Are multiplicative monoids of different rings isomorphic?
$begingroup$
I have some algebraic problem and hope some of you can help me!
The problem is about rings and their multiplicative monoids (semi-groups with neutral element $e$). So $M(mathbb{Z}), M(mathbb{Z}_2[x])$ and $M(mathbb{Q}[[x]])$ denote the multiplicative monoids (factorized Modulo association).
The task is to find out whether these monoids are isomorphic or not.
My first idea was to use the Chinese Remainder Theorem, but it only states something for ideals of one and the same ring, which nothing matching to the given problem here.
Furthermore I tried to exclude at least one possible isomorphism by compairing the cardinal number of the monoids, but they seem to be the same.
What else can I do to find out which of them are isomorphic?
Thank you very much for your help!
abstract-algebra ring-theory monoid ring-isomorphism
asked Dec 2 '18 at 18:10
pcalcpcalc
27518
$endgroup$
add a comment |
$begingroup$
I have some algebraic problem and hope some of you can help me!
The problem is about rings and their multiplicative monoids (semi-groups with neutral element $e$). So $M(mathbb{Z}), M(mathbb{Z}_2[x])$ and $M(mathbb{Q}[[x]])$ denote the multiplicative monoids (factorized Modulo association).
The task is to find out whether these monoids are isomorphic or not.
My first idea was to use the Chinese Remainder Theorem, but it only states something for ideals of one and the same ring, which nothing matching to the given problem here.
Furthermore I tried to exclude at least one possible isomorphism by compairing the cardinal number of the monoids, but they seem to be the same.
What else can I do to find out which of them are isomorphic?
Thank you very much for your help!
abstract-algebra ring-theory monoid ring-isomorphism
asked Dec 2 '18 at 18:10
pcalcpcalc
27518
$endgroup$
$begingroup$
Are you using $M$ to mean “moniod of” or something? It also suggests “matrix ring” but then that raises the question of side lengths.
$endgroup$
– rschwieb
Dec 2 '18 at 20:58
$begingroup$
Hey! Yes, with $M(R)$ I mean monoid of $R$ factorized modulo association
$endgroup$
– pcalc
Dec 2 '18 at 22:09
$begingroup$
where might I find a concise definition of "modulo association"?
$endgroup$
– rschwieb
Dec 3 '18 at 14:00
add a comment |
$begingroup$
I have some algebraic problem and hope some of you can help me!
The problem is about rings and their multiplicative monoids (semi-groups with neutral element $e$). So $M(mathbb{Z}), M(mathbb{Z}_2[x])$ and $M(mathbb{Q}[[x]])$ denote the multiplicative monoids (factorized Modulo association).
The task is to find out whether these monoids are isomorphic or not.
My first idea was to use the Chinese Remainder Theorem, but it only states something for ideals of one and the same ring, which nothing matching to the given problem here.
Furthermore I tried to exclude at least one possible isomorphism by compairing the cardinal number of the monoids, but they seem to be the same.
What else can I do to find out which of them are isomorphic?
Thank you very much for your help!
abstract-algebra ring-theory monoid ring-isomorphism
asked Dec 2 '18 at 18:10
pcalcpcalc
27518
$endgroup$
I have some algebraic problem and hope some of you can help me!
The problem is about rings and their multiplicative monoids (semi-groups with neutral element $e$). So $M(mathbb{Z}), M(mathbb{Z}_2[x])$ and $M(mathbb{Q}[[x]])$ denote the multiplicative monoids (factorized Modulo association).
The task is to find out whether these monoids are isomorphic or not.
My first idea was to use the Chinese Remainder Theorem, but it only states something for ideals of one and the same ring, which nothing matching to the given problem here.
Furthermore I tried to exclude at least one possible isomorphism by compairing the cardinal number of the monoids, but they seem to be the same.
What else can I do to find out which of them are isomorphic?
Thank you very much for your help!
abstract-algebra ring-theory monoid ring-isomorphism
abstract-algebra ring-theory monoid ring-isomorphism
asked Dec 2 '18 at 18:10
pcalcpcalc
27518
asked Dec 2 '18 at 18:10
pcalcpcalc
27518
asked Dec 2 '18 at 18:10
pcalcpcalc
27518
asked Dec 2 '18 at 18:10
pcalcpcalc
27518
asked Dec 2 '18 at 18:10
pcalcpcalc
27518
27518
$begingroup$
Are you using $M$ to mean “moniod of” or something? It also suggests “matrix ring” but then that raises the question of side lengths.
