomorphism or $f$ must fail to be surjective.
If $f$ is not surjective, that means every element of $M$ is invertible so $M$ is a group. Then $M$ must also be simple as a abelian group, so it is cyclic of prime order.
So, the only simple commutative monoids (up to isomorphism) are $L$ and cyclic groups of prime order. In particular, they are all generated by a single element.
answered Jan 7 at 20:55
Eric WofseyEric Wofsey
183k13211338
$endgroup$
add a comment |
$begingroup$
Let $L={0,1}$, considered as a commutative monoid under multiplication. If $M$ is any commutative monoid, there is a homomorphism $f:Mto L$ which sends all invertible elements to $1$ and all non-invertible elements to $0$. If $M$ is simple, then either $f$ must be an isomorphism or $f$ must fail to be surjective.
If $f$ is not surjective, that means every element of $M$ is invertible so $M$ is a group. Then $M$ must also be simple as a abelian group, so it is cyclic of prime order.
So, the only simple commutative monoids (up to isomorphism) are $L$ and cyclic groups of prime order. In particular, they are all generated by a single element.
answered Jan 7 at 20:55
Eric WofseyEric Wofsey
183k13211338
$endgroup$
Let $L={0,1}$, considered as a commutative monoid under multiplication. If $M$ is any commutative monoid, there is a homomorphism $f:Mto L$ which sends all invertible elements to $1$ and all non-invertible elements to $0$. If $M$ is simple, then either $f$ must be an isomorphism or $f$ must fail to be surjective.
If $f$ is not surjective, that means every element of $M$ is invertible so $M$ is a group. Then $M$ must also be simple as a abelian group, so it is cyclic of prime order.
So, the only simple commutative monoids (up to isomorphism) are $L$ and cyclic groups of prime order. In particular, they are all generated by a single element.
answered Jan 7 at 20:55
Eric WofseyEric Wofsey
183k13211338
answered Jan 7 at 20:55
Eric WofseyEric Wofsey
183k13211338
answered Jan 7 at 20:55
Eric WofseyEric Wofsey
183k13211338
answered Jan 7 at 20:55
Eric WofseyEric Wofsey
183k13211338
183k13211338
add a comment |
add a comment |
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