Elements of $R=S^{-1} mathbb{C}[ x,y ]$
I would like to understand how are the elements of $R=S^{-1} mathbb{C}[ x,y ]$ where S is the multiplicative set generated by ${x-a | a in mathbb{C} } cup {y-a | a in mathbb{C} }$.
I would like to understand which is the identity element and this kinds of things in order to try to prove if it is a field or not.
abstract-algebra commutative-algebra modules
asked Nov 13 at 16:42
idriskameni
609
add a comment |
I would like to understand how are the elements of $R=S^{-1} mathbb{C}[ x,y ]$ where S is the multiplicative set generated by ${x-a | a in mathbb{C} } cup {y-a | a in mathbb{C} }$.
I would like to understand which is the identity element and this kinds of things in order to try to prove if it is a field or not.
abstract-algebra commutative-algebra modules
asked Nov 13 at 16:42
idriskameni
609
2
$R$ is the ring of all rational functions in $x$ and $y$ that can be written as fractions of polynomials where the denominator is a product of $x-a$'s and $y-a$'s (for varying $a$). For example, $dfrac{x^3+xy-7y-sqrt{3}}{left(x-1right)^2left(y+2right)} in R$. The identity element is $1 = 1/1$, as always.
– darij grinberg
Nov 13 at 17:47
1
In order to know if it is a field or not what you need to know is which elements are invertible. Since $mathbb{C}[x,y]$ is naturally a subring of $R$ you can try to see which polynomials are invertible in $R$.
– xarles
Nov 13 at 18:25
add a comment |
I would like to understand how are the elements of $R=S^{-1} mathbb{C}[ x,y ]$ where S is the multiplicative set generated by ${x-a | a in mathbb{C} } cup {y-a | a in mathbb{C} }$.
I would like to understand which is the identity element and this kinds of things in order to try to prove if it is a field or not.
abstract-algebra commutative-algebra modules
asked Nov 13 at 16:42
idriskameni
609
I would like to understand how are the elements of $R=S^{-1} mathbb{C}[ x,y ]$ where S is the multiplicative set generated by ${x-a | a in mathbb{C} } cup {y-a | a in mathbb{C} }$.
I would like to understand which is the identity element and this kinds of things in order to try to prove if it is a field or not.
abstract-algebra commutative-algebra modules
abstract-algebra commutative-algebra modules
asked Nov 13 at 16:42
idriskameni
609
asked Nov 13 at 16:42
idriskameni
609
asked Nov 13 at 16:42
idriskameni
609
asked Nov 13 at 16:42
idriskameni
609
asked Nov 13 at 16:42
idriskameni
609
609
2
$R$ is the ring of all rational functions in $x$ and $y$ that can be written as fractions of polynomials where the denominator is a product of $x-a$'s and $y-a$'s (for varying $a$). For example, $dfrac{x^3+xy-7y-sqrt{3}}{left(x-1right)^2left(y+2right)} in R$. The identity element is $1 = 1/1$, as always.
– darij grinberg
Nov 13 at 17:47
1
In order to know if it is a field or not what you need to know is which elements are invertible. Since $mathbb{C}[x,y]$ is naturally a subring of $R$ you can try to see which polynomials are invertible in $R$.
– xarles
Nov 13 at 18:25
add a comment |
2
$R$ is the ring of all rational functions in $x$ and $y$ that can be written as fractions of polynomials where the denominator is a product of $x-a$'s and $y-a$'s (for varying $a$). For example, $dfrac{x^3+xy-7y-sqrt{3}}{left(x-1right)^2left(y+2right)} in R$. The identity element is $1 = 1/1$, as always.
– darij grinberg
Nov 13 at 17:47
1
In order to know if it is a field or not what you need to know is which elements are invertible. Since $mathbb{C}[x,y]$ is naturally a subring of $R$ you can try to see which polynomials are invertible in $R$.
– xarles
Nov 13 at 18:25
2
2
$R$ is the ring of all rational functions in $x$ and $y$ that can be written as fractions of polynomials where the denominator is a product of $x-a$'s and $y-a$'s (for varying $a$). For example, $dfrac{x^3+xy-7y-sqrt{3}}{left(x-1right)^2left(y+2right)} in R$. The identity element is $1 = 1/1$, as always.
– darij grinberg
Nov 13 at 17:47
$R$ is the ring of all rational functions in $x$ and $y$ that can be written as fractions of polynomials where the denominator is a product of $x-a$'s and $y-a$'s (for varying $a$). For example, $dfrac{x^3+xy-7y-sqrt{3}}{left(x-1right)^2left(y+2right)} in R$. The identity element is $1 = 1/1$, as always.
