Mixing
Equality of fractions in localization as quotient
$begingroup$
Let $Ssubset A$ be a subset of a commutative ring. Consider the polynomial algebra $A[x_smid sin S]$ and define the localization $A_S$ as the quotient by the ideal $ (sx_s-1mid sin S)vartriangleleft A[x_smid sin S]$.
Let us write $frac as$ for the equivalence class of $ax_s$ in the quotient $A_S$. I would like to prove that $frac as=frac bt$ iff $exists uin S$ such that $u(at-bs)=0$, like in the construction of localization as a quotient of $Atimes S$.
So we have $frac as-frac bt=0in A_S$ which is equivalent to $ax_s-bx_tin I$. By degree considerations I think we must have $ax_s-bx_t=c(sx_s-1)-c(tx_t-1)$ for some $cin A$, and by comparing coefficients we have $a=cs,b=ct$. Multiplying by $t,s$ respectively we find $at=cst=cts=bs$, whence $at-bs=0$ without any $uin S$.
What is my mistake?
commutative-algebra localization
asked Feb 1 at 0:38
ArrowArrow
5,20831546
$endgroup$
add a comment |
$begingroup$
Let $Ssubset A$ be a subset of a commutative ring. Consider the polynomial algebra $A[x_smid sin S]$ and define the localization $A_S$ as the quotient by the ideal $ (sx_s-1mid sin S)vartriangleleft A[x_smid sin S]$.
Let us write $frac as$ for the equivalence class of $ax_s$ in the quotient $A_S$. I would like to prove that $frac as=frac bt$ iff $exists uin S$ such that $u(at-bs)=0$, like in the construction of localization as a quotient of $Atimes S$.
So we have $frac as-frac bt=0in A_S$ which is equivalent to $ax_s-bx_tin I$. By degree considerations I think we must have $ax_s-bx_t=c(sx_s-1)-c(tx_t-1)$ for some $cin A$, and by comparing coefficients we have $a=cs,b=ct$. Multiplying by $t,s$ respectively we find $at=cst=cts=bs$, whence $at-bs=0$ without any $uin S$.
What is my mistake?
commutative-algebra localization
asked Feb 1 at 0:38
ArrowArrow
5,20831546
$endgroup$
add a comment |
$begingroup$
Let $Ssubset A$ be a subset of a commutative ring. Consider the polynomial algebra $A[x_smid sin S]$ and define the localization $A_S$ as the quotient by the ideal $ (sx_s-1mid sin S)vartriangleleft A[x_smid sin S]$.
Let us write $frac as$ for the equivalence class of $ax_s$ in the quotient $A_S$. I would like to prove that $frac as=frac bt$ iff $exists uin S$ such that $u(at-bs)=0$, like in the construction of localization as a quotient of $Atimes S$.
So we have $frac as-frac bt=0in A_S$ which is equivalent to $ax_s-bx_tin I$. By degree considerations I think we must have $ax_s-bx_t=c(sx_s-1)-c(tx_t-1)$ for some $cin A$, and by comparing coefficients we have $a=cs,b=ct$. Multiplying by $t,s$ respectively we find $at=cst=cts=bs$, whence $at-bs=0$ without any $uin S$.
What is my mistake?
commutative-algebra localization
asked Feb 1 at 0:38
ArrowArrow
5,20831546
$endgroup$
Let $Ssubset A$ be a subset of a commutative ring. Consider the polynomial algebra $A[x_smid sin S]$ and define the localization $A_S$ as the quotient by the ideal $ (sx_s-1mid sin S)vartriangleleft A[x_smid sin S]$.
Let us write $frac as$ for the equivalence class of $ax_s$ in the quotient $A_S$. I would like to prove that $frac as=frac bt$ iff $exists uin S$ such that $u(at-bs)=0$, like in the construction of localization as a quotient of $Atimes S$.
So we have $frac as-frac bt=0in A_S$ which is equivalent to $ax_s-bx_tin I$. By degree considerations I think we must have $ax_s-bx_t=c(sx_s-1)-c(tx_t-1)$ for some $cin A$, and by comparing coefficients we have $a=cs,b=ct$. Multiplying by $t,s$ respectively we find $at=cst=cts=bs$, whence $at-bs=0$ without any $uin S$.
