Mixing
The open sets in the Zariski topology are the complements of finite sets
$begingroup$
What does "that is, on maximal ideals of $k[x]$" mean? Just before that remark it was said that we should work in the Zariski topology on $A^1(k)$, so the remark following is confusing. What exactly is being asked?
Assuming that the question is "show that the open subsets of $A^1(k)$ are the complements of finite sets": an open subset of $A^1(k)$ is of the form $A^1(k)setminus Z(I)$ for an ideal $Isubset k[x]$. It is indeed the complement of the finite set $Z(I)$: since $k$ (hence $k[x]$) is Noetherian, $I$ is finitely generated, and every generator has only finitely many roots. Is that reasoning correct?
Conversely, suppose $S$ is a finite subset of $A^1(k)$. I need to show that $A^1(k)setminus S$ is open. So I need to show that $S$ is an algebraic set, right? Shall I just consider the ideal generated by a polynomial that has as roots the elements of $S$ (e.g. the corresponding Lagrange polynomial)? This should prove that $S$ is the zero set of that ideal, so $A^1(k)setminus S$ is open. But what for do we need that $k$ is algebraically closed?
For the last question, should I prove it for all $n$, or is it suffices to prove it for $n=2$? For $n=2$, this answer suggests to look at $x=y$, but I don't see why this is a counterexample: the zero set of $x-y$ is closed in the Zariski topology, and it's also closed say in $mathbb R^2$ (the product topology coincides with the Euclidean topology, and the line contains all its limit points, hence is closed).
abstract-algebra general-topology algebraic-geometry commutative-algebra
asked Jan 20 at 4:15
user437309user437309
698313
$endgroup$
|
show 2 more comments
$begingroup$
What does "that is, on maximal ideals of $k[x]$" mean? Just before that remark it was said that we should work in the Zariski topology on $A^1(k)$, so the remark following is confusing. What exactly is being asked?
Assuming that the question is "show that the open subsets of $A^1(k)$ are the complements of finite sets": an open subset of $A^1(k)$ is of the form $A^1(k)setminus Z(I)$ for an ideal $Isubset k[x]$. It is indeed the complement of the finite set $Z(I)$: since $k$ (hence $k[x]$) is Noetherian, $I$ is finitely generated, and every generator has only finitely many roots. Is that reasoning correct?
Conversely, suppose $S$ is a finite subset of $A^1(k)$. I need to show that $A^1(k)setminus S$ is open. So I need to show that $S$ is an algebraic set, right? Shall I just consider the ideal generated by a polynomial that has as roots the elements of $S$ (e.g. the corresponding Lagrange polynomial)? This should prove that $S$ is the zero set of that ideal, so $A^1(k)setminus S$ is open. But what for do we need that $k$ is algebraically closed?
For the last question, should I prove it for all $n$, or is it suffices to prove it for $n=2$? For $n=2$, this answer suggests to look at $x=y$, but I don't see why this is a counterexample: the zero set of $x-y$ is closed in the Zariski topology, and it's also closed say in $mathbb R^2$ (the product topology coincides with the Euclidean topology, and the line contains all its limit points, hence is closed).
abstract-algebra general-topology algebraic-geometry commutative-algebra
asked Jan 20 at 4:15
user437309user437309
698313
$endgroup$
1
$begingroup$
When $k$ is algebraically closed, $mathbf A^n(k)$ can be thought of in two ways: first, as the cartesian product $k^n$, and second, as the set of maximal ideals of the ring $R = k[t_1, ... , t_n]$. Explicitly, if $(a_1, ... , a_n) in k^n$, then this identifies with ideal generated by the elements $t_1 - a_1, ... , t_n - a_n$. This is one way of stating the Nullstellensatz.
$endgroup$
– D_S
Jan 20 at 4:29
$begingroup$
"since k (hence k[x]) is Noetherian, I is finitely generated, and every generator has only finitely many roots. Is that reasoning correct?" You're not wrong, but $k[x]$ is even a principal ideal domain, so $I$ would be generated by a single polynomial $f$ which has finitely many roots.
$endgroup$
– D_S
Jan 20 at 4:34
$begingroup$
@D_S So is the problem asking then to prove that the open sets in the space of maximal ideals are the complements of finite sets? (With the topology on the set of maximal ideals being the subset topology of the Zariski topology on the set of prime ideals.) Or is the problem asking what I tried to prove in the question?
