Action of finite group of ring automorphisms
$begingroup$
Let $A$ be commutative ring and $G$ a finite group of ring automorphisms of $A$. We have a finite ring extension $A^G to A$ that induces a surjective map $phi: Spec(A) to Spec(A^G)$. I'd like to prove that given $a in A$ and $b_i in A^G$ such that $$t^n+sum_{i leq n} b_i t^i=prod_{sigma in G}(t-sigma(a)) ,$$ we have $phi(D(a))= cup D(b_i)$ where $D(f)$ is the canonical principal open subset.
I managed to prove one inclusion (namely $subset$) but I've got no idea for the other one.
abstract-algebra algebraic-geometry commutative-algebra affine-schemes
asked Dec 12 '18 at 10:44
Tommaso ScognamiglioTommaso Scognamiglio
492312
$endgroup$
add a comment |
$begingroup$
Let $A$ be commutative ring and $G$ a finite group of ring automorphisms of $A$. We have a finite ring extension $A^G to A$ that induces a surjective map $phi: Spec(A) to Spec(A^G)$. I'd like to prove that given $a in A$ and $b_i in A^G$ such that $$t^n+sum_{i leq n} b_i t^i=prod_{sigma in G}(t-sigma(a)) ,$$ we have $phi(D(a))= cup D(b_i)$ where $D(f)$ is the canonical principal open subset.
I managed to prove one inclusion (namely $subset$) but I've got no idea for the other one.
abstract-algebra algebraic-geometry commutative-algebra affine-schemes
asked Dec 12 '18 at 10:44
Tommaso ScognamiglioTommaso Scognamiglio
492312
$endgroup$
$begingroup$
that looks a lot like an algebraic version of covering space theory, sadly I do not know enough elgebraic geometry to make this precise, but maybe a look on the corresponing proof in topology may help (I even think that the spec you are looking at is actually covering space!)
$endgroup$
– Enkidu
Dec 12 '18 at 12:55
add a comment |
$begingroup$
Let $A$ be commutative ring and $G$ a finite group of ring automorphisms of $A$. We have a finite ring extension $A^G to A$ that induces a surjective map $phi: Spec(A) to Spec(A^G)$. I'd like to prove that given $a in A$ and $b_i in A^G$ such that $$t^n+sum_{i leq n} b_i t^i=prod_{sigma in G}(t-sigma(a)) ,$$ we have $phi(D(a))= cup D(b_i)$ where $D(f)$ is the canonical principal open subset.
I managed to prove one inclusion (namely $subset$) but I've got no idea for the other one.
abstract-algebra algebraic-geometry commutative-algebra affine-schemes
asked Dec 12 '18 at 10:44
Tommaso ScognamiglioTommaso Scognamiglio
492312
$endgroup$
Let $A$ be commutative ring and $G$ a finite group of ring automorphisms of $A$. We have a finite ring extension $A^G to A$ that induces a surjective map $phi: Spec(A) to Spec(A^G)$. I'd like to prove that given $a in A$ and $b_i in A^G$ such that $$t^n+sum_{i leq n} b_i t^i=prod_{sigma in G}(t-sigma(a)) ,$$ we have $phi(D(a))= cup D(b_i)$ where $D(f)$ is the canonical principal open subset.
I managed to prove one inclusion (namely $subset$) but I've got no idea for the other one.
abstract-algebra algebraic-geometry commutative-algebra affine-schemes
abstract-algebra algebraic-geometry commutative-algebra affine-schemes
asked Dec 12 '18 at 10:44
Tommaso ScognamiglioTommaso Scognamiglio
492312
asked Dec 12 '18 at 10:44
Tommaso ScognamiglioTommaso Scognamiglio
492312
edited Dec 12 '18 at 18:38
Tommaso Scognamiglio
asked Dec 12 '18 at 10:44
Tommaso ScognamiglioTommaso Scognamiglio
492312
asked Dec 12 '18 at 10:44
Tommaso ScognamiglioTommaso Scognamiglio
492312
asked Dec 12 '18 at 10:44
Tommaso ScognamiglioTommaso Scognamiglio
492312
492312
$begingroup$
that looks a lot like an algebraic version of covering space theory, sadly I do not know enough elgebraic geometry to make this precise, but maybe a look on the corresponing proof in topology may help (I even think that the spec you are looking at is actually covering space!)
