Ideals invariant under ring automorphisms
I am looking for ideals $Isubset mathbb{F}_2[x,y]$ with the following properties:
$I$ is generated by two homogeneous elements;
$I$ is invariant under the $SL_2(mathbb{F}_2)$-action on $mathbb{F}_2[x,y]$ (given by extending the action on the two dimensional sub vector space spanned by $x,y$).;- The quotient $mathbb{F}_2[x,y]/I$ is finite.
So far the only examples I know are $(x^n,y^n)$ and $(x^3,x^2+xy+y^2)$. The quotient $mathbb{F}_2[x,y]/(x^3,x^2+xy+y^2)$ is the cohomology ring $H^*(S^3/Q_8;mathbb{F}_2)$ for the standard action of the quaternion group on the three sphere.
Are these the only examples. Is it possible to classify all such ideals ?
Edit: Of course $(x^n,y^n)$ is only invariant when $n$ is a power of two (since $xmapsto x, ymapsto x+y$ is also an automorphism). So we have even fewer examples.
rt.representation-theory ac.commutative-algebra ra.rings-and-algebras invariant-theory ideals
asked Dec 15 '18 at 15:16
HenrikRüpingHenrikRüping
4,9562051
add a comment |
I am looking for ideals $Isubset mathbb{F}_2[x,y]$ with the following properties:
$I$ is generated by two homogeneous elements;
$I$ is invariant under the $SL_2(mathbb{F}_2)$-action on $mathbb{F}_2[x,y]$ (given by extending the action on the two dimensional sub vector space spanned by $x,y$).;- The quotient $mathbb{F}_2[x,y]/I$ is finite.
So far the only examples I know are $(x^n,y^n)$ and $(x^3,x^2+xy+y^2)$. The quotient $mathbb{F}_2[x,y]/(x^3,x^2+xy+y^2)$ is the cohomology ring $H^*(S^3/Q_8;mathbb{F}_2)$ for the standard action of the quaternion group on the three sphere.
Are these the only examples. Is it possible to classify all such ideals ?
Edit: Of course $(x^n,y^n)$ is only invariant when $n$ is a power of two (since $xmapsto x, ymapsto x+y$ is also an automorphism). So we have even fewer examples.
rt.representation-theory ac.commutative-algebra ra.rings-and-algebras invariant-theory ideals
asked Dec 15 '18 at 15:16
HenrikRüpingHenrikRüping
4,9562051
1
The ideal $(x^n,y^n)$ is invariant iff $n$ is a power of 2, isn't it?
– YCor
Dec 15 '18 at 16:20
1
Hi, why is $x^3,y^3$ invariant?
– Hailong Dao
Dec 15 '18 at 16:20
Yes indeed. Thanks to you both. I mixed up things quite a lot. I was also thinking about this question only with the action that flips $x$ and $y$ and not the entire $sl_2$-action. Then I mixed things up sorry.
– HenrikRüping
Dec 15 '18 at 16:32
Sorry. Corrected. Thanks.
– HenrikRüping
Dec 15 '18 at 23:29
add a comment |
I am looking for ideals $Isubset mathbb{F}_2[x,y]$ with the following properties:
$I$ is generated by two homogeneous elements;
$I$ is invariant under the $SL_2(mathbb{F}_2)$-action on $mathbb{F}_2[x,y]$ (given by extending the action on the two dimensional sub vector space spanned by $x,y$).;- The quotient $mathbb{F}_2[x,y]/I$ is finite.
So far the only examples I know are $(x^n,y^n)$ and $(x^3,x^2+xy+y^2)$. The quotient $mathbb{F}_2[x,y]/(x^3,x^2+xy+y^2)$ is the cohomology ring $H^*(S^3/Q_8;mathbb{F}_2)$ for the standard action of the quaternion group on the three sphere.
Are these the only examples. Is it possible to classify all such ideals ?