– darij grinberg
Nov 13 at 17:47
1
1
In order to know if it is a field or not what you need to know is which elements are invertible. Since $mathbb{C}[x,y]$ is naturally a subring of $R$ you can try to see which polynomials are invertible in $R$.
– xarles
Nov 13 at 18:25
In order to know if it is a field or not what you need to know is which elements are invertible. Since $mathbb{C}[x,y]$ is naturally a subring of $R$ you can try to see which polynomials are invertible in $R$.
– xarles
Nov 13 at 18:25
add a comment |
1 Answer
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We know $S=langle {x-ahspace{0,1cm}|hspace{0,1cm}ain mathbb{C}} cup {y-ahspace{0,1cm}|hspace{0,1cm}ain mathbb{C}} rangle$, what means that if $sin S$ then it can be written as $s=(x-a_1)cdot ... cdot (x-a_n)(y-b_1)cdot ... cdot (y-b_m)$ for $a_i, b_j in mathbb{C}$ and some $n,m$ positive integers.
Moreover, we know that $$R=S^{-1}mathbb{C}[x,y] = { [f,s]hspace{0,1cm}|hspace{0,1cm} fin mathbb{C}[x,y] , s in S } / sim$$
where $[f,s]sim[f',s']$ if and only if $exists s'' in S$ such that $s''(s'f-sf')=0$. Since $sin S$ implies $sneq0$, we have that $[f,s]sim[f',s']$ if and only if $(s'f-sf')=0$ if and only if $f/s =f'/s'$.
Knowing that, we need to see if all elements in $R$ have their inverse in $R$. This means $forall [f,s] in S^{-1} mathbb{C}[x,y]$, $exists [f',s'] in S^{-1} mathbb{C}[x,y]$ such that $[f,s]cdot[f',s'] sim 1_R $ where $1_R sim [t,t]$ for any $tin S$. This is because $Ssubset mathbb{C}[x,y]$ and we need to have the same element in order to have $t/t = 1_R$.
Now consider $[(x+1)(x+y),(x+1)] in S^{-1} mathbb{C}[x,y]$. This element is clearly in $R$ since $(x+1)$ has the form we have described before and $(x+1)(x+y) in mathbb{C}[x,y]$. The inverse of this element should be $[g,j]$ such that $gin mathbb{C}[x,y]$ and $jin S$. It must satisfy the following expression: $$[(x+1)(x+y),(x+1)] cdot [g,j] sim 1_R$$
That means $(x+1)(x+y)gt=(x+1)jt$, $j=(x+y)g$ where $gin mathbb{C}[x,y]$. But $ nexists jin S$ such that $j=(x+y)g$ with $gin mathbb{C}[x,y]$ since $j$ has the form described above.
The element $[(x+1)(x+y),(x+1)]$ has no inverse in $R$. Hence not all elements have their inverse in $R$. Hence $R$ is not a field.
answered Nov 25 at 7:20
idriskameni
609
add a comment |
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We know $S=langle {x-ahspace{0,1cm}|hspace{0,1cm}ain mathbb{C}} cup {y-ahspace{0,1cm}|hspace{0,1cm}ain mathbb{C}} rangle$, what means that if $sin S$ then it can be written as $s=(x-a_1)cdot ... cdot (x-a_n)(y-b_1)cdot ... cdot (y-b_m)$ for $a_i, b_j in mathbb{C}$ and some $n,m$ positive integers.
Moreover, we know that $$R=S^{-1}mathbb{C}[x,y] = { [f,s]hspace{0,1cm}|hspace{0,1cm} fin mathbb{C}[x,y] , s in S } / sim$$
where $[f,s]sim[f',s']$ if and only if $exists s'' in S$ such that $s''(s'f-sf')=0$. Since $sin S$ implies $sneq0$, we have that $[f,s]sim[f',s']$ if and only if $(s'f-sf')=0$ if and only if $f/s =f'/s'$.
Knowing that, we need to see if all elements in $R$ have their inverse in $R$. This means $forall [f,s] in S^{-1} mathbb{C}[x,y]$, $exists [f',s'] in S^{-1} mathbb{C}[x,y]$ such that $[f,s]cdot[f',s'] sim 1_R $ where $1_R sim [t,t]$ for any $tin S$. This is because $Ssubset mathbb{C}[x,y]$ and we need to have the same element in order to have $t/t = 1_R$.