What is my mistake?
commutative-algebra localization
commutative-algebra localization
asked Feb 1 at 0:38
ArrowArrow
5,20831546
asked Feb 1 at 0:38
ArrowArrow
5,20831546
asked Feb 1 at 0:38
ArrowArrow
5,20831546
asked Feb 1 at 0:38
ArrowArrow
5,20831546
asked Feb 1 at 0:38
ArrowArrow
5,20831546
5,20831546
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think the main mistake is in the statement "by degree considerations..." I did not verify that it is true for integral domains, but these considerations are more complicated in the presence of zerodivisors.
Let's consider a (classical) localization in a ring $A$ with zerodivisors. For example, let $A = mathbb{Z}[y]/(2y)$ and invert $2$. Then in the localization $y=0$.
Now consider your polynomial-ring construction, in which $2x_2 - 1 = 0$ and (multiplying by $y$) $$2x_2 y - y = 0,$$ i.e. $y=0$. This looks like a separate, more minor, problem: it is a relation not of the form $$frac{a}{s}=frac{b}{t}$$ for $s,t in S$ (you might want to take $1 in S$.) I'll assume we took $1in S$, and use a corresponding variable $x_1$ below.
Anyway, $y=0$ and in particular we have for example $$yx_2 - 0x_1 = yx_2 in I.$$ So by your degree considerations, we should get $yx_2 = c(2x_2 - 1) - c(1x_1-1)$ in $A[x_1, x_2]$ for some $cin A$. This is not the case. To exhibit this relation as an element of $I$ you need something like $yx_2 cdot (2x_2 - 1),$ where $yx_2$ is not of degree $0$, hence not in $A$, so the degree considerations fail.
I previously wrote a "correction" to your proof, but it was mainly nonsense. There was some germ of a good idea hidden there, though. Here is a sketch of a correct proof:
Consider a ring $A$ and a module $M$ over it, and take $$(Moplus M) / ((x,sx))$$ for some fixed $sin A$. What is the submodule generated by the first coordinate? If an element $x$ is annihilated by $s$, then $$(x,0) = (x,0) - (x,sx) = (0,0)$$ in the quotient. It can be verified directly that the submodule is just $$frac{M}{text{ker}(s:Mrightarrow M)}.$$ What if we take $$bigoplus_{i,j ge 0} M$$ and mod out by elements of the form $$se_{i+1,j}-e_{i,j},quad te_{i,j+1}-e_{i,j}?$$ I claim the $e_{0,0}$ coordinate then generates a module isomorphic to $$frac{M}{bigcup_n text{ker}(st)^n:Mrightarrow M},$$ et cetera.
To see this, consider the colimit explicitly. If an element of the submodule generated by $M_{0,0}$ in the quotient vanishes, it vanishes already in some f.g. submodule, which is contained in some direct sum of a finite set of $M_{i,j}$, and hence in a "rectangular" submodule $$bigoplus_{0le i,jle n} M_{i,j}.$$ Any element here can be "pushed" to one supported only in $M_{n,n}$, and it is easy to check what it means that it vanishes there. The same argument generalizes to higher "dimensions" $$bigoplus M_{i,j,k}.$$
You want to take $$B:= Aleft[{x_s}_{sin S}right]$$ and consider some ideal generated by elements of the form $(sx_s - 1)$. Considering the polynomial ring as a free $A$-module, the required statement follows from the previous one, once one shows the ideal is generated as an $A$-submodule of $B$ by elements of the same form as above. This is not so bad since $$(sx_s -1)(tx_t - 1) = (stx_sx_t - 1) - (sx_s -1) - (tx_t - 1)$$ is already a sum of three relations, and the same will occur for more complicated monomials in place of $x_s$ and $x_t$. Multiplying a relation by a monomial also gives you a sum of relations in the quotient submodule of $$bigoplus_{i,j}M_i,j$$ above.
answered Feb 1 at 22:40
gyashfegyashfe
963
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3095680%2fequality-of-fractions-in-localization-as-quotient%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think the main mistake is in the statement "by degree considerations..." I did not verify that it is true for integral domains, but these considerations are more complicated in the presence of zerodivisors.
Let's consider a (classical) localization in a ring $A$ with zerodivisors. For example, let $A = mathbb{Z}[y]/(2y)$ and invert $2$. Then in the localization $y=0$.
Now consider your polynomial-ring construction, in which $2x_2 - 1 = 0$ and (multiplying by $y$) $$2x_2 y - y = 0,$$ i.e. $y=0$. This looks like a separate, more minor, problem: it is a relation not of the form $$frac{a}{s}=frac{b}{t}$$ for $s,t in S$ (you might want to take $1 in S$.) I'll assume we took $1in S$, and use a corresponding variable $x_1$ below.