$endgroup$
– user437309
Jan 20 at 14:38
$begingroup$
I can't tell which definition of $mathbf A^n(k)$ they are using, $k^n$ or the maximal ideals, because I don't know what book this is
$endgroup$
– D_S
Jan 20 at 16:33
$begingroup$
@D_S It's Eisenbud's "Commutative Algebra ..." (p.54). Earlier in this exercise he introduced the topology on the prime spectrum. But he explicitly said that this is a topology on $operatorname{Spec}(R)$, whereas in the exercise he is talking about the topology on $A^1(k)$.
$endgroup$
– user437309
Jan 20 at 16:41
|
show 2 more comments
$begingroup$
What does "that is, on maximal ideals of $k[x]$" mean? Just before that remark it was said that we should work in the Zariski topology on $A^1(k)$, so the remark following is confusing. What exactly is being asked?
Assuming that the question is "show that the open subsets of $A^1(k)$ are the complements of finite sets": an open subset of $A^1(k)$ is of the form $A^1(k)setminus Z(I)$ for an ideal $Isubset k[x]$. It is indeed the complement of the finite set $Z(I)$: since $k$ (hence $k[x]$) is Noetherian, $I$ is finitely generated, and every generator has only finitely many roots. Is that reasoning correct?
Conversely, suppose $S$ is a finite subset of $A^1(k)$. I need to show that $A^1(k)setminus S$ is open. So I need to show that $S$ is an algebraic set, right? Shall I just consider the ideal generated by a polynomial that has as roots the elements of $S$ (e.g. the corresponding Lagrange polynomial)? This should prove that $S$ is the zero set of that ideal, so $A^1(k)setminus S$ is open. But what for do we need that $k$ is algebraically closed?
For the last question, should I prove it for all $n$, or is it suffices to prove it for $n=2$? For $n=2$, this answer suggests to look at $x=y$, but I don't see why this is a counterexample: the zero set of $x-y$ is closed in the Zariski topology, and it's also closed say in $mathbb R^2$ (the product topology coincides with the Euclidean topology, and the line contains all its limit points, hence is closed).
abstract-algebra general-topology algebraic-geometry commutative-algebra
asked Jan 20 at 4:15
user437309user437309
698313
$endgroup$
What does "that is, on maximal ideals of $k[x]$" mean? Just before that remark it was said that we should work in the Zariski topology on $A^1(k)$, so the remark following is confusing. What exactly is being asked?
Assuming that the question is "show that the open subsets of $A^1(k)$ are the complements of finite sets": an open subset of $A^1(k)$ is of the form $A^1(k)setminus Z(I)$ for an ideal $Isubset k[x]$. It is indeed the complement of the finite set $Z(I)$: since $k$ (hence $k[x]$) is Noetherian, $I$ is finitely generated, and every generator has only finitely many roots. Is that reasoning correct?
Conversely, suppose $S$ is a finite subset of $A^1(k)$. I need to show that $A^1(k)setminus S$ is open. So I need to show that $S$ is an algebraic set, right? Shall I just consider the ideal generated by a polynomial that has as roots the elements of $S$ (e.g. the corresponding Lagrange polynomial)? This should prove that $S$ is the zero set of that ideal, so $A^1(k)setminus S$ is open. But what for do we need that $k$ is algebraically closed?
For the last question, should I prove it for all $n$, or is it suffices to prove it for $n=2$? For $n=2$, this answer suggests to look at $x=y$, but I don't see why this is a counterexample: the zero set of $x-y$ is closed in the Zariski topology, and it's also closed say in $mathbb R^2$ (the product topology coincides with the Euclidean topology, and the line contains all its limit points, hence is closed).
abstract-algebra general-topology algebraic-geometry commutative-algebra
abstract-algebra general-topology algebraic-geometry commutative-algebra
asked Jan 20 at 4:15
user437309user437309
698313
asked Jan 20 at 4:15
user437309user437309
698313
asked Jan 20 at 4:15
user437309user437309
698313
asked Jan 20 at 4:15
user437309user437309
698313
asked Jan 20 at 4:15
user437309user437309
698313
698313
1
$begingroup$
When $k$ is algebraically closed, $mathbf A^n(k)$ can be thought of in two ways: first, as the cartesian product $k^n$, and second, as the set of maximal ideals of the ring $R = k[t_1, ... , t_n]$. Explicitly, if $(a_1, ... , a_n) in k^n$, then this identifies with ideal generated by the elements $t_1 - a_1, ... , t_n - a_n$. This is one way of stating the Nullstellensatz.
