Given a Multivariate Quotient Ring, can one find a Monoid ring with the same properties?
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Given a Polynomial ring, say $$S = R[x,y]/(x^2+x-1,y^3-y)$$ Is there a Monoid M and a ring $$T = R[M] cong S?$$ If there is such a Monoid what would it be? I am able to find one for univariate S:
$R$ is any ring with unity. $$S = R[x]/(p(x)),qquad n = deg(p)$$ assuming $p(x)$ is monic over $R$. The basis is of $T$ is $B[k]$ with $k in Bbb{Z}/nBbb{Z}$,
$R[Bbb{Z}/nBbb{Z}]$ with the basis action $f(a,b)$ being:
$$sum_{k=0}^{n-1} -c_{k}*B[k+a+b mod n] | a+b < min(a,b)
quad text{or}quad a+bquad text{otherwise}$$
abstract-algebra
asked Dec 28 '18 at 4:55
Malachi WadasMalachi Wadas
112
$endgroup$
|
show 5 more comments
$begingroup$
Given a Polynomial ring, say $$S = R[x,y]/(x^2+x-1,y^3-y)$$ Is there a Monoid M and a ring $$T = R[M] cong S?$$ If there is such a Monoid what would it be? I am able to find one for univariate S:
$R$ is any ring with unity. $$S = R[x]/(p(x)),qquad n = deg(p)$$ assuming $p(x)$ is monic over $R$. The basis is of $T$ is $B[k]$ with $k in Bbb{Z}/nBbb{Z}$,
$R[Bbb{Z}/nBbb{Z}]$ with the basis action $f(a,b)$ being:
$$sum_{k=0}^{n-1} -c_{k}*B[k+a+b mod n] | a+b < min(a,b)
quad text{or}quad a+bquad text{otherwise}$$
abstract-algebra
asked Dec 28 '18 at 4:55
Malachi WadasMalachi Wadas
112
$endgroup$
$begingroup$
The product of basis elements must agree with the operation of the monoid, so I don't think $R[x]/langle p(x)rangle$ works for all polynomials of degree $n$. It does work for $M=Bbb{Z}/nBbb{Z}$ when $p(x)=x^n-1$ (other polynomials may also work depending on the ring $R$),
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:07
$begingroup$
I’ve used sage math to test it my self and it works for any polynomial, the operation is associative as well, with 0 as the unit, I am currently working with R being The Reals or Complexes
$endgroup$
– Malachi Wadas
Dec 28 '18 at 7:13
$begingroup$
In the ring $R[M]$ with basis $B[m], min M$ the product must be $B[m_1]*B[m_2]=B[m_1cdot m_2]$. I don't think you have that. Here $m_1cdot m_2$ is the operation of your monoid $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:18
$begingroup$
I assume that your definition of a monoid ring parallels that of the group ring.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:21
1
$begingroup$
From what you given it seems any polynomials of the form $$x^{p} - x^{q} = 0; p > q $$ Form a Monoid ring.
$endgroup$
– Malachi Wadas
Dec 29 '18 at 3:32
|
show 5 more comments
$begingroup$
Given a Polynomial ring, say $$S = R[x,y]/(x^2+x-1,y^3-y)$$ Is there a Monoid M and a ring $$T = R[M] cong S?$$ If there is such a Monoid what would it be? I am able to find one for univariate S:
$R$ is any ring with unity. $$S = R[x]/(p(x)),qquad n = deg(p)$$ assuming $p(x)$ is monic over $R$. The basis is of $T$ is $B[k]$ with $k in Bbb{Z}/nBbb{Z}$,
$R[Bbb{Z}/nBbb{Z}]$ with the basis action $f(a,b)$ being:
$$sum_{k=0}^{n-1} -c_{k}*B[k+a+b mod n] | a+b < min(a,b)
quad text{or}quad a+bquad text{otherwise}$$
abstract-algebra
asked Dec 28 '18 at 4:55
Malachi WadasMalachi Wadas
112
$endgroup$
Given a Polynomial ring, say $$S = R[x,y]/(x^2+x-1,y^3-y)$$ Is there a Monoid M and a ring $$T = R[M] cong S?$$ If there is such a Monoid what would it be? I am able to find one for univariate S:
$R$ is any ring with unity. $$S = R[x]/(p(x)),qquad n = deg(p)$$ assuming $p(x)$ is monic over $R$. The basis is of $T$ is $B[k]$ with $k in Bbb{Z}/nBbb{Z}$,
$R[Bbb{Z}/nBbb{Z}]$ with the basis action $f(a,b)$ being:
$$sum_{k=0}^{n-1} -c_{k}*B[k+a+b mod n] | a+b < min(a,b)
quad text{or}quad a+bquad text{otherwise}$$
abstract-algebra
abstract-algebra
asked Dec 28 '18 at 4:55
Malachi WadasMalachi Wadas
112
asked Dec 28 '18 at 4:55
Malachi WadasMalachi Wadas
112
edited Dec 28 '18 at 7:37
Malachi Wadas
asked Dec 28 '18 at 4:55
Malachi WadasMalachi Wadas
112
asked Dec 28 '18 at 4:55
Malachi WadasMalachi Wadas
112
asked Dec 28 '18 at 4:55
Malachi WadasMalachi Wadas
112
112
$begingroup$
The product of basis elements must agree with the operation of the monoid, so I don't think $R[x]/langle p(x)rangle$ works for all polynomials of degree $n$. It does work for $M=Bbb{Z}/nBbb{Z}$ when $p(x)=x^n-1$ (other polynomials may also work depending on the ring $R$),
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:07
$begingroup$
I’ve used sage math to test it my self and it works for any polynomial, the operation is associative as well, with 0 as the unit, I am currently working with R being The Reals or Complexes
$endgroup$
– Malachi Wadas
Dec 28 '18 at 7:13
$begingroup$
In the ring $R[M]$ with basis $B[m], min M$ the product must be $B[m_1]*B[m_2]=B[m_1cdot m_2]$. I don't think you have that. Here $m_1cdot m_2$ is the operation of your monoid $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:18
$begingroup$
I assume that your definition of a monoid ring parallels that of the group ring.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:21
1
$begingroup$
From what you given it seems any polynomials of the form $$x^{p} - x^{q} = 0; p > q $$ Form a Monoid ring.
