Action of finite group of ring automorphisms












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 18:38







Tommaso Scognamiglio

















asked Dec 12 '18 at 10:44









Tommaso ScognamiglioTommaso Scognamiglio

487312




487312












  • $begingroup$
    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




$begingroup$
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










1 Answer
1






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 19:36

























answered Dec 12 '18 at 18:08









MaxMax

14.2k11142




14.2k11142












  • $begingroup$
    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










  • $begingroup$
    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


















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