$endgroup$
– D_S
Jan 20 at 4:29
$begingroup$
"since k (hence k[x]) is Noetherian, I is finitely generated, and every generator has only finitely many roots. Is that reasoning correct?" You're not wrong, but $k[x]$ is even a principal ideal domain, so $I$ would be generated by a single polynomial $f$ which has finitely many roots.
$endgroup$
– D_S
Jan 20 at 4:34
$begingroup$
@D_S So is the problem asking then to prove that the open sets in the space of maximal ideals are the complements of finite sets? (With the topology on the set of maximal ideals being the subset topology of the Zariski topology on the set of prime ideals.) Or is the problem asking what I tried to prove in the question?
$endgroup$
– user437309
Jan 20 at 14:38
$begingroup$
I can't tell which definition of $mathbf A^n(k)$ they are using, $k^n$ or the maximal ideals, because I don't know what book this is
$endgroup$
– D_S
Jan 20 at 16:33
$begingroup$
@D_S It's Eisenbud's "Commutative Algebra ..." (p.54). Earlier in this exercise he introduced the topology on the prime spectrum. But he explicitly said that this is a topology on $operatorname{Spec}(R)$, whereas in the exercise he is talking about the topology on $A^1(k)$.
$endgroup$
– user437309
Jan 20 at 16:41
|
show 2 more comments
1
$begingroup$
When $k$ is algebraically closed, $mathbf A^n(k)$ can be thought of in two ways: first, as the cartesian product $k^n$, and second, as the set of maximal ideals of the ring $R = k[t_1, ... , t_n]$. Explicitly, if $(a_1, ... , a_n) in k^n$, then this identifies with ideal generated by the elements $t_1 - a_1, ... , t_n - a_n$. This is one way of stating the Nullstellensatz.
$endgroup$
– D_S
Jan 20 at 4:29
$begingroup$
"since k (hence k[x]) is Noetherian, I is finitely generated, and every generator has only finitely many roots. Is that reasoning correct?" You're not wrong, but $k[x]$ is even a principal ideal domain, so $I$ would be generated by a single polynomial $f$ which has finitely many roots.
$endgroup$
– D_S
Jan 20 at 4:34
$begingroup$
@D_S So is the problem asking then to prove that the open sets in the space of maximal ideals are the complements of finite sets? (With the topology on the set of maximal ideals being the subset topology of the Zariski topology on the set of prime ideals.) Or is the problem asking what I tried to prove in the question?
$endgroup$
– user437309
Jan 20 at 14:38
$begingroup$
I can't tell which definition of $mathbf A^n(k)$ they are using, $k^n$ or the maximal ideals, because I don't know what book this is
$endgroup$
– D_S
Jan 20 at 16:33
$begingroup$
@D_S It's Eisenbud's "Commutative Algebra ..." (p.54). Earlier in this exercise he introduced the topology on the prime spectrum. But he explicitly said that this is a topology on $operatorname{Spec}(R)$, whereas in the exercise he is talking about the topology on $A^1(k)$.
$endgroup$
– user437309
Jan 20 at 16:41
1
1
$begingroup$
When $k$ is algebraically closed, $mathbf A^n(k)$ can be thought of in two ways: first, as the cartesian product $k^n$, and second, as the set of maximal ideals of the ring $R = k[t_1, ... , t_n]$. Explicitly, if $(a_1, ... , a_n) in k^n$, then this identifies with ideal generated by the elements $t_1 - a_1, ... , t_n - a_n$. This is one way of stating the Nullstellensatz.
$endgroup$
– D_S
Jan 20 at 4:29
$begingroup$
When $k$ is algebraically closed, $mathbf A^n(k)$ can be thought of in two ways: first, as the cartesian product $k^n$, and second, as the set of maximal ideals of the ring $R = k[t_1, ... , t_n]$. Explicitly, if $(a_1, ... , a_n) in k^n$, then this identifies with ideal generated by the elements $t_1 - a_1, ... , t_n - a_n$. This is one way of stating the Nullstellensatz.
