Question regarding localization of polynomial ring
$begingroup$
I made an exercise that went as follows:
Suppose $R = frac{mathbb{R}[x,y]}{(xy)}$. Define the multiplicative set $$ S = left{ 1 + (xy), x + (xy), x^2 + (xy), ldots right}.$$
In the exercise, I showed that $S^{-1}R cong mathbb{R}[x,x^{-1}].$
My question is as follows: Why don't we just define $R = mathbb{R}[x]$ and consider the multiplicative set $S = left{ 1,x,x^2, ldotsright}$? Wouldn't we also then get $S^{-1}R cong mathbb{R}[x,x^{-1}]$? If so, what is the advantage one gets with the definition used in the exercise, if any?
abstract-algebra commutative-algebra localization
asked Dec 22 '18 at 20:41
GregGreg
183112
$endgroup$
add a comment |
$begingroup$
I made an exercise that went as follows:
Suppose $R = frac{mathbb{R}[x,y]}{(xy)}$. Define the multiplicative set $$ S = left{ 1 + (xy), x + (xy), x^2 + (xy), ldots right}.$$
In the exercise, I showed that $S^{-1}R cong mathbb{R}[x,x^{-1}].$
My question is as follows: Why don't we just define $R = mathbb{R}[x]$ and consider the multiplicative set $S = left{ 1,x,x^2, ldotsright}$? Wouldn't we also then get $S^{-1}R cong mathbb{R}[x,x^{-1}]$? If so, what is the advantage one gets with the definition used in the exercise, if any?
abstract-algebra commutative-algebra localization
asked Dec 22 '18 at 20:41
GregGreg
183112
$endgroup$
1
$begingroup$
No advantage. It's an exercise where you have to notice that $x$ becomes invertible in the ring of fractions, and therefore the ideal generated by $xy$ equals the ideal generated by $y$ (which eventually goes to $0$).
$endgroup$
– user26857
Dec 22 '18 at 21:40
$begingroup$
That's what I assumed. Just thought it was strange (and thus perhaps useful) to throw in something that gets annihilated later on. Thanks !
$endgroup$
– Greg
Dec 22 '18 at 21:42
2
$begingroup$
If the problem is geometric in origin, you might have been originally dealing with functions on two-dimensional real affine space, and then wanted to restrict to only the axis (this corresponds to taking the quotient $Bbb R[x,y]to R$). Then, you might have been interested in functions on these spaces which are invertible on a particular locus (this would correspond to localizing). In the end you have a simpler description of the ring you're working with, but you may want to remember the original perspective when thinking about what the ring "means."
$endgroup$
– Stahl
Dec 22 '18 at 22:05
add a comment |
$begingroup$
I made an exercise that went as follows:
Suppose $R = frac{mathbb{R}[x,y]}{(xy)}$. Define the multiplicative set $$ S = left{ 1 + (xy), x + (xy), x^2 + (xy), ldots right}.$$
In the exercise, I showed that $S^{-1}R cong mathbb{R}[x,x^{-1}].$
My question is as follows: Why don't we just define $R = mathbb{R}[x]$ and consider the multiplicative set $S = left{ 1,x,x^2, ldotsright}$? Wouldn't we also then get $S^{-1}R cong mathbb{R}[x,x^{-1}]$? If so, what is the advantage one gets with the definition used in the exercise, if any?
abstract-algebra commutative-algebra localization
asked Dec 22 '18 at 20:41
GregGreg
183112
$endgroup$
I made an exercise that went as follows:
Suppose $R = frac{mathbb{R}[x,y]}{(xy)}$. Define the multiplicative set $$ S = left{ 1 + (xy), x + (xy), x^2 + (xy), ldots right}.$$
In the exercise, I showed that $S^{-1}R cong mathbb{R}[x,x^{-1}].$
My question is as follows: Why don't we just define $R = mathbb{R}[x]$ and consider the multiplicative set $S = left{ 1,x,x^2, ldotsright}$? Wouldn't we also then get $S^{-1}R cong mathbb{R}[x,x^{-1}]$? If so, what is the advantage one gets with the definition used in the exercise, if any?
abstract-algebra commutative-algebra localization
abstract-algebra commutative-algebra localization
asked Dec 22 '18 at 20:41
GregGreg
183112
asked Dec 22 '18 at 20:41
GregGreg
183112
edited Dec 22 '18 at 20:54
Greg
asked Dec 22 '18 at 20:41
GregGreg
183112
asked Dec 22 '18 at 20:41
GregGreg
183112
asked Dec 22 '18 at 20:41
GregGreg
183112
183112
1
$begingroup$
No advantage. It's an exercise where you have to notice that $x$ becomes invertible in the ring of fractions, and therefore the ideal generated by $xy$ equals the ideal generated by $y$ (which eventually goes to $0$).
