set of closed points is dense in Spec $A$
$begingroup$
I have a question regarding the above topic.
Let $k$ be a (not neccesarily algebraically closed) field, $A$ a finitely generated $k$-Algebra. Show that the set of the closed points is dense in Spec $A$.
So far I have got the following. I was not quite sure, how to use the fact, that $A$ is a finitely generated $k$-Algebra.
But we know that the closed points in $A$ are the maximal ideals. Any topological space is the union of its irreducible components. So we have Spec $A = bigcup_{i} U_i$ for some closed, irreducible subsets $U_i$. Each $U_i$ has a unique generic point $mathfrak p_i$, so that $overline{{mathfrak p_i}} = U_i$. Now, since $A$ is a finitely generated $k$-Algebra, we have for any prime ideal $mathfrak p$:
$$mathfrak p = bigcap_{mathfrak m supset mathfrak p} mathfrak m,$$
where $mathfrak m$ is a maximal ideal. So it follows, that
$$ text{Spec }A = bigcup_i U_i = bigcup_i overline{{mathfrak p_i}} = bigcup_i overline{left({bigcap_{mathfrak m supset mathfrak p} mathfrak m}right)_i} = overline{bigcup_i left({bigcap_{mathfrak m supset mathfrak p} mathfrak m}right)_i}$$
So it follows, that the closure of the set of maximal ideal equals Spec $A$.
I would appreciate any comments, where the proof is mistaken and what a correct proof would look like.
abstract-algebra general-topology algebraic-geometry maximal-and-prime-ideals
edited Nov 24 '18 at 12:08
Scientifica
6,37641335
asked Nov 24 '18 at 11:58
Jonas W.Jonas W.
616
$endgroup$
add a comment |
$begingroup$
I have a question regarding the above topic.
Let $k$ be a (not neccesarily algebraically closed) field, $A$ a finitely generated $k$-Algebra. Show that the set of the closed points is dense in Spec $A$.
So far I have got the following. I was not quite sure, how to use the fact, that $A$ is a finitely generated $k$-Algebra.
But we know that the closed points in $A$ are the maximal ideals. Any topological space is the union of its irreducible components. So we have Spec $A = bigcup_{i} U_i$ for some closed, irreducible subsets $U_i$. Each $U_i$ has a unique generic point $mathfrak p_i$, so that $overline{{mathfrak p_i}} = U_i$. Now, since $A$ is a finitely generated $k$-Algebra, we have for any prime ideal $mathfrak p$:
$$mathfrak p = bigcap_{mathfrak m supset mathfrak p} mathfrak m,$$
where $mathfrak m$ is a maximal ideal. So it follows, that
$$ text{Spec }A = bigcup_i U_i = bigcup_i overline{{mathfrak p_i}} = bigcup_i overline{left({bigcap_{mathfrak m supset mathfrak p} mathfrak m}right)_i} = overline{bigcup_i left({bigcap_{mathfrak m supset mathfrak p} mathfrak m}right)_i}$$
So it follows, that the closure of the set of maximal ideal equals Spec $A$.
I would appreciate any comments, where the proof is mistaken and what a correct proof would look like.
abstract-algebra general-topology algebraic-geometry maximal-and-prime-ideals
edited Nov 24 '18 at 12:08
Scientifica
6,37641335
asked Nov 24 '18 at 11:58
Jonas W.Jonas W.
616
$endgroup$
$begingroup$
"it follows": it doesn't, at least not without a few words of explanation. What about the following thought : if $xin A$ is not a unit, then $x$ belongs to some maximal ideal ?
$endgroup$
– Max
Nov 24 '18 at 13:00
$begingroup$
Alright. Then how does it follow, if it does?
$endgroup$
– Jonas W.
Nov 24 '18 at 13:09
$begingroup$
You must use @[name of the user] so that the user in question gets notified. If someone comments below the post then op will always get notified but not other users. I thought you are new here and didn't know this.
$endgroup$
– random123
Nov 25 '18 at 6:48
add a comment |
$begingroup$
I have a question regarding the above topic.
Let $k$ be a (not neccesarily algebraically closed) field, $A$ a finitely generated $k$-Algebra. Show that the set of the closed points is dense in Spec $A$.
So far I have got the following. I was not quite sure, how to use the fact, that $A$ is a finitely generated $k$-Algebra.
But we know that the closed points in $A$ are the maximal ideals. Any topological space is the union of its irreducible components. So we have Spec $A = bigcup_{i} U_i$ for some closed, irreducible subsets $U_i$. Each $U_i$ has a unique generic point $mathfrak p_i$, so that $overline{{mathfrak p_i}} = U_i$. Now, since $A$ is a finitely generated $k$-Algebra, we have for any prime ideal $mathfrak p$:
$$mathfrak p = bigcap_{mathfrak m supset mathfrak p} mathfrak m,$$
where $mathfrak m$ is a maximal ideal. So it follows, that
$$ text{Spec }A = bigcup_i U_i = bigcup_i overline{{mathfrak p_i}} = bigcup_i overline{left({bigcap_{mathfrak m supset mathfrak p} mathfrak m}right)_i} = overline{bigcup_i left({bigcap_{mathfrak m supset mathfrak p} mathfrak m}right)_i}$$
So it follows, that the closure of the set of maximal ideal equals Spec $A$.
I would appreciate any comments, where the proof is mistaken and what a correct proof would look like.
abstract-algebra general-topology algebraic-geometry maximal-and-prime-ideals
edited Nov 24 '18 at 12:08
Scientifica
6,37641335
asked Nov 24 '18 at 11:58
Jonas W.Jonas W.