Edit: Of course $(x^n,y^n)$ is only invariant when $n$ is a power of two (since $xmapsto x, ymapsto x+y$ is also an automorphism). So we have even fewer examples.
rt.representation-theory ac.commutative-algebra ra.rings-and-algebras invariant-theory ideals
asked Dec 15 '18 at 15:16
HenrikRüpingHenrikRüping
4,9562051
I am looking for ideals $Isubset mathbb{F}_2[x,y]$ with the following properties:
$I$ is generated by two homogeneous elements;
$I$ is invariant under the $SL_2(mathbb{F}_2)$-action on $mathbb{F}_2[x,y]$ (given by extending the action on the two dimensional sub vector space spanned by $x,y$).;- The quotient $mathbb{F}_2[x,y]/I$ is finite.
So far the only examples I know are $(x^n,y^n)$ and $(x^3,x^2+xy+y^2)$. The quotient $mathbb{F}_2[x,y]/(x^3,x^2+xy+y^2)$ is the cohomology ring $H^*(S^3/Q_8;mathbb{F}_2)$ for the standard action of the quaternion group on the three sphere.
Are these the only examples. Is it possible to classify all such ideals ?
Edit: Of course $(x^n,y^n)$ is only invariant when $n$ is a power of two (since $xmapsto x, ymapsto x+y$ is also an automorphism). So we have even fewer examples.
rt.representation-theory ac.commutative-algebra ra.rings-and-algebras invariant-theory ideals
rt.representation-theory ac.commutative-algebra ra.rings-and-algebras invariant-theory ideals
asked Dec 15 '18 at 15:16
HenrikRüpingHenrikRüping
4,9562051
asked Dec 15 '18 at 15:16
HenrikRüpingHenrikRüping
4,9562051
edited Dec 15 '18 at 23:27
HenrikRüping
asked Dec 15 '18 at 15:16
HenrikRüpingHenrikRüping
4,9562051
asked Dec 15 '18 at 15:16
HenrikRüpingHenrikRüping
4,9562051
asked Dec 15 '18 at 15:16
HenrikRüpingHenrikRüping
4,9562051
4,9562051
1
The ideal $(x^n,y^n)$ is invariant iff $n$ is a power of 2, isn't it?
– YCor
Dec 15 '18 at 16:20
1
Hi, why is $x^3,y^3$ invariant?
– Hailong Dao
Dec 15 '18 at 16:20
Yes indeed. Thanks to you both. I mixed up things quite a lot. I was also thinking about this question only with the action that flips $x$ and $y$ and not the entire $sl_2$-action. Then I mixed things up sorry.
– HenrikRüping
Dec 15 '18 at 16:32
Sorry. Corrected. Thanks.
– HenrikRüping
Dec 15 '18 at 23:29
add a comment |
1
The ideal $(x^n,y^n)$ is invariant iff $n$ is a power of 2, isn't it?
– YCor
Dec 15 '18 at 16:20
1
Hi, why is $x^3,y^3$ invariant?
– Hailong Dao
Dec 15 '18 at 16:20
Yes indeed. Thanks to you both. I mixed up things quite a lot. I was also thinking about this question only with the action that flips $x$ and $y$ and not the entire $sl_2$-action. Then I mixed things up sorry.
– HenrikRüping
Dec 15 '18 at 16:32
Sorry. Corrected. Thanks.
– HenrikRüping
Dec 15 '18 at 23:29
1
1
The ideal $(x^n,y^n)$ is invariant iff $n$ is a power of 2, isn't it?
– YCor
Dec 15 '18 at 16:20
The ideal $(x^n,y^n)$ is invariant iff $n$ is a power of 2, isn't it?
– YCor
Dec 15 '18 at 16:20
1
1
Hi, why is $x^3,y^3$ invariant?
– Hailong Dao
Dec 15 '18 at 16:20
Hi, why is $x^3,y^3$ invariant?
– Hailong Dao
Dec 15 '18 at 16:20
Yes indeed. Thanks to you both. I mixed up things quite a lot. I was also thinking about this question only with the action that flips $x$ and $y$ and not the entire $sl_2$-action. Then I mixed things up sorry.