Anyway, $y=0$ and in particular we have for example $$yx_2 - 0x_1 = yx_2 in I.$$ So by your degree considerations, we should get $yx_2 = c(2x_2 - 1) - c(1x_1-1)$ in $A[x_1, x_2]$ for some $cin A$. This is not the case. To exhibit this relation as an element of $I$ you need something like $yx_2 cdot (2x_2 - 1),$ where $yx_2$ is not of degree $0$, hence not in $A$, so the degree considerations fail.
I previously wrote a "correction" to your proof, but it was mainly nonsense. There was some germ of a good idea hidden there, though. Here is a sketch of a correct proof:
Consider a ring $A$ and a module $M$ over it, and take $$(Moplus M) / ((x,sx))$$ for some fixed $sin A$. What is the submodule generated by the first coordinate? If an element $x$ is annihilated by $s$, then $$(x,0) = (x,0) - (x,sx) = (0,0)$$ in the quotient. It can be verified directly that the submodule is just $$frac{M}{text{ker}(s:Mrightarrow M)}.$$ What if we take $$bigoplus_{i,j ge 0} M$$ and mod out by elements of the form $$se_{i+1,j}-e_{i,j},quad te_{i,j+1}-e_{i,j}?$$ I claim the $e_{0,0}$ coordinate then generates a module isomorphic to $$frac{M}{bigcup_n text{ker}(st)^n:Mrightarrow M},$$ et cetera.
To see this, consider the colimit explicitly. If an element of the submodule generated by $M_{0,0}$ in the quotient vanishes, it vanishes already in some f.g. submodule, which is contained in some direct sum of a finite set of $M_{i,j}$, and hence in a "rectangular" submodule $$bigoplus_{0le i,jle n} M_{i,j}.$$ Any element here can be "pushed" to one supported only in $M_{n,n}$, and it is easy to check what it means that it vanishes there. The same argument generalizes to higher "dimensions" $$bigoplus M_{i,j,k}.$$
You want to take $$B:= Aleft[{x_s}_{sin S}right]$$ and consider some ideal generated by elements of the form $(sx_s - 1)$. Considering the polynomial ring as a free $A$-module, the required statement follows from the previous one, once one shows the ideal is generated as an $A$-submodule of $B$ by elements of the same form as above. This is not so bad since $$(sx_s -1)(tx_t - 1) = (stx_sx_t - 1) - (sx_s -1) - (tx_t - 1)$$ is already a sum of three relations, and the same will occur for more complicated monomials in place of $x_s$ and $x_t$. Multiplying a relation by a monomial also gives you a sum of relations in the quotient submodule of $$bigoplus_{i,j}M_i,j$$ above.
answered Feb 1 at 22:40
gyashfegyashfe
963
$endgroup$
add a comment |
$begingroup$
I think the main mistake is in the statement "by degree considerations..." I did not verify that it is true for integral domains, but these considerations are more complicated in the presence of zerodivisors.
Let's consider a (classical) localization in a ring $A$ with zerodivisors. For example, let $A = mathbb{Z}[y]/(2y)$ and invert $2$. Then in the localization $y=0$.
Now consider your polynomial-ring construction, in which $2x_2 - 1 = 0$ and (multiplying by $y$) $$2x_2 y - y = 0,$$ i.e. $y=0$. This looks like a separate, more minor, problem: it is a relation not of the form $$frac{a}{s}=frac{b}{t}$$ for $s,t in S$ (you might want to take $1 in S$.) I'll assume we took $1in S$, and use a corresponding variable $x_1$ below.
Anyway, $y=0$ and in particular we have for example $$yx_2 - 0x_1 = yx_2 in I.$$ So by your degree considerations, we should get $yx_2 = c(2x_2 - 1) - c(1x_1-1)$ in $A[x_1, x_2]$ for some $cin A$. This is not the case. To exhibit this relation as an element of $I$ you need something like $yx_2 cdot (2x_2 - 1),$ where $yx_2$ is not of degree $0$, hence not in $A$, so the degree considerations fail.
I previously wrote a "correction" to your proof, but it was mainly nonsense. There was some germ of a good idea hidden there, though. Here is a sketch of a correct proof:
Consider a ring $A$ and a module $M$ over it, and take $$(Moplus M) / ((x,sx))$$ for some fixed $sin A$. What is the submodule generated by the first coordinate? If an element $x$ is annihilated by $s$, then $$(x,0) = (x,0) - (x,sx) = (0,0)$$ in the quotient. It can be verified directly that the submodule is just $$frac{M}{text{ker}(s:Mrightarrow M)}.$$ What if we take $$bigoplus_{i,j ge 0} M$$ and mod out by elements of the form $$se_{i+1,j}-e_{i,j},quad te_{i,j+1}-e_{i,j}?$$ I claim the $e_{0,0}$ coordinate then generates a module isomorphic to $$frac{M}{bigcup_n text{ker}(st)^n:Mrightarrow M},$$ et cetera.