$endgroup$
– rschwieb
Dec 2 '18 at 20:58
$begingroup$
Hey! Yes, with $M(R)$ I mean monoid of $R$ factorized modulo association
$endgroup$
– pcalc
Dec 2 '18 at 22:09
$begingroup$
where might I find a concise definition of "modulo association"?
$endgroup$
– rschwieb
Dec 3 '18 at 14:00
add a comment |
$begingroup$
Are you using $M$ to mean “moniod of” or something? It also suggests “matrix ring” but then that raises the question of side lengths.
$endgroup$
– rschwieb
Dec 2 '18 at 20:58
$begingroup$
Hey! Yes, with $M(R)$ I mean monoid of $R$ factorized modulo association
$endgroup$
– pcalc
Dec 2 '18 at 22:09
$begingroup$
where might I find a concise definition of "modulo association"?
$endgroup$
– rschwieb
Dec 3 '18 at 14:00
$begingroup$
Are you using $M$ to mean “moniod of” or something? It also suggests “matrix ring” but then that raises the question of side lengths.
$endgroup$
– rschwieb
Dec 2 '18 at 20:58
$begingroup$
Are you using $M$ to mean “moniod of” or something? It also suggests “matrix ring” but then that raises the question of side lengths.
$endgroup$
– rschwieb
Dec 2 '18 at 20:58
$begingroup$
Hey! Yes, with $M(R)$ I mean monoid of $R$ factorized modulo association
$endgroup$
– pcalc
Dec 2 '18 at 22:09
$begingroup$
Hey! Yes, with $M(R)$ I mean monoid of $R$ factorized modulo association
$endgroup$
– pcalc
Dec 2 '18 at 22:09
$begingroup$
where might I find a concise definition of "modulo association"?
$endgroup$
– rschwieb
Dec 3 '18 at 14:00
$begingroup$
where might I find a concise definition of "modulo association"?
$endgroup$
– rschwieb
Dec 3 '18 at 14:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's find what's $M(mathbb{Q}[[x]])$. The invertible elements in $mathbb{Q}[[x]]$ are precisely the power series with nonzero constant term. So, modulo association, we only get
$$
[0],quad [1],quad [x],quad [x^2],quad dotsc
$$
which is the free monoid over one generator with the addition of an absorbing element $[0]$. (With $[t]$ I'll denote the equivalence class modulo association.)
The monoid $M(mathbb{Z})$ instead is the free monoid on a countable basis (the primes) with the addition of an absorbing element (fundamental theorem of arithmetic).
Similarly, $M(mathbb{Z}_2[x])$ is the free monoid on a countable basis with the addition of an absorbing element, because $mathbb{Z}_2[x]$ is a principal ideal domain, hence a unique factorization domain, with a countable number of irreducible elements (up to association), because there is at least an irreducible monic polynomial for every positive degree.
Actually, the last mentioned fact is not necessary: what we need is to prove that there are infinitely many irreducible monic polynomials in $mathbb{Z}[x]$. The proof is essentially the same as Euclid's proof for the integers: if $p_1(x), dots,p_n(x)$ are monic irreducible polynomials, then
$$
P(x)=p_1(x)p_2(x)dots p_n(x)+1
$$
is a monic polynomial that has positive degree, so it is divisible by an irreducible (monic) polynomial, which cannot be in the list $p_1(x),dots,p_n(x)$. Thus no finite list of irreducible polynomials contains all irreducible monic polynomials. Consequently there are infinitely many irreducible monic polynomials. Being monic, they are pairwise not associate. Since the ring $mathbb{Z}_2[x]$ is countable, we have proved the statement that $M(mathbb{Z}_2[x])$ is the free monoid on a countable basis with an absorbing element added.
answered Dec 2 '18 at 21:52
egregegreg
180k1485202
$endgroup$
$begingroup$
Hi and thank you very much for your respond! I think I can follow your explaination about $M(mathbb{Z}_2[x]$ and $mathbb{Z}$. My conclusio would be, that both of them have an infinite but countable basis elements in the basis, over which they are free, so they are isomorphic (?). I didn't get your first argument, because I don't see why the elements with nonzero constant term are the invertible ones. Could you explain this fact in more detail? (or have I overlooked something, that is quite obvious?)