616
$endgroup$
I have a question regarding the above topic.
Let $k$ be a (not neccesarily algebraically closed) field, $A$ a finitely generated $k$-Algebra. Show that the set of the closed points is dense in Spec $A$.
So far I have got the following. I was not quite sure, how to use the fact, that $A$ is a finitely generated $k$-Algebra.
But we know that the closed points in $A$ are the maximal ideals. Any topological space is the union of its irreducible components. So we have Spec $A = bigcup_{i} U_i$ for some closed, irreducible subsets $U_i$. Each $U_i$ has a unique generic point $mathfrak p_i$, so that $overline{{mathfrak p_i}} = U_i$. Now, since $A$ is a finitely generated $k$-Algebra, we have for any prime ideal $mathfrak p$:
$$mathfrak p = bigcap_{mathfrak m supset mathfrak p} mathfrak m,$$
where $mathfrak m$ is a maximal ideal. So it follows, that
$$ text{Spec }A = bigcup_i U_i = bigcup_i overline{{mathfrak p_i}} = bigcup_i overline{left({bigcap_{mathfrak m supset mathfrak p} mathfrak m}right)_i} = overline{bigcup_i left({bigcap_{mathfrak m supset mathfrak p} mathfrak m}right)_i}$$
So it follows, that the closure of the set of maximal ideal equals Spec $A$.
I would appreciate any comments, where the proof is mistaken and what a correct proof would look like.
abstract-algebra general-topology algebraic-geometry maximal-and-prime-ideals
abstract-algebra general-topology algebraic-geometry maximal-and-prime-ideals
edited Nov 24 '18 at 12:08
Scientifica
6,37641335
asked Nov 24 '18 at 11:58
Jonas W.Jonas W.
616
edited Nov 24 '18 at 12:08
Scientifica
6,37641335
asked Nov 24 '18 at 11:58
Jonas W.Jonas W.
616
edited Nov 24 '18 at 12:08
Scientifica
6,37641335
edited Nov 24 '18 at 12:08
Scientifica
6,37641335
edited Nov 24 '18 at 12:08
Scientifica
6,37641335
6,37641335
asked Nov 24 '18 at 11:58
Jonas W.Jonas W.
616
asked Nov 24 '18 at 11:58
Jonas W.Jonas W.
616
asked Nov 24 '18 at 11:58
Jonas W.Jonas W.
616
616
$begingroup$
"it follows": it doesn't, at least not without a few words of explanation. What about the following thought : if $xin A$ is not a unit, then $x$ belongs to some maximal ideal ?
$endgroup$
– Max
Nov 24 '18 at 13:00
$begingroup$
Alright. Then how does it follow, if it does?
$endgroup$
– Jonas W.
Nov 24 '18 at 13:09
$begingroup$
You must use @[name of the user] so that the user in question gets notified. If someone comments below the post then op will always get notified but not other users. I thought you are new here and didn't know this.
$endgroup$
– random123
Nov 25 '18 at 6:48
add a comment |
$begingroup$
"it follows": it doesn't, at least not without a few words of explanation. What about the following thought : if $xin A$ is not a unit, then $x$ belongs to some maximal ideal ?
$endgroup$
– Max
Nov 24 '18 at 13:00
$begingroup$
Alright. Then how does it follow, if it does?
$endgroup$
– Jonas W.
Nov 24 '18 at 13:09
$begingroup$
You must use @[name of the user] so that the user in question gets notified. If someone comments below the post then op will always get notified but not other users. I thought you are new here and didn't know this.
$endgroup$
– random123
Nov 25 '18 at 6:48
$begingroup$
"it follows": it doesn't, at least not without a few words of explanation. What about the following thought : if $xin A$ is not a unit, then $x$ belongs to some maximal ideal ?
$endgroup$
– Max
Nov 24 '18 at 13:00
$begingroup$
"it follows": it doesn't, at least not without a few words of explanation. What about the following thought : if $xin A$ is not a unit, then $x$ belongs to some maximal ideal ?
$endgroup$
– Max
Nov 24 '18 at 13:00
$begingroup$
Alright. Then how does it follow, if it does?
$endgroup$
– Jonas W.
Nov 24 '18 at 13:09
$begingroup$
Alright. Then how does it follow, if it does?
$endgroup$
– Jonas W.
Nov 24 '18 at 13:09
$begingroup$
You must use @[name of the user] so that the user in question gets notified. If someone comments below the post then op will always get notified but not other users. I thought you are new here and didn't know this.
$endgroup$
– random123
Nov 25 '18 at 6:48
$begingroup$
You must use @[name of the user] so that the user in question gets notified. If someone comments below the post then op will always get notified but not other users. I thought you are new here and didn't know this.
$endgroup$
– random123
Nov 25 '18 at 6:48
add a comment |
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$begingroup$
"it follows": it doesn't, at least not without a few words of explanation. What about the following thought : if $xin A$ is not a unit, then $x$ belongs to some maximal ideal ?
$endgroup$
– Max
Nov 24 '18 at 13:00
$begingroup$
Alright. Then how does it follow, if it does?
$endgroup$
– Jonas W.
Nov 24 '18 at 13:09
$begingroup$
You must use @[name of the user] so that the user in question gets notified. If someone comments below the post then op will always get notified but not other users. I thought you are new here and didn't know this.
$endgroup$
– random123
Nov 25 '18 at 6:48