$endgroup$
– D_S
Jan 20 at 4:29
$begingroup$
"since k (hence k[x]) is Noetherian, I is finitely generated, and every generator has only finitely many roots. Is that reasoning correct?" You're not wrong, but $k[x]$ is even a principal ideal domain, so $I$ would be generated by a single polynomial $f$ which has finitely many roots.
$endgroup$
– D_S
Jan 20 at 4:34
$begingroup$
"since k (hence k[x]) is Noetherian, I is finitely generated, and every generator has only finitely many roots. Is that reasoning correct?" You're not wrong, but $k[x]$ is even a principal ideal domain, so $I$ would be generated by a single polynomial $f$ which has finitely many roots.
$endgroup$
– D_S
Jan 20 at 4:34
$begingroup$
@D_S So is the problem asking then to prove that the open sets in the space of maximal ideals are the complements of finite sets? (With the topology on the set of maximal ideals being the subset topology of the Zariski topology on the set of prime ideals.) Or is the problem asking what I tried to prove in the question?
$endgroup$
– user437309
Jan 20 at 14:38
$begingroup$
@D_S So is the problem asking then to prove that the open sets in the space of maximal ideals are the complements of finite sets? (With the topology on the set of maximal ideals being the subset topology of the Zariski topology on the set of prime ideals.) Or is the problem asking what I tried to prove in the question?
$endgroup$
– user437309
Jan 20 at 14:38
$begingroup$
I can't tell which definition of $mathbf A^n(k)$ they are using, $k^n$ or the maximal ideals, because I don't know what book this is
$endgroup$
– D_S
Jan 20 at 16:33
$begingroup$
I can't tell which definition of $mathbf A^n(k)$ they are using, $k^n$ or the maximal ideals, because I don't know what book this is
$endgroup$
– D_S
Jan 20 at 16:33
$begingroup$
@D_S It's Eisenbud's "Commutative Algebra ..." (p.54). Earlier in this exercise he introduced the topology on the prime spectrum. But he explicitly said that this is a topology on $operatorname{Spec}(R)$, whereas in the exercise he is talking about the topology on $A^1(k)$.
$endgroup$
– user437309
Jan 20 at 16:41
$begingroup$
@D_S It's Eisenbud's "Commutative Algebra ..." (p.54). Earlier in this exercise he introduced the topology on the prime spectrum. But he explicitly said that this is a topology on $operatorname{Spec}(R)$, whereas in the exercise he is talking about the topology on $A^1(k)$.
$endgroup$
– user437309
Jan 20 at 16:41
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Assume $k$ is algebraically closed. Let's use the definition that $mathbf A^1(k)$ consists of all maximal ideals of the ring $k[t]$, with the induced topology from $operatorname{Spec} k[t]$. Since $k$ is algebraically closed, every maximal ideal is of the form $mathfrak m = (t - a)$ for a unique $a in k$. It follows from the definition of the Zariski topology that the closed sets in $mathbf A^1(k)$ are those of the form
$$Z(I) = { mathfrak m in mathbf A^1(k) : I subseteq mathfrak m }$$
where $I$ is any ideal of $k[t]$. Since $k[t]$ is a principal ideal domain, $I$ is generated by a polynomial $f(t) = (t - a_1)^{m_1} cdots (t - a_n)^{m_n}$ for $a_i in k$. For a maximal ideal $mathfrak m = (t - a)$ of $k[t]$, check that $I subseteq mathfrak m$ if and only if $a$ is one of the $a_i$. It follows that
$$Z(I) = { (t - a_1), ... , (t - a_n) }$$
and is in particular a finite set.
answered Jan 20 at 18:39
D_SD_S
13.7k61552
$endgroup$
$begingroup$
Thanks! It turned out to be very easy. Regarding the last question of the problem, is it still asking about the set of maximal ideals in $k[t]$, or is it now asking about the "usual" $A^n(k)$ since it's emphasized there that $A^n(k)=k^n$?