$endgroup$
– Malachi Wadas
Dec 29 '18 at 3:32
|
show 5 more comments
$begingroup$
The product of basis elements must agree with the operation of the monoid, so I don't think $R[x]/langle p(x)rangle$ works for all polynomials of degree $n$. It does work for $M=Bbb{Z}/nBbb{Z}$ when $p(x)=x^n-1$ (other polynomials may also work depending on the ring $R$),
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:07
$begingroup$
I’ve used sage math to test it my self and it works for any polynomial, the operation is associative as well, with 0 as the unit, I am currently working with R being The Reals or Complexes
$endgroup$
– Malachi Wadas
Dec 28 '18 at 7:13
$begingroup$
In the ring $R[M]$ with basis $B[m], min M$ the product must be $B[m_1]*B[m_2]=B[m_1cdot m_2]$. I don't think you have that. Here $m_1cdot m_2$ is the operation of your monoid $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:18
$begingroup$
I assume that your definition of a monoid ring parallels that of the group ring.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:21
1
$begingroup$
From what you given it seems any polynomials of the form $$x^{p} - x^{q} = 0; p > q $$ Form a Monoid ring.
$endgroup$
– Malachi Wadas
Dec 29 '18 at 3:32
$begingroup$
The product of basis elements must agree with the operation of the monoid, so I don't think $R[x]/langle p(x)rangle$ works for all polynomials of degree $n$. It does work for $M=Bbb{Z}/nBbb{Z}$ when $p(x)=x^n-1$ (other polynomials may also work depending on the ring $R$),
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:07
$begingroup$
The product of basis elements must agree with the operation of the monoid, so I don't think $R[x]/langle p(x)rangle$ works for all polynomials of degree $n$. It does work for $M=Bbb{Z}/nBbb{Z}$ when $p(x)=x^n-1$ (other polynomials may also work depending on the ring $R$),
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:07
$begingroup$
I’ve used sage math to test it my self and it works for any polynomial, the operation is associative as well, with 0 as the unit, I am currently working with R being The Reals or Complexes
$endgroup$
– Malachi Wadas
Dec 28 '18 at 7:13
$begingroup$
I’ve used sage math to test it my self and it works for any polynomial, the operation is associative as well, with 0 as the unit, I am currently working with R being The Reals or Complexes
$endgroup$
– Malachi Wadas
Dec 28 '18 at 7:13
$begingroup$
In the ring $R[M]$ with basis $B[m], min M$ the product must be $B[m_1]*B[m_2]=B[m_1cdot m_2]$. I don't think you have that. Here $m_1cdot m_2$ is the operation of your monoid $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:18
$begingroup$
In the ring $R[M]$ with basis $B[m], min M$ the product must be $B[m_1]*B[m_2]=B[m_1cdot m_2]$. I don't think you have that. Here $m_1cdot m_2$ is the operation of your monoid $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:18
$begingroup$
I assume that your definition of a monoid ring parallels that of the group ring.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:21
$begingroup$
I assume that your definition of a monoid ring parallels that of the group ring.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:21
1
1
$begingroup$
From what you given it seems any polynomials of the form $$x^{p} - x^{q} = 0; p > q $$ Form a Monoid ring.
$endgroup$
– Malachi Wadas
Dec 29 '18 at 3:32
$begingroup$
From what you given it seems any polynomials of the form $$x^{p} - x^{q} = 0; p > q $$ Form a Monoid ring.