$endgroup$
– Enkidu
Dec 12 '18 at 12:55
add a comment |
$begingroup$
that looks a lot like an algebraic version of covering space theory, sadly I do not know enough elgebraic geometry to make this precise, but maybe a look on the corresponing proof in topology may help (I even think that the spec you are looking at is actually covering space!)
$endgroup$
– Enkidu
Dec 12 '18 at 12:55
$begingroup$
that looks a lot like an algebraic version of covering space theory, sadly I do not know enough elgebraic geometry to make this precise, but maybe a look on the corresponing proof in topology may help (I even think that the spec you are looking at is actually covering space!)
$endgroup$
– Enkidu
Dec 12 '18 at 12:55
$begingroup$
that looks a lot like an algebraic version of covering space theory, sadly I do not know enough elgebraic geometry to make this precise, but maybe a look on the corresponing proof in topology may help (I even think that the spec you are looking at is actually covering space!)
$endgroup$
– Enkidu
Dec 12 '18 at 12:55
add a comment |
1 Answer
1
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oldest
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$begingroup$
Take $A=mathbb{Z}$, $G=1$.
Then $5^3 = 125 = 3times 25 + 10 times 5 = 3times 5^2 + 10times 5$; $3,10in A^G$; but $D(5)neq D(3)cup D(10)$ (note that $phi = id$). So you need some more conditions for it to work, if it can work.
With the new condition : For $i=0$, $b_0 = displaystyleprod_{sigmain G}sigma(a)$, thus if $p$ is a prime ideal of $A^G$ and $phi(q) = p$, and $ain q$, then $b_0 in qcap A^G = p$, thus $pnotin D(b_0)$. Thus $pin D(b_0)implies qin D(a) implies pin D(phi(a))$.
More generally, if $G= {id=sigma_0, sigma_1,...,sigma_n}$ we have $b_i = displaystylesum_{i_0<...<i_l} prod_{k=0}^l sigma_{i_k}(a)$ for $l=n-i$. Thus if $p$ is prime in $A^G, p=qcap A^G$ with $q$ prime in $A$, and $b_inotin p$, then $b_inotin q$, so for some $i_0<...<i_l$, $displaystyleprod_{k=0}^l sigma_{i_k}(a)notin q$. Applying $g=sigma_{i_0}^{-1}$ to both sides we get in particular $anotin gcdot q$. But $phi(gcdot q) = phi(q) = p$, so $gcdot q in D(a)$ and $p=phi(gcdot q)in phi(D(a))$, thus $D(b_i)subset phi(D(a))$.
Thus $cup_i D(b_i)subset phi(D(a))$. The other inclusion is easy as you pointed out.
answered Dec 12 '18 at 18:08
MaxMax
14.4k11142
$endgroup$
$begingroup$
You're right. I corrected now. I had way more strict condition.
$endgroup$
– Tommaso Scognamiglio
Dec 12 '18 at 18:39
$begingroup$
I added a solution for the corrected question.
$endgroup$
– Max
Dec 12 '18 at 19:36
add a comment |
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$begingroup$
Take $A=mathbb{Z}$, $G=1$.
Then $5^3 = 125 = 3times 25 + 10 times 5 = 3times 5^2 + 10times 5$; $3,10in A^G$; but $D(5)neq D(3)cup D(10)$ (note that $phi = id$). So you need some more conditions for it to work, if it can work.
With the new condition : For $i=0$, $b_0 = displaystyleprod_{sigmain G}sigma(a)$, thus if $p$ is a prime ideal of $A^G$ and $phi(q) = p$, and $ain q$, then $b_0 in qcap A^G = p$, thus $pnotin D(b_0)$. Thus $pin D(b_0)implies qin D(a) implies pin D(phi(a))$.