$endgroup$
– user437309
Jan 20 at 18:47
$begingroup$
I would assume the $k^n$ definition, but they are basically equivalent, since whenever $A$ and $B$ are finitely generated $k$-algebras over $k$ algebraically closed, you can identify (as sets, not as topological spaces) $$operatorname{m-Spec}(A otimes_k B) = operatorname{m-Spec} A times operatorname{m-Spec} B$$ and $$k[t_1, ... , t_n] otimes_k [t_1, ... , t_m] = k[t_1, ... , t_{n+m}]$$
$endgroup$
– D_S
Jan 20 at 18:49
$begingroup$
The empty set must be open too. But it is the complement of a finite set iff $MaxSpec(k[t])$ is finite. Is it always the case? Is this case covered by your argument?
$endgroup$
– user437309
Jan 20 at 21:59
$begingroup$
Take $I$ to be the zero ideal, then $Z(I)$ will be the whole space.
$endgroup$
– D_S
Jan 21 at 1:10
$begingroup$
That's true but we have to show that the open sets are the complemets of finite sets (equivalently, the closed sets are the finite sets), but the fact that $Z(I)$ is the whole space doesn't say that $Z(I)$ is finite.
$endgroup$
– user437309
Jan 21 at 1:13
|
show 3 more comments
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1 Answer
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$begingroup$
Assume $k$ is algebraically closed. Let's use the definition that $mathbf A^1(k)$ consists of all maximal ideals of the ring $k[t]$, with the induced topology from $operatorname{Spec} k[t]$. Since $k$ is algebraically closed, every maximal ideal is of the form $mathfrak m = (t - a)$ for a unique $a in k$. It follows from the definition of the Zariski topology that the closed sets in $mathbf A^1(k)$ are those of the form
$$Z(I) = { mathfrak m in mathbf A^1(k) : I subseteq mathfrak m }$$
where $I$ is any ideal of $k[t]$. Since $k[t]$ is a principal ideal domain, $I$ is generated by a polynomial $f(t) = (t - a_1)^{m_1} cdots (t - a_n)^{m_n}$ for $a_i in k$. For a maximal ideal $mathfrak m = (t - a)$ of $k[t]$, check that $I subseteq mathfrak m$ if and only if $a$ is one of the $a_i$. It follows that
$$Z(I) = { (t - a_1), ... , (t - a_n) }$$
and is in particular a finite set.
answered Jan 20 at 18:39
D_SD_S
13.7k61552
$endgroup$
$begingroup$
Thanks! It turned out to be very easy. Regarding the last question of the problem, is it still asking about the set of maximal ideals in $k[t]$, or is it now asking about the "usual" $A^n(k)$ since it's emphasized there that $A^n(k)=k^n$?
$endgroup$
– user437309
Jan 20 at 18:47
$begingroup$
I would assume the $k^n$ definition, but they are basically equivalent, since whenever $A$ and $B$ are finitely generated $k$-algebras over $k$ algebraically closed, you can identify (as sets, not as topological spaces) $$operatorname{m-Spec}(A otimes_k B) = operatorname{m-Spec} A times operatorname{m-Spec} B$$ and $$k[t_1, ... , t_n] otimes_k [t_1, ... , t_m] = k[t_1, ... , t_{n+m}]$$
$endgroup$
– D_S
Jan 20 at 18:49
$begingroup$
The empty set must be open too. But it is the complement of a finite set iff $MaxSpec(k[t])$ is finite. Is it always the case? Is this case covered by your argument?
$endgroup$
– user437309
Jan 20 at 21:59
$begingroup$
Take $I$ to be the zero ideal, then $Z(I)$ will be the whole space.
$endgroup$
– D_S
Jan 21 at 1:10
$begingroup$
That's true but we have to show that the open sets are the complemets of finite sets (equivalently, the closed sets are the finite sets), but the fact that $Z(I)$ is the whole space doesn't say that $Z(I)$ is finite.