$endgroup$
– Malachi Wadas
Dec 29 '18 at 3:32
|
show 5 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Multiplicatively, its the monoid (ring extension of $R$) $R[bar x,bar y]$, where $bar x^3 + bar x - 1 = bar 0$ and $bar y^3 - bar y = bar 0$.
answered Dec 28 '18 at 7:50
WuestenfuxWuestenfux
4,7291513
$endgroup$
$begingroup$
This is not what a monoid ring is. If $M$ is a (multiplicative) monoid, the monoid ring $R[M]$ is the ring of formal linear combinations $sum_{min M} r_m m$ with $r_min R$ for all $min M$, componentwise addition, and multiplication extending $R$-linearly the multiplication in $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 9:07
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Multiplicatively, its the monoid (ring extension of $R$) $R[bar x,bar y]$, where $bar x^3 + bar x - 1 = bar 0$ and $bar y^3 - bar y = bar 0$.
answered Dec 28 '18 at 7:50
WuestenfuxWuestenfux
4,7291513
$endgroup$
$begingroup$
This is not what a monoid ring is. If $M$ is a (multiplicative) monoid, the monoid ring $R[M]$ is the ring of formal linear combinations $sum_{min M} r_m m$ with $r_min R$ for all $min M$, componentwise addition, and multiplication extending $R$-linearly the multiplication in $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 9:07
add a comment |
$begingroup$
Multiplicatively, its the monoid (ring extension of $R$) $R[bar x,bar y]$, where $bar x^3 + bar x - 1 = bar 0$ and $bar y^3 - bar y = bar 0$.
answered Dec 28 '18 at 7:50
WuestenfuxWuestenfux
4,7291513
$endgroup$
$begingroup$
This is not what a monoid ring is. If $M$ is a (multiplicative) monoid, the monoid ring $R[M]$ is the ring of formal linear combinations $sum_{min M} r_m m$ with $r_min R$ for all $min M$, componentwise addition, and multiplication extending $R$-linearly the multiplication in $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 9:07
add a comment |
$begingroup$
Multiplicatively, its the monoid (ring extension of $R$) $R[bar x,bar y]$, where $bar x^3 + bar x - 1 = bar 0$ and $bar y^3 - bar y = bar 0$.
answered Dec 28 '18 at 7:50
WuestenfuxWuestenfux
4,7291513
$endgroup$
Multiplicatively, its the monoid (ring extension of $R$) $R[bar x,bar y]$, where $bar x^3 + bar x - 1 = bar 0$ and $bar y^3 - bar y = bar 0$.
answered Dec 28 '18 at 7:50
WuestenfuxWuestenfux
4,7291513
edited Dec 28 '18 at 8:25
answered Dec 28 '18 at 7:50
WuestenfuxWuestenfux
4,7291513
answered Dec 28 '18 at 7:50
WuestenfuxWuestenfux
4,7291513
answered Dec 28 '18 at 7:50
WuestenfuxWuestenfux
4,7291513
4,7291513
$begingroup$
This is not what a monoid ring is. If $M$ is a (multiplicative) monoid, the monoid ring $R[M]$ is the ring of formal linear combinations $sum_{min M} r_m m$ with $r_min R$ for all $min M$, componentwise addition, and multiplication extending $R$-linearly the multiplication in $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 9:07
add a comment |
$begingroup$
This is not what a monoid ring is. If $M$ is a (multiplicative) monoid, the monoid ring $R[M]$ is the ring of formal linear combinations $sum_{min M} r_m m$ with $r_min R$ for all $min M$, componentwise addition, and multiplication extending $R$-linearly the multiplication in $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 9:07
$begingroup$
This is not what a monoid ring is. If $M$ is a (multiplicative) monoid, the monoid ring $R[M]$ is the ring of formal linear combinations $sum_{min M} r_m m$ with $r_min R$ for all $min M$, componentwise addition, and multiplication extending $R$-linearly the multiplication in $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 9:07
$begingroup$
This is not what a monoid ring is. If $M$ is a (multiplicative) monoid, the monoid ring $R[M]$ is the ring of formal linear combinations $sum_{min M} r_m m$ with $r_min R$ for all $min M$, componentwise addition, and multiplication extending $R$-linearly the multiplication in $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 9:07
add a comment |
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$begingroup$
The product of basis elements must agree with the operation of the monoid, so I don't think $R[x]/langle p(x)rangle$ works for all polynomials of degree $n$. It does work for $M=Bbb{Z}/nBbb{Z}$ when $p(x)=x^n-1$ (other polynomials may also work depending on the ring $R$),
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:07
$begingroup$
I’ve used sage math to test it my self and it works for any polynomial, the operation is associative as well, with 0 as the unit, I am currently working with R being The Reals or Complexes
$endgroup$
– Malachi Wadas
Dec 28 '18 at 7:13
$begingroup$
In the ring $R[M]$ with basis $B[m], min M$ the product must be $B[m_1]*B[m_2]=B[m_1cdot m_2]$. I don't think you have that. Here $m_1cdot m_2$ is the operation of your monoid $M$.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:18
$begingroup$
I assume that your definition of a monoid ring parallels that of the group ring.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 7:21
1
$begingroup$
From what you given it seems any polynomials of the form $$x^{p} - x^{q} = 0; p > q $$ Form a Monoid ring.
$endgroup$
– Malachi Wadas
Dec 29 '18 at 3:32