$endgroup$
– pcalc
Dec 2 '18 at 22:20
$begingroup$
@pcalc Yes, two free monoids on a countable basis are isomorphic (the absorbing element does not pose problems). For the invertible power series, see math.stackexchange.com/questions/245325
$endgroup$
– egreg
Dec 2 '18 at 22:28
$begingroup$
aah thanks! So basically all of them are isomorphic, because of their basis' cardinal number?
$endgroup$
– pcalc
Dec 2 '18 at 22:32
$begingroup$
@pcalc No, the monoid $M(mathbb{Q}[[x]])$ is free over one element: the basis is $[x]$.
$endgroup$
– egreg
Dec 2 '18 at 22:32
$begingroup$
Hm, interesting! I missed the bit about modulo association. Even after searching the internet I'm still not quite clear what it means.
$endgroup$
– rschwieb
Dec 3 '18 at 1:51
|
show 5 more comments
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's find what's $M(mathbb{Q}[[x]])$. The invertible elements in $mathbb{Q}[[x]]$ are precisely the power series with nonzero constant term. So, modulo association, we only get
$$
[0],quad [1],quad [x],quad [x^2],quad dotsc
$$
which is the free monoid over one generator with the addition of an absorbing element $[0]$. (With $[t]$ I'll denote the equivalence class modulo association.)
The monoid $M(mathbb{Z})$ instead is the free monoid on a countable basis (the primes) with the addition of an absorbing element (fundamental theorem of arithmetic).
Similarly, $M(mathbb{Z}_2[x])$ is the free monoid on a countable basis with the addition of an absorbing element, because $mathbb{Z}_2[x]$ is a principal ideal domain, hence a unique factorization domain, with a countable number of irreducible elements (up to association), because there is at least an irreducible monic polynomial for every positive degree.
Actually, the last mentioned fact is not necessary: what we need is to prove that there are infinitely many irreducible monic polynomials in $mathbb{Z}[x]$. The proof is essentially the same as Euclid's proof for the integers: if $p_1(x), dots,p_n(x)$ are monic irreducible polynomials, then
$$
P(x)=p_1(x)p_2(x)dots p_n(x)+1
$$
is a monic polynomial that has positive degree, so it is divisible by an irreducible (monic) polynomial, which cannot be in the list $p_1(x),dots,p_n(x)$. Thus no finite list of irreducible polynomials contains all irreducible monic polynomials. Consequently there are infinitely many irreducible monic polynomials. Being monic, they are pairwise not associate. Since the ring $mathbb{Z}_2[x]$ is countable, we have proved the statement that $M(mathbb{Z}_2[x])$ is the free monoid on a countable basis with an absorbing element added.
answered Dec 2 '18 at 21:52
egregegreg
180k1485202
$endgroup$
$begingroup$
Hi and thank you very much for your respond! I think I can follow your explaination about $M(mathbb{Z}_2[x]$ and $mathbb{Z}$. My conclusio would be, that both of them have an infinite but countable basis elements in the basis, over which they are free, so they are isomorphic (?). I didn't get your first argument, because I don't see why the elements with nonzero constant term are the invertible ones. Could you explain this fact in more detail? (or have I overlooked something, that is quite obvious?)
$endgroup$
– pcalc
Dec 2 '18 at 22:20
$begingroup$
@pcalc Yes, two free monoids on a countable basis are isomorphic (the absorbing element does not pose problems). For the invertible power series, see math.stackexchange.com/questions/245325
$endgroup$
– egreg
Dec 2 '18 at 22:28
$begingroup$
aah thanks! So basically all of them are isomorphic, because of their basis' cardinal number?
$endgroup$
– pcalc
Dec 2 '18 at 22:32
$begingroup$
@pcalc No, the monoid $M(mathbb{Q}[[x]])$ is free over one element: the basis is $[x]$.
$endgroup$
– egreg
Dec 2 '18 at 22:32
$begingroup$
Hm, interesting! I missed the bit about modulo association. Even after searching the internet I'm still not quite clear what it means.
