A problem about the basis of the Zariski topology
Let $A$ be a commutative ring with $1_A$. Let $X= mathrm{Spec}(A)$ be a topological space equipped with the Zariski topology. Let $U subseteq X$ be an open set and choose $x_1, ldots, x_n in U$. Prove that there exists $f in A$ such that $${ x_1, ldots, x_n } subset D(f) subseteq U.$$
I'm stuck with proving the existence of such $f$. Any help will be appreciated.
Note: For $x in X=mathrm{Spec}(A)$, we shall write an ideal with $j_x$. Then $$V(f)= {x in X : f in j_x} text{and} D(f)=X-V(f).$$ Further, we know that $D(f)$ forms a basis for the open sets of $X$.
algebraic-geometry zariski-topology
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Let $A$ be a commutative ring with $1_A$. Let $X= mathrm{Spec}(A)$ be a topological space equipped with the Zariski topology. Let $U subseteq X$ be an open set and choose $x_1, ldots, x_n in U$. Prove that there exists $f in A$ such that $${ x_1, ldots, x_n } subset D(f) subseteq U.$$
I'm stuck with proving the existence of such $f$. Any help will be appreciated.
Note: For $x in X=mathrm{Spec}(A)$, we shall write an ideal with $j_x$. Then $$V(f)= {x in X : f in j_x} text{and} D(f)=X-V(f).$$ Further, we know that $D(f)$ forms a basis for the open sets of $X$.
algebraic-geometry zariski-topology
Is $U$ open in $operatorname{Spec}A $?
– Bernard
Dec 1 at 10:32
Thanks, I fixed it.
– user621469
Dec 1 at 10:48
add a comment |
Let $A$ be a commutative ring with $1_A$. Let $X= mathrm{Spec}(A)$ be a topological space equipped with the Zariski topology. Let $U subseteq X$ be an open set and choose $x_1, ldots, x_n in U$. Prove that there exists $f in A$ such that $${ x_1, ldots, x_n } subset D(f) subseteq U.$$
I'm stuck with proving the existence of such $f$. Any help will be appreciated.
Note: For $x in X=mathrm{Spec}(A)$, we shall write an ideal with $j_x$. Then $$V(f)= {x in X : f in j_x} text{and} D(f)=X-V(f).$$ Further, we know that $D(f)$ forms a basis for the open sets of $X$.
algebraic-geometry zariski-topology
Let $A$ be a commutative ring with $1_A$. Let $X= mathrm{Spec}(A)$ be a topological space equipped with the Zariski topology. Let $U subseteq X$ be an open set and choose $x_1, ldots, x_n in U$. Prove that there exists $f in A$ such that $${ x_1, ldots, x_n } subset D(f) subseteq U.$$
I'm stuck with proving the existence of such $f$. Any help will be appreciated.
Note: For $x in X=mathrm{Spec}(A)$, we shall write an ideal with $j_x$. Then $$V(f)= {x in X : f in j_x} text{and} D(f)=X-V(f).$$ Further, we know that $D(f)$ forms a basis for the open sets of $X$.
algebraic-geometry zariski-topology
algebraic-geometry zariski-topology
edited Dec 1 at 10:48
asked Dec 1 at 10:27
user621469
Is $U$ open in $operatorname{Spec}A $?
– Bernard
Dec 1 at 10:32
Thanks, I fixed it.
– user621469
Dec 1 at 10:48
add a comment |
Is $U$ open in $operatorname{Spec}A $?
– Bernard
Dec 1 at 10:32
Thanks, I fixed it.
– user621469
Dec 1 at 10:48
Is $U$ open in $operatorname{Spec}A $?
– Bernard
Dec 1 at 10:32
Is $U$ open in $operatorname{Spec}A $?
– Bernard
Dec 1 at 10:32
Thanks, I fixed it.
– user621469
Dec 1 at 10:48
Thanks, I fixed it.
– user621469
Dec 1 at 10:48
add a comment |
1 Answer
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The key is to use the prime avoidance lemma, which says that if an ideal $I$ is contained in a finite union of prime ideals, then it is contained in one of them.