$endgroup$
– user26857
Dec 22 '18 at 21:40
$begingroup$
That's what I assumed. Just thought it was strange (and thus perhaps useful) to throw in something that gets annihilated later on. Thanks !
$endgroup$
– Greg
Dec 22 '18 at 21:42
2
$begingroup$
If the problem is geometric in origin, you might have been originally dealing with functions on two-dimensional real affine space, and then wanted to restrict to only the axis (this corresponds to taking the quotient $Bbb R[x,y]to R$). Then, you might have been interested in functions on these spaces which are invertible on a particular locus (this would correspond to localizing). In the end you have a simpler description of the ring you're working with, but you may want to remember the original perspective when thinking about what the ring "means."
$endgroup$
– Stahl
Dec 22 '18 at 22:05
add a comment |
1
$begingroup$
No advantage. It's an exercise where you have to notice that $x$ becomes invertible in the ring of fractions, and therefore the ideal generated by $xy$ equals the ideal generated by $y$ (which eventually goes to $0$).
$endgroup$
– user26857
Dec 22 '18 at 21:40
$begingroup$
That's what I assumed. Just thought it was strange (and thus perhaps useful) to throw in something that gets annihilated later on. Thanks !
$endgroup$
– Greg
Dec 22 '18 at 21:42
2
$begingroup$
If the problem is geometric in origin, you might have been originally dealing with functions on two-dimensional real affine space, and then wanted to restrict to only the axis (this corresponds to taking the quotient $Bbb R[x,y]to R$). Then, you might have been interested in functions on these spaces which are invertible on a particular locus (this would correspond to localizing). In the end you have a simpler description of the ring you're working with, but you may want to remember the original perspective when thinking about what the ring "means."
$endgroup$
– Stahl
Dec 22 '18 at 22:05
1
1
$begingroup$
No advantage. It's an exercise where you have to notice that $x$ becomes invertible in the ring of fractions, and therefore the ideal generated by $xy$ equals the ideal generated by $y$ (which eventually goes to $0$).
$endgroup$
– user26857
Dec 22 '18 at 21:40
$begingroup$
No advantage. It's an exercise where you have to notice that $x$ becomes invertible in the ring of fractions, and therefore the ideal generated by $xy$ equals the ideal generated by $y$ (which eventually goes to $0$).
$endgroup$
– user26857
Dec 22 '18 at 21:40
$begingroup$
That's what I assumed. Just thought it was strange (and thus perhaps useful) to throw in something that gets annihilated later on. Thanks !
$endgroup$
– Greg
Dec 22 '18 at 21:42
$begingroup$
That's what I assumed. Just thought it was strange (and thus perhaps useful) to throw in something that gets annihilated later on. Thanks !
$endgroup$
– Greg
Dec 22 '18 at 21:42
2
2
$begingroup$
If the problem is geometric in origin, you might have been originally dealing with functions on two-dimensional real affine space, and then wanted to restrict to only the axis (this corresponds to taking the quotient $Bbb R[x,y]to R$). Then, you might have been interested in functions on these spaces which are invertible on a particular locus (this would correspond to localizing). In the end you have a simpler description of the ring you're working with, but you may want to remember the original perspective when thinking about what the ring "means."
$endgroup$
– Stahl
Dec 22 '18 at 22:05
$begingroup$
If the problem is geometric in origin, you might have been originally dealing with functions on two-dimensional real affine space, and then wanted to restrict to only the axis (this corresponds to taking the quotient $Bbb R[x,y]to R$). Then, you might have been interested in functions on these spaces which are invertible on a particular locus (this would correspond to localizing). In the end you have a simpler description of the ring you're working with, but you may want to remember the original perspective when thinking about what the ring "means."
$endgroup$
– Stahl
Dec 22 '18 at 22:05
add a comment |
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$begingroup$
No advantage. It's an exercise where you have to notice that $x$ becomes invertible in the ring of fractions, and therefore the ideal generated by $xy$ equals the ideal generated by $y$ (which eventually goes to $0$).
$endgroup$
– user26857
Dec 22 '18 at 21:40
$begingroup$
That's what I assumed. Just thought it was strange (and thus perhaps useful) to throw in something that gets annihilated later on. Thanks !
$endgroup$
– Greg
Dec 22 '18 at 21:42
2
$begingroup$
If the problem is geometric in origin, you might have been originally dealing with functions on two-dimensional real affine space, and then wanted to restrict to only the axis (this corresponds to taking the quotient $Bbb R[x,y]to R$). Then, you might have been interested in functions on these spaces which are invertible on a particular locus (this would correspond to localizing). In the end you have a simpler description of the ring you're working with, but you may want to remember the original perspective when thinking about what the ring "means."
$endgroup$
– Stahl
Dec 22 '18 at 22:05