To see this, consider the colimit explicitly. If an element of the submodule generated by $M_{0,0}$ in the quotient vanishes, it vanishes already in some f.g. submodule, which is contained in some direct sum of a finite set of $M_{i,j}$, and hence in a "rectangular" submodule $$bigoplus_{0le i,jle n} M_{i,j}.$$ Any element here can be "pushed" to one supported only in $M_{n,n}$, and it is easy to check what it means that it vanishes there. The same argument generalizes to higher "dimensions" $$bigoplus M_{i,j,k}.$$
You want to take $$B:= Aleft[{x_s}_{sin S}right]$$ and consider some ideal generated by elements of the form $(sx_s - 1)$. Considering the polynomial ring as a free $A$-module, the required statement follows from the previous one, once one shows the ideal is generated as an $A$-submodule of $B$ by elements of the same form as above. This is not so bad since $$(sx_s -1)(tx_t - 1) = (stx_sx_t - 1) - (sx_s -1) - (tx_t - 1)$$ is already a sum of three relations, and the same will occur for more complicated monomials in place of $x_s$ and $x_t$. Multiplying a relation by a monomial also gives you a sum of relations in the quotient submodule of $$bigoplus_{i,j}M_i,j$$ above.
answered Feb 1 at 22:40
gyashfegyashfe
963
$endgroup$
add a comment |
$begingroup$
I think the main mistake is in the statement "by degree considerations..." I did not verify that it is true for integral domains, but these considerations are more complicated in the presence of zerodivisors.
Let's consider a (classical) localization in a ring $A$ with zerodivisors. For example, let $A = mathbb{Z}[y]/(2y)$ and invert $2$. Then in the localization $y=0$.
Now consider your polynomial-ring construction, in which $2x_2 - 1 = 0$ and (multiplying by $y$) $$2x_2 y - y = 0,$$ i.e. $y=0$. This looks like a separate, more minor, problem: it is a relation not of the form $$frac{a}{s}=frac{b}{t}$$ for $s,t in S$ (you might want to take $1 in S$.) I'll assume we took $1in S$, and use a corresponding variable $x_1$ below.
Anyway, $y=0$ and in particular we have for example $$yx_2 - 0x_1 = yx_2 in I.$$ So by your degree considerations, we should get $yx_2 = c(2x_2 - 1) - c(1x_1-1)$ in $A[x_1, x_2]$ for some $cin A$. This is not the case. To exhibit this relation as an element of $I$ you need something like $yx_2 cdot (2x_2 - 1),$ where $yx_2$ is not of degree $0$, hence not in $A$, so the degree considerations fail.
I previously wrote a "correction" to your proof, but it was mainly nonsense. There was some germ of a good idea hidden there, though. Here is a sketch of a correct proof:
Consider a ring $A$ and a module $M$ over it, and take $$(Moplus M) / ((x,sx))$$ for some fixed $sin A$. What is the submodule generated by the first coordinate? If an element $x$ is annihilated by $s$, then $$(x,0) = (x,0) - (x,sx) = (0,0)$$ in the quotient. It can be verified directly that the submodule is just $$frac{M}{text{ker}(s:Mrightarrow M)}.$$ What if we take $$bigoplus_{i,j ge 0} M$$ and mod out by elements of the form $$se_{i+1,j}-e_{i,j},quad te_{i,j+1}-e_{i,j}?$$ I claim the $e_{0,0}$ coordinate then generates a module isomorphic to $$frac{M}{bigcup_n text{ker}(st)^n:Mrightarrow M},$$ et cetera.
