Prime Avoidance Application

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I have a question about a step in the proof of following statement:



Let $A$ be an integrally closed domain with the field of fractions $K, L$ a finite normal extension of $K, B$ the integral closure of $A$ in $L$. Then the group $G=operatorname Gal(L/K)$ acts transitively on each fiber of $operatorname SpecBto operatorname SpecA$.



Here one considers the prime ideals $p_1$ and $p_2$, and $sigma in operatorname Gal(L/K)$



Suppose $p_2neq sigma(p_1)$. Why then the prime avoidance provides that there is an element $x in p_2$ such that $sigma (x)not in mathfrak p_1 $ for any $sigma$? Thats not clear to me.



The statement of prime avoidance is that if an ideal $I$ in a commutative ring $R$ is contained in a union of finitely many prime ideals $P_i$'s, then it is contained in $P_i$ for some $i$.



Source: https://en.wikipedia.org/wiki/Integral_element#Integral_extensions




commutative-algebra




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  • See Exercise 5.13 of Atiyah-Macdonald (answer on the next page).
    – Kenny Lau
    Aug 2 at 19:26











  • What you say is in general false. Take $p_2=0$ and $p_1neq 0$.
    – Mohan
    Aug 2 at 22:09














up vote
0
down vote

favorite












I have a question about a step in the proof of following statement:



Let $A$ be an integrally closed domain with the field of fractions $K, L$ a finite normal extension of $K, B$ the integral closure of $A$ in $L$. Then the group $G=operatorname Gal(L/K)$ acts transitively on each fiber of $operatorname SpecBto operatorname SpecA$.



Here one considers the prime ideals $p_1$ and $p_2$, and $sigma in operatorname Gal(L/K)$



Suppose $p_2neq sigma(p_1)$. Why then the prime avoidance provides that there is an element $x in p_2$ such that $sigma (x)not in mathfrak p_1 $ for any $sigma$? Thats not clear to me.



The statement of prime avoidance is that if an ideal $I$ in a commutative ring $R$ is contained in a union of finitely many prime ideals $P_i$'s, then it is contained in $P_i$ for some $i$.



Source: https://en.wikipedia.org/wiki/Integral_element#Integral_extensions




commutative-algebra




share|cite|improve this question



















  • See Exercise 5.13 of Atiyah-Macdonald (answer on the next page).
    – Kenny Lau
    Aug 2 at 19:26











  • What you say is in general false. Take $p_2=0$ and $p_1neq 0$.
    – Mohan
    Aug 2 at 22:09












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have a question about a step in the proof of following statement:



Let $A$ be an integrally closed domain with the field of fractions $K, L$ a finite normal extension of $K, B$ the integral closure of $A$ in $L$. Then the group $G=operatorname Gal(L/K)$ acts transitively on each fiber of $operatorname SpecBto operatorname SpecA$.



Here one considers the prime ideals $p_1$ and $p_2$, and $sigma in operatorname Gal(L/K)$



Suppose $p_2neq sigma(p_1)$. Why then the prime avoidance provides that there is an element $x in p_2$ such that $sigma (x)not in mathfrak p_1 $ for any $sigma$? Thats not clear to me.



The statement of prime avoidance is that if an ideal $I$ in a commutative ring $R$ is contained in a union of finitely many prime ideals $P_i$'s, then it is contained in $P_i$ for some $i$.



Source: https://en.wikipedia.org/wiki/Integral_element#Integral_extensions




commutative-algebra




share|cite|improve this question











I have a question about a step in the proof of following statement:



Let $A$ be an integrally closed domain with the field of fractions $K, L$ a finite normal extension of $K, B$ the integral closure of $A$ in $L$. Then the group $G=operatorname Gal(L/K)$ acts transitively on each fiber of $operatorname SpecBto operatorname SpecA$.



Here one considers the prime ideals $p_1$ and $p_2$, and $sigma in operatorname Gal(L/K)$



Suppose $p_2neq sigma(p_1)$. Why then the prime avoidance provides that there is an element $x in p_2$ such that $sigma (x)not in mathfrak p_1 $ for any $sigma$? Thats not clear to me.



The statement of prime avoidance is that if an ideal $I$ in a commutative ring $R$ is contained in a union of finitely many prime ideals $P_i$'s, then it is contained in $P_i$ for some $i$.



Source: https://en.wikipedia.org/wiki/Integral_element#Integral_extensions





commutative-algebra





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asked Aug 2 at 19:19









KarlPeter

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  • See Exercise 5.13 of Atiyah-Macdonald (answer on the next page).
    – Kenny Lau
    Aug 2 at 19:26











  • What you say is in general false. Take $p_2=0$ and $p_1neq 0$.
    – Mohan
    Aug 2 at 22:09
















  • See Exercise 5.13 of Atiyah-Macdonald (answer on the next page).
    – Kenny Lau
    Aug 2 at 19:26











  • What you say is in general false. Take $p_2=0$ and $p_1neq 0$.
    – Mohan
    Aug 2 at 22:09















See Exercise 5.13 of Atiyah-Macdonald (answer on the next page).
– Kenny Lau
Aug 2 at 19:26





See Exercise 5.13 of Atiyah-Macdonald (answer on the next page).
– Kenny Lau
Aug 2 at 19:26













What you say is in general false. Take $p_2=0$ and $p_1neq 0$.
– Mohan
Aug 2 at 22:09




What you say is in general false. Take $p_2=0$ and $p_1neq 0$.
– Mohan
Aug 2 at 22:09















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