$endgroup$
– user437309
Jan 21 at 1:13
|
show 3 more comments
$begingroup$
Assume $k$ is algebraically closed. Let's use the definition that $mathbf A^1(k)$ consists of all maximal ideals of the ring $k[t]$, with the induced topology from $operatorname{Spec} k[t]$. Since $k$ is algebraically closed, every maximal ideal is of the form $mathfrak m = (t - a)$ for a unique $a in k$. It follows from the definition of the Zariski topology that the closed sets in $mathbf A^1(k)$ are those of the form
$$Z(I) = { mathfrak m in mathbf A^1(k) : I subseteq mathfrak m }$$
where $I$ is any ideal of $k[t]$. Since $k[t]$ is a principal ideal domain, $I$ is generated by a polynomial $f(t) = (t - a_1)^{m_1} cdots (t - a_n)^{m_n}$ for $a_i in k$. For a maximal ideal $mathfrak m = (t - a)$ of $k[t]$, check that $I subseteq mathfrak m$ if and only if $a$ is one of the $a_i$. It follows that
$$Z(I) = { (t - a_1), ... , (t - a_n) }$$
and is in particular a finite set.
answered Jan 20 at 18:39
D_SD_S
13.7k61552
$endgroup$
$begingroup$
Thanks! It turned out to be very easy. Regarding the last question of the problem, is it still asking about the set of maximal ideals in $k[t]$, or is it now asking about the "usual" $A^n(k)$ since it's emphasized there that $A^n(k)=k^n$?
$endgroup$
– user437309
Jan 20 at 18:47
$begingroup$
I would assume the $k^n$ definition, but they are basically equivalent, since whenever $A$ and $B$ are finitely generated $k$-algebras over $k$ algebraically closed, you can identify (as sets, not as topological spaces) $$operatorname{m-Spec}(A otimes_k B) = operatorname{m-Spec} A times operatorname{m-Spec} B$$ and $$k[t_1, ... , t_n] otimes_k [t_1, ... , t_m] = k[t_1, ... , t_{n+m}]$$
$endgroup$
– D_S
Jan 20 at 18:49
$begingroup$
The empty set must be open too. But it is the complement of a finite set iff $MaxSpec(k[t])$ is finite. Is it always the case? Is this case covered by your argument?
$endgroup$
– user437309
Jan 20 at 21:59
$begingroup$
Take $I$ to be the zero ideal, then $Z(I)$ will be the whole space.
$endgroup$
– D_S
Jan 21 at 1:10
$begingroup$
That's true but we have to show that the open sets are the complemets of finite sets (equivalently, the closed sets are the finite sets), but the fact that $Z(I)$ is the whole space doesn't say that $Z(I)$ is finite.
$endgroup$
– user437309
Jan 21 at 1:13
|
show 3 more comments
$begingroup$
Assume $k$ is algebraically closed. Let's use the definition that $mathbf A^1(k)$ consists of all maximal ideals of the ring $k[t]$, with the induced topology from $operatorname{Spec} k[t]$. Since $k$ is algebraically closed, every maximal ideal is of the form $mathfrak m = (t - a)$ for a unique $a in k$. It follows from the definition of the Zariski topology that the closed sets in $mathbf A^1(k)$ are those of the form
$$Z(I) = { mathfrak m in mathbf A^1(k) : I subseteq mathfrak m }$$
where $I$ is any ideal of $k[t]$. Since $k[t]$ is a principal ideal domain, $I$ is generated by a polynomial $f(t) = (t - a_1)^{m_1} cdots (t - a_n)^{m_n}$ for $a_i in k$. For a maximal ideal $mathfrak m = (t - a)$ of $k[t]$, check that $I subseteq mathfrak m$ if and only if $a$ is one of the $a_i$. It follows that
$$Z(I) = { (t - a_1), ... , (t - a_n) }$$
and is in particular a finite set.
answered Jan 20 at 18:39
D_SD_S
13.7k61552
$endgroup$
Assume $k$ is algebraically closed. Let's use the definition that $mathbf A^1(k)$ consists of all maximal ideals of the ring $k[t]$, with the induced topology from $operatorname{Spec} k[t]$. Since $k$ is algebraically closed, every maximal ideal is of the form $mathfrak m = (t - a)$ for a unique $a in k$. It follows from the definition of the Zariski topology that the closed sets in $mathbf A^1(k)$ are those of the form
$$Z(I) = { mathfrak m in mathbf A^1(k) : I subseteq mathfrak m }$$
where $I$ is any ideal of $k[t]$. Since $k[t]$ is a principal ideal domain, $I$ is generated by a polynomial $f(t) = (t - a_1)^{m_1} cdots (t - a_n)^{m_n}$ for $a_i in k$. For a maximal ideal $mathfrak m = (t - a)$ of $k[t]$, check that $I subseteq mathfrak m$ if and only if $a$ is one of the $a_i$. It follows that
$$Z(I) = { (t - a_1), ... , (t - a_n) }$$
and is in particular a finite set.