More generally, if $G= {id=sigma_0, sigma_1,...,sigma_n}$ we have $b_i = displaystylesum_{i_0<...<i_l} prod_{k=0}^l sigma_{i_k}(a)$ for $l=n-i$. Thus if $p$ is prime in $A^G, p=qcap A^G$ with $q$ prime in $A$, and $b_inotin p$, then $b_inotin q$, so for some $i_0<...<i_l$, $displaystyleprod_{k=0}^l sigma_{i_k}(a)notin q$. Applying $g=sigma_{i_0}^{-1}$ to both sides we get in particular $anotin gcdot q$. But $phi(gcdot q) = phi(q) = p$, so $gcdot q in D(a)$ and $p=phi(gcdot q)in phi(D(a))$, thus $D(b_i)subset phi(D(a))$.
Thus $cup_i D(b_i)subset phi(D(a))$. The other inclusion is easy as you pointed out.
answered Dec 12 '18 at 18:08
MaxMax
14.4k11142
$endgroup$
$begingroup$
You're right. I corrected now. I had way more strict condition.
$endgroup$
– Tommaso Scognamiglio
Dec 12 '18 at 18:39
$begingroup$
I added a solution for the corrected question.
$endgroup$
– Max
Dec 12 '18 at 19:36
add a comment |
$begingroup$
Take $A=mathbb{Z}$, $G=1$.
Then $5^3 = 125 = 3times 25 + 10 times 5 = 3times 5^2 + 10times 5$; $3,10in A^G$; but $D(5)neq D(3)cup D(10)$ (note that $phi = id$). So you need some more conditions for it to work, if it can work.
With the new condition : For $i=0$, $b_0 = displaystyleprod_{sigmain G}sigma(a)$, thus if $p$ is a prime ideal of $A^G$ and $phi(q) = p$, and $ain q$, then $b_0 in qcap A^G = p$, thus $pnotin D(b_0)$. Thus $pin D(b_0)implies qin D(a) implies pin D(phi(a))$.
More generally, if $G= {id=sigma_0, sigma_1,...,sigma_n}$ we have $b_i = displaystylesum_{i_0<...<i_l} prod_{k=0}^l sigma_{i_k}(a)$ for $l=n-i$. Thus if $p$ is prime in $A^G, p=qcap A^G$ with $q$ prime in $A$, and $b_inotin p$, then $b_inotin q$, so for some $i_0<...<i_l$, $displaystyleprod_{k=0}^l sigma_{i_k}(a)notin q$. Applying $g=sigma_{i_0}^{-1}$ to both sides we get in particular $anotin gcdot q$. But $phi(gcdot q) = phi(q) = p$, so $gcdot q in D(a)$ and $p=phi(gcdot q)in phi(D(a))$, thus $D(b_i)subset phi(D(a))$.
Thus $cup_i D(b_i)subset phi(D(a))$. The other inclusion is easy as you pointed out.
answered Dec 12 '18 at 18:08
MaxMax
14.4k11142
$endgroup$
$begingroup$
You're right. I corrected now. I had way more strict condition.
$endgroup$
– Tommaso Scognamiglio
Dec 12 '18 at 18:39
$begingroup$
I added a solution for the corrected question.
$endgroup$
– Max
Dec 12 '18 at 19:36
add a comment |
$begingroup$
Take $A=mathbb{Z}$, $G=1$.
Then $5^3 = 125 = 3times 25 + 10 times 5 = 3times 5^2 + 10times 5$; $3,10in A^G$; but $D(5)neq D(3)cup D(10)$ (note that $phi = id$). So you need some more conditions for it to work, if it can work.
With the new condition : For $i=0$, $b_0 = displaystyleprod_{sigmain G}sigma(a)$, thus if $p$ is a prime ideal of $A^G$ and $phi(q) = p$, and $ain q$, then $b_0 in qcap A^G = p$, thus $pnotin D(b_0)$. Thus $pin D(b_0)implies qin D(a) implies pin D(phi(a))$.