Now consider $[(x+1)(x+y),(x+1)] in S^{-1} mathbb{C}[x,y]$. This element is clearly in $R$ since $(x+1)$ has the form we have described before and $(x+1)(x+y) in mathbb{C}[x,y]$. The inverse of this element should be $[g,j]$ such that $gin mathbb{C}[x,y]$ and $jin S$. It must satisfy the following expression: $$[(x+1)(x+y),(x+1)] cdot [g,j] sim 1_R$$
That means $(x+1)(x+y)gt=(x+1)jt$, $j=(x+y)g$ where $gin mathbb{C}[x,y]$. But $ nexists jin S$ such that $j=(x+y)g$ with $gin mathbb{C}[x,y]$ since $j$ has the form described above.
The element $[(x+1)(x+y),(x+1)]$ has no inverse in $R$. Hence not all elements have their inverse in $R$. Hence $R$ is not a field.
answered Nov 25 at 7:20
idriskameni
609
add a comment |
We know $S=langle {x-ahspace{0,1cm}|hspace{0,1cm}ain mathbb{C}} cup {y-ahspace{0,1cm}|hspace{0,1cm}ain mathbb{C}} rangle$, what means that if $sin S$ then it can be written as $s=(x-a_1)cdot ... cdot (x-a_n)(y-b_1)cdot ... cdot (y-b_m)$ for $a_i, b_j in mathbb{C}$ and some $n,m$ positive integers.
Moreover, we know that $$R=S^{-1}mathbb{C}[x,y] = { [f,s]hspace{0,1cm}|hspace{0,1cm} fin mathbb{C}[x,y] , s in S } / sim$$
where $[f,s]sim[f',s']$ if and only if $exists s'' in S$ such that $s''(s'f-sf')=0$. Since $sin S$ implies $sneq0$, we have that $[f,s]sim[f',s']$ if and only if $(s'f-sf')=0$ if and only if $f/s =f'/s'$.
Knowing that, we need to see if all elements in $R$ have their inverse in $R$. This means $forall [f,s] in S^{-1} mathbb{C}[x,y]$, $exists [f',s'] in S^{-1} mathbb{C}[x,y]$ such that $[f,s]cdot[f',s'] sim 1_R $ where $1_R sim [t,t]$ for any $tin S$. This is because $Ssubset mathbb{C}[x,y]$ and we need to have the same element in order to have $t/t = 1_R$.
Now consider $[(x+1)(x+y),(x+1)] in S^{-1} mathbb{C}[x,y]$. This element is clearly in $R$ since $(x+1)$ has the form we have described before and $(x+1)(x+y) in mathbb{C}[x,y]$. The inverse of this element should be $[g,j]$ such that $gin mathbb{C}[x,y]$ and $jin S$. It must satisfy the following expression: $$[(x+1)(x+y),(x+1)] cdot [g,j] sim 1_R$$
That means $(x+1)(x+y)gt=(x+1)jt$, $j=(x+y)g$ where $gin mathbb{C}[x,y]$. But $ nexists jin S$ such that $j=(x+y)g$ with $gin mathbb{C}[x,y]$ since $j$ has the form described above.
The element $[(x+1)(x+y),(x+1)]$ has no inverse in $R$. Hence not all elements have their inverse in $R$. Hence $R$ is not a field.
answered Nov 25 at 7:20
idriskameni
609
add a comment |
We know $S=langle {x-ahspace{0,1cm}|hspace{0,1cm}ain mathbb{C}} cup {y-ahspace{0,1cm}|hspace{0,1cm}ain mathbb{C}} rangle$, what means that if $sin S$ then it can be written as $s=(x-a_1)cdot ... cdot (x-a_n)(y-b_1)cdot ... cdot (y-b_m)$ for $a_i, b_j in mathbb{C}$ and some $n,m$ positive integers.
Moreover, we know that $$R=S^{-1}mathbb{C}[x,y] = { [f,s]hspace{0,1cm}|hspace{0,1cm} fin mathbb{C}[x,y] , s in S } / sim$$
where $[f,s]sim[f',s']$ if and only if $exists s'' in S$ such that $s''(s'f-sf')=0$. Since $sin S$ implies $sneq0$, we have that $[f,s]sim[f',s']$ if and only if $(s'f-sf')=0$ if and only if $f/s =f'/s'$.