Let $I$ be an ideal in $A$ such that $U=Xsetminus V(I)$. By definition, $x_1,ldots,x_n$ are primes of $A$ that don't contain $I$, since they are not in $V(I)$. Then we use prime avoidance to choose $$fin Isetminus bigcup_{i=1}^n x_i$$
Then $fin I$, so if a prime contains $I$, it contains $f$. Hence $V(f)supseteq V(I)$, so $D(f)subseteq U$. More importantly, $fnotin x_i$ for any $i$, so $x_inotin V(f)$, which means that $x_iin D(f)$ for all $i$, as desired.
answered Dec 1 at 19:06
jgon
12.4k21940
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The key is to use the prime avoidance lemma, which says that if an ideal $I$ is contained in a finite union of prime ideals, then it is contained in one of them.
Let $I$ be an ideal in $A$ such that $U=Xsetminus V(I)$. By definition, $x_1,ldots,x_n$ are primes of $A$ that don't contain $I$, since they are not in $V(I)$. Then we use prime avoidance to choose $$fin Isetminus bigcup_{i=1}^n x_i$$
Then $fin I$, so if a prime contains $I$, it contains $f$. Hence $V(f)supseteq V(I)$, so $D(f)subseteq U$. More importantly, $fnotin x_i$ for any $i$, so $x_inotin V(f)$, which means that $x_iin D(f)$ for all $i$, as desired.
answered Dec 1 at 19:06
jgon
12.4k21940
add a comment |
The key is to use the prime avoidance lemma, which says that if an ideal $I$ is contained in a finite union of prime ideals, then it is contained in one of them.
Let $I$ be an ideal in $A$ such that $U=Xsetminus V(I)$. By definition, $x_1,ldots,x_n$ are primes of $A$ that don't contain $I$, since they are not in $V(I)$. Then we use prime avoidance to choose $$fin Isetminus bigcup_{i=1}^n x_i$$
Then $fin I$, so if a prime contains $I$, it contains $f$. Hence $V(f)supseteq V(I)$, so $D(f)subseteq U$. More importantly, $fnotin x_i$ for any $i$, so $x_inotin V(f)$, which means that $x_iin D(f)$ for all $i$, as desired.
answered Dec 1 at 19:06
jgon
12.4k21940
add a comment |
The key is to use the prime avoidance lemma, which says that if an ideal $I$ is contained in a finite union of prime ideals, then it is contained in one of them.
Let $I$ be an ideal in $A$ such that $U=Xsetminus V(I)$. By definition, $x_1,ldots,x_n$ are primes of $A$ that don't contain $I$, since they are not in $V(I)$. Then we use prime avoidance to choose $$fin Isetminus bigcup_{i=1}^n x_i$$
Then $fin I$, so if a prime contains $I$, it contains $f$. Hence $V(f)supseteq V(I)$, so $D(f)subseteq U$. More importantly, $fnotin x_i$ for any $i$, so $x_inotin V(f)$, which means that $x_iin D(f)$ for all $i$, as desired.
answered Dec 1 at 19:06
jgon
12.4k21940
The key is to use the prime avoidance lemma, which says that if an ideal $I$ is contained in a finite union of prime ideals, then it is contained in one of them.
Let $I$ be an ideal in $A$ such that $U=Xsetminus V(I)$. By definition, $x_1,ldots,x_n$ are primes of $A$ that don't contain $I$, since they are not in $V(I)$. Then we use prime avoidance to choose $$fin Isetminus bigcup_{i=1}^n x_i$$
Then $fin I$, so if a prime contains $I$, it contains $f$. Hence $V(f)supseteq V(I)$, so $D(f)subseteq U$. More importantly, $fnotin x_i$ for any $i$, so $x_inotin V(f)$, which means that $x_iin D(f)$ for all $i$, as desired.
answered Dec 1 at 19:06
jgon
12.4k21940
answered Dec 1 at 19:06
jgon
12.4k21940
answered Dec 1 at 19:06
jgon
12.4k21940
answered Dec 1 at 19:06
jgon
12.4k21940
12.4k21940
add a comment |
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Is $U$ open in $operatorname{Spec}A $?
– Bernard
Dec 1 at 10:32
Thanks, I fixed it.
– user621469
Dec 1 at 10:48