– HenrikRüping
Dec 15 '18 at 16:32
Yes indeed. Thanks to you both. I mixed up things quite a lot. I was also thinking about this question only with the action that flips $x$ and $y$ and not the entire $sl_2$-action. Then I mixed things up sorry.
– HenrikRüping
Dec 15 '18 at 16:32
Sorry. Corrected. Thanks.
– HenrikRüping
Dec 15 '18 at 23:29
Sorry. Corrected. Thanks.
– HenrikRüping
Dec 15 '18 at 23:29
add a comment |
2 Answers
2
active
oldest
votes
There are tons of examples. Put $u = x^2+xy+y^2$ and $v = xy(x+y)$. Then $u$ and $v$ are $SL_2(mathbb{F}_2)$ invariant. If $f(s,t)$ and $g(s,t)$ are homogenous polynomials with respect to the grading $deg s = 2$, $deg t=3$, then $langle f(u,v), g(u,v) rangle$ is a $SL_2(mathbb{F}_2)$-invariant generated in the manner you describe. I claim that, furthermore, if $mathbb{F}_2[s,t]/langle f, g rangle$ is finite then so is $mathbb{F}_2[x,y] / langle f(u,v), g(u,v) rangle$.
For this claim, we just need to show that $mathbb{F}_2[x,y]$ is module finite over $mathbb{F}_2[u,v]$. It is enough to show that the generators $x$ and $y$ are integral over $mathbb{F}_2[u,v]$. Indeed, they are both roots of the monic polynomial
$$z^3 + u z + v.$$
(I found $u$ and $v$ by multiplying out the polynomial $(z+x)(z+y)(z+x+y)$.)
edited Dec 15 '18 at 23:14
Matthieu Romagny
2,0821722
answered Dec 15 '18 at 17:01
David E SpeyerDavid E Speyer
105k8273534
add a comment |
One can see that $I=(f_1,dots, f_s)$ is $G$-invariant iff $alphacdot f_i in I$ for each $alpha in G$ and $i$. From this one can prove easily that if $I,J$ are $G$-invariant, then so is $IJ$ and $I+J$. Thus, for example, $(x,y)^n$ are invariant for all $ngeq 0$. One also have that $I=(x^{2^k}, y^{2^k})$ is invariant as pointed out by YCor. So you can generate many other examples. A complete classification seems difficult though.
For the case of two-generated ideals, one could use the above remark and the fact that "Frobenius commutes with linear change of variables" to show:
1) If $I=(f,g)$ is invariant then $J= (f^{2^k}, g^{2^k})$ is invariant.
2) If $I=(f,g)$ is invariant and $deg(f)> deg(g)$ then $J=(f^{2^k}, g^l)$ is invariant with $lleq 2^k$.
For instance the ideal $(x^6, x^2+xy+y^2)$ is invariant.
answered Dec 15 '18 at 21:42
Hailong DaoHailong Dao
19.6k374127
Note that these ideals might not be generated by two elements anymore.
– HenrikRüping
Dec 15 '18 at 23:32
@HenrikRüping: Ah, yes, I was thinking about general ideals, not just two-generated ones. I added a few comments about this case.
– Hailong Dao
Dec 17 '18 at 0:26
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: "504"
};
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%2fmathoverflow.net%2fquestions%2f318736%2fideals-invariant-under-ring-automorphisms%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
There are tons of examples. Put $u = x^2+xy+y^2$ and $v = xy(x+y)$. Then $u$ and $v$ are $SL_2(mathbb{F}_2)$ invariant. If $f(s,t)$ and $g(s,t)$ are homogenous polynomials with respect to the grading $deg s = 2$, $deg t=3$, then $langle f(u,v), g(u,v) rangle$ is a $SL_2(mathbb{F}_2)$-invariant generated in the manner you describe. I claim that, furthermore, if $mathbb{F}_2[s,t]/langle f, g rangle$ is finite then so is $mathbb{F}_2[x,y] / langle f(u,v), g(u,v) rangle$.