answered Jan 20 at 18:39
D_SD_S
13.7k61552
answered Jan 20 at 18:39
D_SD_S
13.7k61552
answered Jan 20 at 18:39
D_SD_S
13.7k61552
answered Jan 20 at 18:39
D_SD_S
13.7k61552
13.7k61552
$begingroup$
Thanks! It turned out to be very easy. Regarding the last question of the problem, is it still asking about the set of maximal ideals in $k[t]$, or is it now asking about the "usual" $A^n(k)$ since it's emphasized there that $A^n(k)=k^n$?
$endgroup$
– user437309
Jan 20 at 18:47
$begingroup$
I would assume the $k^n$ definition, but they are basically equivalent, since whenever $A$ and $B$ are finitely generated $k$-algebras over $k$ algebraically closed, you can identify (as sets, not as topological spaces) $$operatorname{m-Spec}(A otimes_k B) = operatorname{m-Spec} A times operatorname{m-Spec} B$$ and $$k[t_1, ... , t_n] otimes_k [t_1, ... , t_m] = k[t_1, ... , t_{n+m}]$$
$endgroup$
– D_S
Jan 20 at 18:49
$begingroup$
The empty set must be open too. But it is the complement of a finite set iff $MaxSpec(k[t])$ is finite. Is it always the case? Is this case covered by your argument?
$endgroup$
– user437309
Jan 20 at 21:59
$begingroup$
Take $I$ to be the zero ideal, then $Z(I)$ will be the whole space.
$endgroup$
– D_S
Jan 21 at 1:10
$begingroup$
That's true but we have to show that the open sets are the complemets of finite sets (equivalently, the closed sets are the finite sets), but the fact that $Z(I)$ is the whole space doesn't say that $Z(I)$ is finite.
$endgroup$
– user437309
Jan 21 at 1:13
|
show 3 more comments
$begingroup$
Thanks! It turned out to be very easy. Regarding the last question of the problem, is it still asking about the set of maximal ideals in $k[t]$, or is it now asking about the "usual" $A^n(k)$ since it's emphasized there that $A^n(k)=k^n$?
$endgroup$
– user437309
Jan 20 at 18:47
$begingroup$
I would assume the $k^n$ definition, but they are basically equivalent, since whenever $A$ and $B$ are finitely generated $k$-algebras over $k$ algebraically closed, you can identify (as sets, not as topological spaces) $$operatorname{m-Spec}(A otimes_k B) = operatorname{m-Spec} A times operatorname{m-Spec} B$$ and $$k[t_1, ... , t_n] otimes_k [t_1, ... , t_m] = k[t_1, ... , t_{n+m}]$$
$endgroup$
– D_S
Jan 20 at 18:49
$begingroup$
The empty set must be open too. But it is the complement of a finite set iff $MaxSpec(k[t])$ is finite. Is it always the case? Is this case covered by your argument?
$endgroup$
– user437309
Jan 20 at 21:59
$begingroup$
Take $I$ to be the zero ideal, then $Z(I)$ will be the whole space.
$endgroup$
– D_S
Jan 21 at 1:10
$begingroup$
That's true but we have to show that the open sets are the complemets of finite sets (equivalently, the closed sets are the finite sets), but the fact that $Z(I)$ is the whole space doesn't say that $Z(I)$ is finite.
$endgroup$
– user437309
Jan 21 at 1:13
$begingroup$
Thanks! It turned out to be very easy. Regarding the last question of the problem, is it still asking about the set of maximal ideals in $k[t]$, or is it now asking about the "usual" $A^n(k)$ since it's emphasized there that $A^n(k)=k^n$?
$endgroup$
– user437309
Jan 20 at 18:47
$begingroup$
Thanks! It turned out to be very easy. Regarding the last question of the problem, is it still asking about the set of maximal ideals in $k[t]$, or is it now asking about the "usual" $A^n(k)$ since it's emphasized there that $A^n(k)=k^n$?