To see this, consider the colimit explicitly. If an element of the submodule generated by $M_{0,0}$ in the quotient vanishes, it vanishes already in some f.g. submodule, which is contained in some direct sum of a finite set of $M_{i,j}$, and hence in a "rectangular" submodule $$bigoplus_{0le i,jle n} M_{i,j}.$$ Any element here can be "pushed" to one supported only in $M_{n,n}$, and it is easy to check what it means that it vanishes there. The same argument generalizes to higher "dimensions" $$bigoplus M_{i,j,k}.$$
You want to take $$B:= Aleft[{x_s}_{sin S}right]$$ and consider some ideal generated by elements of the form $(sx_s - 1)$. Considering the polynomial ring as a free $A$-module, the required statement follows from the previous one, once one shows the ideal is generated as an $A$-submodule of $B$ by elements of the same form as above. This is not so bad since $$(sx_s -1)(tx_t - 1) = (stx_sx_t - 1) - (sx_s -1) - (tx_t - 1)$$ is already a sum of three relations, and the same will occur for more complicated monomials in place of $x_s$ and $x_t$. Multiplying a relation by a monomial also gives you a sum of relations in the quotient submodule of $$bigoplus_{i,j}M_i,j$$ above.
answered Feb 1 at 22:40
gyashfegyashfe
963
$endgroup$
I think the main mistake is in the statement "by degree considerations..." I did not verify that it is true for integral domains, but these considerations are more complicated in the presence of zerodivisors.
Let's consider a (classical) localization in a ring $A$ with zerodivisors. For example, let $A = mathbb{Z}[y]/(2y)$ and invert $2$. Then in the localization $y=0$.
Now consider your polynomial-ring construction, in which $2x_2 - 1 = 0$ and (multiplying by $y$) $$2x_2 y - y = 0,$$ i.e. $y=0$. This looks like a separate, more minor, problem: it is a relation not of the form $$frac{a}{s}=frac{b}{t}$$ for $s,t in S$ (you might want to take $1 in S$.) I'll assume we took $1in S$, and use a corresponding variable $x_1$ below.
Anyway, $y=0$ and in particular we have for example $$yx_2 - 0x_1 = yx_2 in I.$$ So by your degree considerations, we should get $yx_2 = c(2x_2 - 1) - c(1x_1-1)$ in $A[x_1, x_2]$ for some $cin A$. This is not the case. To exhibit this relation as an element of $I$ you need something like $yx_2 cdot (2x_2 - 1),$ where $yx_2$ is not of degree $0$, hence not in $A$, so the degree considerations fail.
I previously wrote a "correction" to your proof, but it was mainly nonsense. There was some germ of a good idea hidden there, though. Here is a sketch of a correct proof:
Consider a ring $A$ and a module $M$ over it, and take $$(Moplus M) / ((x,sx))$$ for some fixed $sin A$. What is the submodule generated by the first coordinate? If an element $x$ is annihilated by $s$, then $$(x,0) = (x,0) - (x,sx) = (0,0)$$ in the quotient. It can be verified directly that the submodule is just $$frac{M}{text{ker}(s:Mrightarrow M)}.$$ What if we take $$bigoplus_{i,j ge 0} M$$ and mod out by elements of the form $$se_{i+1,j}-e_{i,j},quad te_{i,j+1}-e_{i,j}?$$ I claim the $e_{0,0}$ coordinate then generates a module isomorphic to $$frac{M}{bigcup_n text{ker}(st)^n:Mrightarrow M},$$ et cetera.
To see this, consider the colimit explicitly. If an element of the submodule generated by $M_{0,0}$ in the quotient vanishes, it vanishes already in some f.g. submodule, which is contained in some direct sum of a finite set of $M_{i,j}$, and hence in a "rectangular" submodule $$bigoplus_{0le i,jle n} M_{i,j}.$$ Any element here can be "pushed" to one supported only in $M_{n,n}$, and it is easy to check what it means that it vanishes there. The same argument generalizes to higher "dimensions" $$bigoplus M_{i,j,k}.$$
You want to take $$B:= Aleft[{x_s}_{sin S}right]$$ and consider some ideal generated by elements of the form $(sx_s - 1)$. Considering the polynomial ring as a free $A$-module, the required statement follows from the previous one, once one shows the ideal is generated as an $A$-submodule of $B$ by elements of the same form as above. This is not so bad since $$(sx_s -1)(tx_t - 1) = (stx_sx_t - 1) - (sx_s -1) - (tx_t - 1)$$ is already a sum of three relations, and the same will occur for more complicated monomials in place of $x_s$ and $x_t$. Multiplying a relation by a monomial also gives you a sum of relations in the quotient submodule of $$bigoplus_{i,j}M_i,j$$ above.
answered Feb 1 at 22:40
gyashfegyashfe
963
edited Feb 3 at 13:49
answered Feb 1 at 22:40
gyashfegyashfe
963
answered Feb 1 at 22:40
gyashfegyashfe
963
answered Feb 1 at 22:40
gyashfegyashfe
963
963
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3095680%2fequality-of-fractions-in-localization-as-quotient%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