For this claim, we just need to show that $mathbb{F}_2[x,y]$ is module finite over $mathbb{F}_2[u,v]$. It is enough to show that the generators $x$ and $y$ are integral over $mathbb{F}_2[u,v]$. Indeed, they are both roots of the monic polynomial
$$z^3 + u z + v.$$
(I found $u$ and $v$ by multiplying out the polynomial $(z+x)(z+y)(z+x+y)$.)
edited Dec 15 '18 at 23:14
Matthieu Romagny
2,0821722
answered Dec 15 '18 at 17:01
David E SpeyerDavid E Speyer
105k8273534
add a comment |
There are tons of examples. Put $u = x^2+xy+y^2$ and $v = xy(x+y)$. Then $u$ and $v$ are $SL_2(mathbb{F}_2)$ invariant. If $f(s,t)$ and $g(s,t)$ are homogenous polynomials with respect to the grading $deg s = 2$, $deg t=3$, then $langle f(u,v), g(u,v) rangle$ is a $SL_2(mathbb{F}_2)$-invariant generated in the manner you describe. I claim that, furthermore, if $mathbb{F}_2[s,t]/langle f, g rangle$ is finite then so is $mathbb{F}_2[x,y] / langle f(u,v), g(u,v) rangle$.
For this claim, we just need to show that $mathbb{F}_2[x,y]$ is module finite over $mathbb{F}_2[u,v]$. It is enough to show that the generators $x$ and $y$ are integral over $mathbb{F}_2[u,v]$. Indeed, they are both roots of the monic polynomial
$$z^3 + u z + v.$$
(I found $u$ and $v$ by multiplying out the polynomial $(z+x)(z+y)(z+x+y)$.)
edited Dec 15 '18 at 23:14
Matthieu Romagny
2,0821722
answered Dec 15 '18 at 17:01
David E SpeyerDavid E Speyer
105k8273534
add a comment |
There are tons of examples. Put $u = x^2+xy+y^2$ and $v = xy(x+y)$. Then $u$ and $v$ are $SL_2(mathbb{F}_2)$ invariant. If $f(s,t)$ and $g(s,t)$ are homogenous polynomials with respect to the grading $deg s = 2$, $deg t=3$, then $langle f(u,v), g(u,v) rangle$ is a $SL_2(mathbb{F}_2)$-invariant generated in the manner you describe. I claim that, furthermore, if $mathbb{F}_2[s,t]/langle f, g rangle$ is finite then so is $mathbb{F}_2[x,y] / langle f(u,v), g(u,v) rangle$.
For this claim, we just need to show that $mathbb{F}_2[x,y]$ is module finite over $mathbb{F}_2[u,v]$. It is enough to show that the generators $x$ and $y$ are integral over $mathbb{F}_2[u,v]$. Indeed, they are both roots of the monic polynomial
$$z^3 + u z + v.$$
(I found $u$ and $v$ by multiplying out the polynomial $(z+x)(z+y)(z+x+y)$.)
edited Dec 15 '18 at 23:14
Matthieu Romagny
2,0821722
answered Dec 15 '18 at 17:01
David E SpeyerDavid E Speyer
105k8273534
There are tons of examples. Put $u = x^2+xy+y^2$ and $v = xy(x+y)$. Then $u$ and $v$ are $SL_2(mathbb{F}_2)$ invariant. If $f(s,t)$ and $g(s,t)$ are homogenous polynomials with respect to the grading $deg s = 2$, $deg t=3$, then $langle f(u,v), g(u,v) rangle$ is a $SL_2(mathbb{F}_2)$-invariant generated in the manner you describe. I claim that, furthermore, if $mathbb{F}_2[s,t]/langle f, g rangle$ is finite then so is $mathbb{F}_2[x,y] / langle f(u,v), g(u,v) rangle$.