$endgroup$
– rschwieb
Dec 3 '18 at 1:51
|
show 5 more comments
$begingroup$
Let's find what's $M(mathbb{Q}[[x]])$. The invertible elements in $mathbb{Q}[[x]]$ are precisely the power series with nonzero constant term. So, modulo association, we only get
$$
[0],quad [1],quad [x],quad [x^2],quad dotsc
$$
which is the free monoid over one generator with the addition of an absorbing element $[0]$. (With $[t]$ I'll denote the equivalence class modulo association.)
The monoid $M(mathbb{Z})$ instead is the free monoid on a countable basis (the primes) with the addition of an absorbing element (fundamental theorem of arithmetic).
Similarly, $M(mathbb{Z}_2[x])$ is the free monoid on a countable basis with the addition of an absorbing element, because $mathbb{Z}_2[x]$ is a principal ideal domain, hence a unique factorization domain, with a countable number of irreducible elements (up to association), because there is at least an irreducible monic polynomial for every positive degree.
Actually, the last mentioned fact is not necessary: what we need is to prove that there are infinitely many irreducible monic polynomials in $mathbb{Z}[x]$. The proof is essentially the same as Euclid's proof for the integers: if $p_1(x), dots,p_n(x)$ are monic irreducible polynomials, then
$$
P(x)=p_1(x)p_2(x)dots p_n(x)+1
$$
is a monic polynomial that has positive degree, so it is divisible by an irreducible (monic) polynomial, which cannot be in the list $p_1(x),dots,p_n(x)$. Thus no finite list of irreducible polynomials contains all irreducible monic polynomials. Consequently there are infinitely many irreducible monic polynomials. Being monic, they are pairwise not associate. Since the ring $mathbb{Z}_2[x]$ is countable, we have proved the statement that $M(mathbb{Z}_2[x])$ is the free monoid on a countable basis with an absorbing element added.
answered Dec 2 '18 at 21:52
egregegreg
180k1485202
$endgroup$
$begingroup$
Hi and thank you very much for your respond! I think I can follow your explaination about $M(mathbb{Z}_2[x]$ and $mathbb{Z}$. My conclusio would be, that both of them have an infinite but countable basis elements in the basis, over which they are free, so they are isomorphic (?). I didn't get your first argument, because I don't see why the elements with nonzero constant term are the invertible ones. Could you explain this fact in more detail? (or have I overlooked something, that is quite obvious?)
$endgroup$
– pcalc
Dec 2 '18 at 22:20
$begingroup$
@pcalc Yes, two free monoids on a countable basis are isomorphic (the absorbing element does not pose problems). For the invertible power series, see math.stackexchange.com/questions/245325
$endgroup$
– egreg
Dec 2 '18 at 22:28
$begingroup$
aah thanks! So basically all of them are isomorphic, because of their basis' cardinal number?
$endgroup$
– pcalc
Dec 2 '18 at 22:32
$begingroup$
@pcalc No, the monoid $M(mathbb{Q}[[x]])$ is free over one element: the basis is $[x]$.
$endgroup$
– egreg
Dec 2 '18 at 22:32
$begingroup$
Hm, interesting! I missed the bit about modulo association. Even after searching the internet I'm still not quite clear what it means.
$endgroup$
– rschwieb
Dec 3 '18 at 1:51
|
show 5 more comments
$begingroup$
Let's find what's $M(mathbb{Q}[[x]])$. The invertible elements in $mathbb{Q}[[x]]$ are precisely the power series with nonzero constant term. So, modulo association, we only get
$$
[0],quad [1],quad [x],quad [x^2],quad dotsc
$$
which is the free monoid over one generator with the addition of an absorbing element $[0]$. (With $[t]$ I'll denote the equivalence class modulo association.)
The monoid $M(mathbb{Z})$ instead is the free monoid on a countable basis (the primes) with the addition of an absorbing element (fundamental theorem of arithmetic).
Similarly, $M(mathbb{Z}_2[x])$ is the free monoid on a countable basis with the addition of an absorbing element, because $mathbb{Z}_2[x]$ is a principal ideal domain, hence a unique factorization domain, with a countable number of irreducible elements (up to association), because there is at least an irreducible monic polynomial for every positive degree.