More generally, if $G= {id=sigma_0, sigma_1,...,sigma_n}$ we have $b_i = displaystylesum_{i_0<...<i_l} prod_{k=0}^l sigma_{i_k}(a)$ for $l=n-i$. Thus if $p$ is prime in $A^G, p=qcap A^G$ with $q$ prime in $A$, and $b_inotin p$, then $b_inotin q$, so for some $i_0<...<i_l$, $displaystyleprod_{k=0}^l sigma_{i_k}(a)notin q$. Applying $g=sigma_{i_0}^{-1}$ to both sides we get in particular $anotin gcdot q$. But $phi(gcdot q) = phi(q) = p$, so $gcdot q in D(a)$ and $p=phi(gcdot q)in phi(D(a))$, thus $D(b_i)subset phi(D(a))$.
Thus $cup_i D(b_i)subset phi(D(a))$. The other inclusion is easy as you pointed out.
answered Dec 12 '18 at 18:08
MaxMax
14.4k11142
$endgroup$
Take $A=mathbb{Z}$, $G=1$.
Then $5^3 = 125 = 3times 25 + 10 times 5 = 3times 5^2 + 10times 5$; $3,10in A^G$; but $D(5)neq D(3)cup D(10)$ (note that $phi = id$). So you need some more conditions for it to work, if it can work.
With the new condition : For $i=0$, $b_0 = displaystyleprod_{sigmain G}sigma(a)$, thus if $p$ is a prime ideal of $A^G$ and $phi(q) = p$, and $ain q$, then $b_0 in qcap A^G = p$, thus $pnotin D(b_0)$. Thus $pin D(b_0)implies qin D(a) implies pin D(phi(a))$.
More generally, if $G= {id=sigma_0, sigma_1,...,sigma_n}$ we have $b_i = displaystylesum_{i_0<...<i_l} prod_{k=0}^l sigma_{i_k}(a)$ for $l=n-i$. Thus if $p$ is prime in $A^G, p=qcap A^G$ with $q$ prime in $A$, and $b_inotin p$, then $b_inotin q$, so for some $i_0<...<i_l$, $displaystyleprod_{k=0}^l sigma_{i_k}(a)notin q$. Applying $g=sigma_{i_0}^{-1}$ to both sides we get in particular $anotin gcdot q$. But $phi(gcdot q) = phi(q) = p$, so $gcdot q in D(a)$ and $p=phi(gcdot q)in phi(D(a))$, thus $D(b_i)subset phi(D(a))$.
Thus $cup_i D(b_i)subset phi(D(a))$. The other inclusion is easy as you pointed out.
answered Dec 12 '18 at 18:08
MaxMax
14.4k11142
edited Dec 12 '18 at 19:36
answered Dec 12 '18 at 18:08
MaxMax
14.4k11142
answered Dec 12 '18 at 18:08
MaxMax
14.4k11142
answered Dec 12 '18 at 18:08
MaxMax
14.4k11142
14.4k11142
$begingroup$
You're right. I corrected now. I had way more strict condition.
$endgroup$
– Tommaso Scognamiglio
Dec 12 '18 at 18:39
$begingroup$
I added a solution for the corrected question.
$endgroup$
– Max
Dec 12 '18 at 19:36
add a comment |
$begingroup$
You're right. I corrected now. I had way more strict condition.
$endgroup$
– Tommaso Scognamiglio
Dec 12 '18 at 18:39
$begingroup$
I added a solution for the corrected question.
$endgroup$
– Max
Dec 12 '18 at 19:36
$begingroup$
You're right. I corrected now. I had way more strict condition.
$endgroup$
– Tommaso Scognamiglio
Dec 12 '18 at 18:39
$begingroup$
You're right. I corrected now. I had way more strict condition.
$endgroup$
– Tommaso Scognamiglio
Dec 12 '18 at 18:39
$begingroup$
I added a solution for the corrected question.
$endgroup$
– Max
Dec 12 '18 at 19:36
$begingroup$
I added a solution for the corrected question.
$endgroup$
– Max
Dec 12 '18 at 19:36
add a comment |
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$begingroup$
that looks a lot like an algebraic version of covering space theory, sadly I do not know enough elgebraic geometry to make this precise, but maybe a look on the corresponing proof in topology may help (I even think that the spec you are looking at is actually covering space!)
$endgroup$
– Enkidu
Dec 12 '18 at 12:55