For this claim, we just need to show that $mathbb{F}_2[x,y]$ is module finite over $mathbb{F}_2[u,v]$. It is enough to show that the generators $x$ and $y$ are integral over $mathbb{F}_2[u,v]$. Indeed, they are both roots of the monic polynomial
$$z^3 + u z + v.$$
(I found $u$ and $v$ by multiplying out the polynomial $(z+x)(z+y)(z+x+y)$.)
edited Dec 15 '18 at 23:14
Matthieu Romagny
2,0821722
answered Dec 15 '18 at 17:01
David E SpeyerDavid E Speyer
105k8273534
edited Dec 15 '18 at 23:14
Matthieu Romagny
2,0821722
edited Dec 15 '18 at 23:14
Matthieu Romagny
2,0821722
edited Dec 15 '18 at 23:14
Matthieu Romagny
2,0821722
2,0821722
answered Dec 15 '18 at 17:01
David E SpeyerDavid E Speyer
105k8273534
answered Dec 15 '18 at 17:01
David E SpeyerDavid E Speyer
105k8273534
answered Dec 15 '18 at 17:01
David E SpeyerDavid E Speyer
105k8273534
105k8273534
add a comment |
add a comment |
One can see that $I=(f_1,dots, f_s)$ is $G$-invariant iff $alphacdot f_i in I$ for each $alpha in G$ and $i$. From this one can prove easily that if $I,J$ are $G$-invariant, then so is $IJ$ and $I+J$. Thus, for example, $(x,y)^n$ are invariant for all $ngeq 0$. One also have that $I=(x^{2^k}, y^{2^k})$ is invariant as pointed out by YCor. So you can generate many other examples. A complete classification seems difficult though.
For the case of two-generated ideals, one could use the above remark and the fact that "Frobenius commutes with linear change of variables" to show:
1) If $I=(f,g)$ is invariant then $J= (f^{2^k}, g^{2^k})$ is invariant.
2) If $I=(f,g)$ is invariant and $deg(f)> deg(g)$ then $J=(f^{2^k}, g^l)$ is invariant with $lleq 2^k$.
For instance the ideal $(x^6, x^2+xy+y^2)$ is invariant.
answered Dec 15 '18 at 21:42
Hailong DaoHailong Dao
19.6k374127
Note that these ideals might not be generated by two elements anymore.
– HenrikRüping
Dec 15 '18 at 23:32
@HenrikRüping: Ah, yes, I was thinking about general ideals, not just two-generated ones. I added a few comments about this case.
– Hailong Dao
Dec 17 '18 at 0:26
add a comment |
One can see that $I=(f_1,dots, f_s)$ is $G$-invariant iff $alphacdot f_i in I$ for each $alpha in G$ and $i$. From this one can prove easily that if $I,J$ are $G$-invariant, then so is $IJ$ and $I+J$. Thus, for example, $(x,y)^n$ are invariant for all $ngeq 0$. One also have that $I=(x^{2^k}, y^{2^k})$ is invariant as pointed out by YCor. So you can generate many other examples. A complete classification seems difficult though.
For the case of two-generated ideals, one could use the above remark and the fact that "Frobenius commutes with linear change of variables" to show:
1) If $I=(f,g)$ is invariant then $J= (f^{2^k}, g^{2^k})$ is invariant.
2) If $I=(f,g)$ is invariant and $deg(f)> deg(g)$ then $J=(f^{2^k}, g^l)$ is invariant with $lleq 2^k$.
For instance the ideal $(x^6, x^2+xy+y^2)$ is invariant.
answered Dec 15 '18 at 21:42
Hailong DaoHailong Dao
19.6k374127
Note that these ideals might not be generated by two elements anymore.
– HenrikRüping
Dec 15 '18 at 23:32
@HenrikRüping: Ah, yes, I was thinking about general ideals, not just two-generated ones. I added a few comments about this case.
– Hailong Dao
Dec 17 '18 at 0:26
add a comment |
One can see that $I=(f_1,dots, f_s)$ is $G$-invariant iff $alphacdot f_i in I$ for each $alpha in G$ and $i$. From this one can prove easily that if $I,J$ are $G$-invariant, then so is $IJ$ and $I+J$. Thus, for example, $(x,y)^n$ are invariant for all $ngeq 0$. One also have that $I=(x^{2^k}, y^{2^k})$ is invariant as pointed out by YCor. So you can generate many other examples. A complete classification seems difficult though.