Actually, the last mentioned fact is not necessary: what we need is to prove that there are infinitely many irreducible monic polynomials in $mathbb{Z}[x]$. The proof is essentially the same as Euclid's proof for the integers: if $p_1(x), dots,p_n(x)$ are monic irreducible polynomials, then
$$
P(x)=p_1(x)p_2(x)dots p_n(x)+1
$$
is a monic polynomial that has positive degree, so it is divisible by an irreducible (monic) polynomial, which cannot be in the list $p_1(x),dots,p_n(x)$. Thus no finite list of irreducible polynomials contains all irreducible monic polynomials. Consequently there are infinitely many irreducible monic polynomials. Being monic, they are pairwise not associate. Since the ring $mathbb{Z}_2[x]$ is countable, we have proved the statement that $M(mathbb{Z}_2[x])$ is the free monoid on a countable basis with an absorbing element added.
answered Dec 2 '18 at 21:52
egregegreg
180k1485202
$endgroup$
Let's find what's $M(mathbb{Q}[[x]])$. The invertible elements in $mathbb{Q}[[x]]$ are precisely the power series with nonzero constant term. So, modulo association, we only get
$$
[0],quad [1],quad [x],quad [x^2],quad dotsc
$$
which is the free monoid over one generator with the addition of an absorbing element $[0]$. (With $[t]$ I'll denote the equivalence class modulo association.)
The monoid $M(mathbb{Z})$ instead is the free monoid on a countable basis (the primes) with the addition of an absorbing element (fundamental theorem of arithmetic).
Similarly, $M(mathbb{Z}_2[x])$ is the free monoid on a countable basis with the addition of an absorbing element, because $mathbb{Z}_2[x]$ is a principal ideal domain, hence a unique factorization domain, with a countable number of irreducible elements (up to association), because there is at least an irreducible monic polynomial for every positive degree.
Actually, the last mentioned fact is not necessary: what we need is to prove that there are infinitely many irreducible monic polynomials in $mathbb{Z}[x]$. The proof is essentially the same as Euclid's proof for the integers: if $p_1(x), dots,p_n(x)$ are monic irreducible polynomials, then
$$
P(x)=p_1(x)p_2(x)dots p_n(x)+1
$$
is a monic polynomial that has positive degree, so it is divisible by an irreducible (monic) polynomial, which cannot be in the list $p_1(x),dots,p_n(x)$. Thus no finite list of irreducible polynomials contains all irreducible monic polynomials. Consequently there are infinitely many irreducible monic polynomials. Being monic, they are pairwise not associate. Since the ring $mathbb{Z}_2[x]$ is countable, we have proved the statement that $M(mathbb{Z}_2[x])$ is the free monoid on a countable basis with an absorbing element added.
answered Dec 2 '18 at 21:52
egregegreg
180k1485202
edited Dec 2 '18 at 22:05
answered Dec 2 '18 at 21:52
egregegreg
180k1485202
answered Dec 2 '18 at 21:52
egregegreg
180k1485202
answered Dec 2 '18 at 21:52
egregegreg
180k1485202
180k1485202
$begingroup$
Hi and thank you very much for your respond! I think I can follow your explaination about $M(mathbb{Z}_2[x]$ and $mathbb{Z}$. My conclusio would be, that both of them have an infinite but countable basis elements in the basis, over which they are free, so they are isomorphic (?). I didn't get your first argument, because I don't see why the elements with nonzero constant term are the invertible ones. Could you explain this fact in more detail? (or have I overlooked something, that is quite obvious?)
$endgroup$
– pcalc
Dec 2 '18 at 22:20
$begingroup$
@pcalc Yes, two free monoids on a countable basis are isomorphic (the absorbing element does not pose problems). For the invertible power series, see math.stackexchange.com/questions/245325
$endgroup$
– egreg
Dec 2 '18 at 22:28
$begingroup$
aah thanks! So basically all of them are isomorphic, because of their basis' cardinal number?
$endgroup$
– pcalc
Dec 2 '18 at 22:32
$begingroup$
@pcalc No, the monoid $M(mathbb{Q}[[x]])$ is free over one element: the basis is $[x]$.
$endgroup$
– egreg
Dec 2 '18 at 22:32
$begingroup$
Hm, interesting! I missed the bit about modulo association. Even after searching the internet I'm still not quite clear what it means.
$endgroup$
– rschwieb
Dec 3 '18 at 1:51
|
show 5 more comments
$begingroup$
Hi and thank you very much for your respond! I think I can follow your explaination about $M(mathbb{Z}_2[x]$ and $mathbb{Z}$. My conclusio would be, that both of them have an infinite but countable basis elements in the basis, over which they are free, so they are isomorphic (?). I didn't get your first argument, because I don't see why the elements with nonzero constant term are the invertible ones. Could you explain this fact in more detail? (or have I overlooked something, that is quite obvious?)