$endgroup$
– user437309
Jan 20 at 18:47
$begingroup$
I would assume the $k^n$ definition, but they are basically equivalent, since whenever $A$ and $B$ are finitely generated $k$-algebras over $k$ algebraically closed, you can identify (as sets, not as topological spaces) $$operatorname{m-Spec}(A otimes_k B) = operatorname{m-Spec} A times operatorname{m-Spec} B$$ and $$k[t_1, ... , t_n] otimes_k [t_1, ... , t_m] = k[t_1, ... , t_{n+m}]$$
$endgroup$
– D_S
Jan 20 at 18:49
$begingroup$
I would assume the $k^n$ definition, but they are basically equivalent, since whenever $A$ and $B$ are finitely generated $k$-algebras over $k$ algebraically closed, you can identify (as sets, not as topological spaces) $$operatorname{m-Spec}(A otimes_k B) = operatorname{m-Spec} A times operatorname{m-Spec} B$$ and $$k[t_1, ... , t_n] otimes_k [t_1, ... , t_m] = k[t_1, ... , t_{n+m}]$$
$endgroup$
– D_S
Jan 20 at 18:49
$begingroup$
The empty set must be open too. But it is the complement of a finite set iff $MaxSpec(k[t])$ is finite. Is it always the case? Is this case covered by your argument?
$endgroup$
– user437309
Jan 20 at 21:59
$begingroup$
The empty set must be open too. But it is the complement of a finite set iff $MaxSpec(k[t])$ is finite. Is it always the case? Is this case covered by your argument?
$endgroup$
– user437309
Jan 20 at 21:59
$begingroup$
Take $I$ to be the zero ideal, then $Z(I)$ will be the whole space.
$endgroup$
– D_S
Jan 21 at 1:10
$begingroup$
Take $I$ to be the zero ideal, then $Z(I)$ will be the whole space.
$endgroup$
– D_S
Jan 21 at 1:10
$begingroup$
That's true but we have to show that the open sets are the complemets of finite sets (equivalently, the closed sets are the finite sets), but the fact that $Z(I)$ is the whole space doesn't say that $Z(I)$ is finite.
$endgroup$
– user437309
Jan 21 at 1:13
$begingroup$
That's true but we have to show that the open sets are the complemets of finite sets (equivalently, the closed sets are the finite sets), but the fact that $Z(I)$ is the whole space doesn't say that $Z(I)$ is finite.
$endgroup$
– user437309
Jan 21 at 1:13
|
show 3 more comments
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$begingroup$
When $k$ is algebraically closed, $mathbf A^n(k)$ can be thought of in two ways: first, as the cartesian product $k^n$, and second, as the set of maximal ideals of the ring $R = k[t_1, ... , t_n]$. Explicitly, if $(a_1, ... , a_n) in k^n$, then this identifies with ideal generated by the elements $t_1 - a_1, ... , t_n - a_n$. This is one way of stating the Nullstellensatz.
$endgroup$
– D_S
Jan 20 at 4:29
$begingroup$
"since k (hence k[x]) is Noetherian, I is finitely generated, and every generator has only finitely many roots. Is that reasoning correct?" You're not wrong, but $k[x]$ is even a principal ideal domain, so $I$ would be generated by a single polynomial $f$ which has finitely many roots.
$endgroup$
– D_S
Jan 20 at 4:34
$begingroup$
@D_S So is the problem asking then to prove that the open sets in the space of maximal ideals are the complements of finite sets? (With the topology on the set of maximal ideals being the subset topology of the Zariski topology on the set of prime ideals.) Or is the problem asking what I tried to prove in the question?
$endgroup$
– user437309
Jan 20 at 14:38
$begingroup$
I can't tell which definition of $mathbf A^n(k)$ they are using, $k^n$ or the maximal ideals, because I don't know what book this is
$endgroup$
– D_S
Jan 20 at 16:33
$begingroup$
@D_S It's Eisenbud's "Commutative Algebra ..." (p.54). Earlier in this exercise he introduced the topology on the prime spectrum. But he explicitly said that this is a topology on $operatorname{Spec}(R)$, whereas in the exercise he is talking about the topology on $A^1(k)$.
$endgroup$
– user437309
Jan 20 at 16:41