For the case of two-generated ideals, one could use the above remark and the fact that "Frobenius commutes with linear change of variables" to show:
1) If $I=(f,g)$ is invariant then $J= (f^{2^k}, g^{2^k})$ is invariant.
2) If $I=(f,g)$ is invariant and $deg(f)> deg(g)$ then $J=(f^{2^k}, g^l)$ is invariant with $lleq 2^k$.
For instance the ideal $(x^6, x^2+xy+y^2)$ is invariant.
answered Dec 15 '18 at 21:42
Hailong DaoHailong Dao
19.6k374127
One can see that $I=(f_1,dots, f_s)$ is $G$-invariant iff $alphacdot f_i in I$ for each $alpha in G$ and $i$. From this one can prove easily that if $I,J$ are $G$-invariant, then so is $IJ$ and $I+J$. Thus, for example, $(x,y)^n$ are invariant for all $ngeq 0$. One also have that $I=(x^{2^k}, y^{2^k})$ is invariant as pointed out by YCor. So you can generate many other examples. A complete classification seems difficult though.
For the case of two-generated ideals, one could use the above remark and the fact that "Frobenius commutes with linear change of variables" to show:
1) If $I=(f,g)$ is invariant then $J= (f^{2^k}, g^{2^k})$ is invariant.
2) If $I=(f,g)$ is invariant and $deg(f)> deg(g)$ then $J=(f^{2^k}, g^l)$ is invariant with $lleq 2^k$.
For instance the ideal $(x^6, x^2+xy+y^2)$ is invariant.
answered Dec 15 '18 at 21:42
Hailong DaoHailong Dao
19.6k374127
edited Dec 16 '18 at 5:35
answered Dec 15 '18 at 21:42
Hailong DaoHailong Dao
19.6k374127
answered Dec 15 '18 at 21:42
Hailong DaoHailong Dao
19.6k374127
answered Dec 15 '18 at 21:42
Hailong DaoHailong Dao
19.6k374127
19.6k374127
Note that these ideals might not be generated by two elements anymore.
– HenrikRüping
Dec 15 '18 at 23:32
@HenrikRüping: Ah, yes, I was thinking about general ideals, not just two-generated ones. I added a few comments about this case.
– Hailong Dao
Dec 17 '18 at 0:26
add a comment |
Note that these ideals might not be generated by two elements anymore.
– HenrikRüping
Dec 15 '18 at 23:32
@HenrikRüping: Ah, yes, I was thinking about general ideals, not just two-generated ones. I added a few comments about this case.
– Hailong Dao
Dec 17 '18 at 0:26
Note that these ideals might not be generated by two elements anymore.
– HenrikRüping
Dec 15 '18 at 23:32
Note that these ideals might not be generated by two elements anymore.
– HenrikRüping
Dec 15 '18 at 23:32
@HenrikRüping: Ah, yes, I was thinking about general ideals, not just two-generated ones. I added a few comments about this case.
– Hailong Dao
Dec 17 '18 at 0:26
@HenrikRüping: Ah, yes, I was thinking about general ideals, not just two-generated ones. I added a few comments about this case.
– Hailong Dao
Dec 17 '18 at 0:26
add a comment |
Thanks for contributing an answer to MathOverflow!
- 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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2fmathoverflow.net%2fquestions%2f318736%2fideals-invariant-under-ring-automorphisms%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
1
The ideal $(x^n,y^n)$ is invariant iff $n$ is a power of 2, isn't it?
– YCor
Dec 15 '18 at 16:20
1
Hi, why is $x^3,y^3$ invariant?
– Hailong Dao
Dec 15 '18 at 16:20
Yes indeed. Thanks to you both. I mixed up things quite a lot. I was also thinking about this question only with the action that flips $x$ and $y$ and not the entire $sl_2$-action. Then I mixed things up sorry.
– HenrikRüping
Dec 15 '18 at 16:32
Sorry. Corrected. Thanks.
– HenrikRüping
Dec 15 '18 at 23:29