Knowing that, we need to see if all elements in $R$ have their inverse in $R$. This means $forall [f,s] in S^{-1} mathbb{C}[x,y]$, $exists [f',s'] in S^{-1} mathbb{C}[x,y]$ such that $[f,s]cdot[f',s'] sim 1_R $ where $1_R sim [t,t]$ for any $tin S$. This is because $Ssubset mathbb{C}[x,y]$ and we need to have the same element in order to have $t/t = 1_R$.
Now consider $[(x+1)(x+y),(x+1)] in S^{-1} mathbb{C}[x,y]$. This element is clearly in $R$ since $(x+1)$ has the form we have described before and $(x+1)(x+y) in mathbb{C}[x,y]$. The inverse of this element should be $[g,j]$ such that $gin mathbb{C}[x,y]$ and $jin S$. It must satisfy the following expression: $$[(x+1)(x+y),(x+1)] cdot [g,j] sim 1_R$$
That means $(x+1)(x+y)gt=(x+1)jt$, $j=(x+y)g$ where $gin mathbb{C}[x,y]$. But $ nexists jin S$ such that $j=(x+y)g$ with $gin mathbb{C}[x,y]$ since $j$ has the form described above.
The element $[(x+1)(x+y),(x+1)]$ has no inverse in $R$. Hence not all elements have their inverse in $R$. Hence $R$ is not a field.
answered Nov 25 at 7:20
idriskameni
609
We know $S=langle {x-ahspace{0,1cm}|hspace{0,1cm}ain mathbb{C}} cup {y-ahspace{0,1cm}|hspace{0,1cm}ain mathbb{C}} rangle$, what means that if $sin S$ then it can be written as $s=(x-a_1)cdot ... cdot (x-a_n)(y-b_1)cdot ... cdot (y-b_m)$ for $a_i, b_j in mathbb{C}$ and some $n,m$ positive integers.
Moreover, we know that $$R=S^{-1}mathbb{C}[x,y] = { [f,s]hspace{0,1cm}|hspace{0,1cm} fin mathbb{C}[x,y] , s in S } / sim$$
where $[f,s]sim[f',s']$ if and only if $exists s'' in S$ such that $s''(s'f-sf')=0$. Since $sin S$ implies $sneq0$, we have that $[f,s]sim[f',s']$ if and only if $(s'f-sf')=0$ if and only if $f/s =f'/s'$.
Knowing that, we need to see if all elements in $R$ have their inverse in $R$. This means $forall [f,s] in S^{-1} mathbb{C}[x,y]$, $exists [f',s'] in S^{-1} mathbb{C}[x,y]$ such that $[f,s]cdot[f',s'] sim 1_R $ where $1_R sim [t,t]$ for any $tin S$. This is because $Ssubset mathbb{C}[x,y]$ and we need to have the same element in order to have $t/t = 1_R$.
Now consider $[(x+1)(x+y),(x+1)] in S^{-1} mathbb{C}[x,y]$. This element is clearly in $R$ since $(x+1)$ has the form we have described before and $(x+1)(x+y) in mathbb{C}[x,y]$. The inverse of this element should be $[g,j]$ such that $gin mathbb{C}[x,y]$ and $jin S$. It must satisfy the following expression: $$[(x+1)(x+y),(x+1)] cdot [g,j] sim 1_R$$
That means $(x+1)(x+y)gt=(x+1)jt$, $j=(x+y)g$ where $gin mathbb{C}[x,y]$. But $ nexists jin S$ such that $j=(x+y)g$ with $gin mathbb{C}[x,y]$ since $j$ has the form described above.
The element $[(x+1)(x+y),(x+1)]$ has no inverse in $R$. Hence not all elements have their inverse in $R$. Hence $R$ is not a field.
answered Nov 25 at 7:20
idriskameni
609
edited Nov 27 at 17:51
answered Nov 25 at 7:20
idriskameni
609
answered Nov 25 at 7:20
idriskameni
609
answered Nov 25 at 7:20
idriskameni
609
609
add a comment |
add a comment |
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$R$ is the ring of all rational functions in $x$ and $y$ that can be written as fractions of polynomials where the denominator is a product of $x-a$'s and $y-a$'s (for varying $a$). For example, $dfrac{x^3+xy-7y-sqrt{3}}{left(x-1right)^2left(y+2right)} in R$. The identity element is $1 = 1/1$, as always.
– darij grinberg
Nov 13 at 17:47
1
In order to know if it is a field or not what you need to know is which elements are invertible. Since $mathbb{C}[x,y]$ is naturally a subring of $R$ you can try to see which polynomials are invertible in $R$.
– xarles
Nov 13 at 18:25