$endgroup$
– pcalc
Dec 2 '18 at 22:20
$begingroup$
@pcalc Yes, two free monoids on a countable basis are isomorphic (the absorbing element does not pose problems). For the invertible power series, see math.stackexchange.com/questions/245325
$endgroup$
– egreg
Dec 2 '18 at 22:28
$begingroup$
aah thanks! So basically all of them are isomorphic, because of their basis' cardinal number?
$endgroup$
– pcalc
Dec 2 '18 at 22:32
$begingroup$
@pcalc No, the monoid $M(mathbb{Q}[[x]])$ is free over one element: the basis is $[x]$.
$endgroup$
– egreg
Dec 2 '18 at 22:32
$begingroup$
Hm, interesting! I missed the bit about modulo association. Even after searching the internet I'm still not quite clear what it means.
$endgroup$
– rschwieb
Dec 3 '18 at 1:51
$begingroup$
Hi and thank you very much for your respond! I think I can follow your explaination about $M(mathbb{Z}_2[x]$ and $mathbb{Z}$. My conclusio would be, that both of them have an infinite but countable basis elements in the basis, over which they are free, so they are isomorphic (?). I didn't get your first argument, because I don't see why the elements with nonzero constant term are the invertible ones. Could you explain this fact in more detail? (or have I overlooked something, that is quite obvious?)
$endgroup$
– pcalc
Dec 2 '18 at 22:20
$begingroup$
Hi and thank you very much for your respond! I think I can follow your explaination about $M(mathbb{Z}_2[x]$ and $mathbb{Z}$. My conclusio would be, that both of them have an infinite but countable basis elements in the basis, over which they are free, so they are isomorphic (?). I didn't get your first argument, because I don't see why the elements with nonzero constant term are the invertible ones. Could you explain this fact in more detail? (or have I overlooked something, that is quite obvious?)
$endgroup$
– pcalc
Dec 2 '18 at 22:20
$begingroup$
@pcalc Yes, two free monoids on a countable basis are isomorphic (the absorbing element does not pose problems). For the invertible power series, see math.stackexchange.com/questions/245325
$endgroup$
– egreg
Dec 2 '18 at 22:28
$begingroup$
@pcalc Yes, two free monoids on a countable basis are isomorphic (the absorbing element does not pose problems). For the invertible power series, see math.stackexchange.com/questions/245325
$endgroup$
– egreg
Dec 2 '18 at 22:28
$begingroup$
aah thanks! So basically all of them are isomorphic, because of their basis' cardinal number?
$endgroup$
– pcalc
Dec 2 '18 at 22:32
$begingroup$
aah thanks! So basically all of them are isomorphic, because of their basis' cardinal number?
$endgroup$
– pcalc
Dec 2 '18 at 22:32
$begingroup$
@pcalc No, the monoid $M(mathbb{Q}[[x]])$ is free over one element: the basis is $[x]$.
$endgroup$
– egreg
Dec 2 '18 at 22:32
$begingroup$
@pcalc No, the monoid $M(mathbb{Q}[[x]])$ is free over one element: the basis is $[x]$.
$endgroup$
– egreg
Dec 2 '18 at 22:32
$begingroup$
Hm, interesting! I missed the bit about modulo association. Even after searching the internet I'm still not quite clear what it means.
$endgroup$
– rschwieb
Dec 3 '18 at 1:51
$begingroup$
Hm, interesting! I missed the bit about modulo association. Even after searching the internet I'm still not quite clear what it means.
$endgroup$
– rschwieb
Dec 3 '18 at 1:51
|
show 5 more comments
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$begingroup$
Are you using $M$ to mean “moniod of” or something? It also suggests “matrix ring” but then that raises the question of side lengths.
$endgroup$
– rschwieb
Dec 2 '18 at 20:58
$begingroup$
Hey! Yes, with $M(R)$ I mean monoid of $R$ factorized modulo association
$endgroup$
– pcalc
Dec 2 '18 at 22:09
$begingroup$
where might I find a concise definition of "modulo association"?
$endgroup$
– rschwieb
Dec 3 '18 at 14:00