Geometrically irreducible/integral
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zheng xudong
Nov 4, 2010, 1:39:55 AM11/4/10
to Prof. Shiu-chun Wong's Cocktail Seminar
Dear seniors,
While I am reading Hartshorne, a notion, namely 'geometrically
something' as in the title, comes about which I am not quite clear.
Basically, statements are contained in Ch.II, Ex. 3.15.
Intuitively, say in the affine case, consider a variety over a field $k
$, if we work over an extension of the base field, we will get more
points, since it is saying we are looking for zero loci for the same
ideal in a larger field. But how can this be seen from the definition
of the product notion?
What topological information can we get from base extension?
Thanks a lot.
Xudong
While I am reading Hartshorne, a notion, namely 'geometrically
something' as in the title, comes about which I am not quite clear.
Basically, statements are contained in Ch.II, Ex. 3.15.
Intuitively, say in the affine case, consider a variety over a field $k
$, if we work over an extension of the base field, we will get more
points, since it is saying we are looking for zero loci for the same
ideal in a larger field. But how can this be seen from the definition
of the product notion?
What topological information can we get from base extension?
Thanks a lot.
Xudong
1987xk47
Nov 4, 2010, 5:48:40 AM11/4/10
to cocktai...@googlegroups.com
Hi Xudong:
I'm also working on Hartshorne now.
About the "geometric something", my opinion is the following.
$X \times_k k^\bar$ is a sheme of finite type over an algebraically closed field, if it also
integral and separated then it comes from some variety (the funtor $t$ in section 2). So
scheme over an algebraically closed field looks more like varieties than other shemes,
i.e., they are more "geometric". So we say $X$ is "geometrically something" if $X \times_k k^\bar$
is "something".
First, if $X$ is affine, then any closed point of $X \times_k k^\bar$ maps to a closed point of $X$, i.e.,
a maximal ideal $m$ of $k^\bar [...]$ contracts to a maximal ideal $n$ of $k[...]$. We have
maps $k \longrightarrow k[...]/n \longrightarrow k^\bar [...]/m$. ($k^\bar [...]/m$ is just $k^\bar$).
Since $k^\bar$ is algebraic closure, $k[...]/n$ is integral over $k$, hence a field itself, so $n$ is maximal.
Conversely, for any maximal ideal $n$ of $k[...]$, if the extension of $n$ is not the whole ring (I believe so but I haven't found a rigorous proof), then it is the contraction of $m$ which contains $n^e$.
So $X \times_k k^\bar$ has more closed points than $X$. General case follows from affine case.
In general, I think the underlying set of product of two schemes $X$ and $Y$ is not easy to describe (it is definitely not the set-theoretic product of underlying sets of $X$ and $Y$), because generic points of
the product do not look like $(x, y)$. In some cases, when $x$ and $y$ are closed points, we can say
$(x, y)$ is closed point of $X \times Y$.
Best
Ke
I'm also working on Hartshorne now.
About the "geometric something", my opinion is the following.
$X \times_k k^\bar$ is a sheme of finite type over an algebraically closed field, if it also
integral and separated then it comes from some variety (the funtor $t$ in section 2). So
scheme over an algebraically closed field looks more like varieties than other shemes,
i.e., they are more "geometric". So we say $X$ is "geometrically something" if $X \times_k k^\bar$
is "something".
First, if $X$ is affine, then any closed point of $X \times_k k^\bar$ maps to a closed point of $X$, i.e.,
a maximal ideal $m$ of $k^\bar [...]$ contracts to a maximal ideal $n$ of $k[...]$. We have
maps $k \longrightarrow k[...]/n \longrightarrow k^\bar [...]/m$. ($k^\bar [...]/m$ is just $k^\bar$).
Since $k^\bar$ is algebraic closure, $k[...]/n$ is integral over $k$, hence a field itself, so $n$ is maximal.
Conversely, for any maximal ideal $n$ of $k[...]$, if the extension of $n$ is not the whole ring (I believe so but I haven't found a rigorous proof), then it is the contraction of $m$ which contains $n^e$.
So $X \times_k k^\bar$ has more closed points than $X$. General case follows from affine case.
In general, I think the underlying set of product of two schemes $X$ and $Y$ is not easy to describe (it is definitely not the set-theoretic product of underlying sets of $X$ and $Y$), because generic points of
the product do not look like $(x, y)$. In some cases, when $x$ and $y$ are closed points, we can say
$(x, y)$ is closed point of $X \times Y$.
Best
Ke
> -----原始邮件----- > 发件人: "zheng xudong" <hacken...@gmail.com> > 发送时间: 2010年11月4日 星期四 > 收件人: "Prof. Shiu-chun Wong's Cocktail Seminar" <cocktai...@googlegroups.com> > 抄送: > 主题: Geometrically irreducible/integral > > Dear seniors, > > While I am reading Hartshorne, a notion, namely 'geometrically > something' as in the title, comes about which I am not quite clear. > Basically, statements are contained in Ch.II, Ex. 3.15. > > Intuitively, say in the affine case, consider a variety over a field $k > $, if we work over an extension of the base field, we will get more > points, since it is saying we are looking for zero loci for the same > ideal in a larger field. But how can this be seen from the definition > of the product notion? > > What topological information can we get from base extension? > > Thanks a lot. > Xudong > > -- > You received this message because you are subscribed to the Google Groups "Prof. Shiu-chun Wong's Cocktail Seminar" group. > To post to this group, send email to cocktai...@googlegroups.com. > To unsubscribe from this group, send email to cocktailsemin...@googlegroups.com. > For more options, visit this group at http://groups.google.com/group/cocktailseminar?hl=en. >
Yi Zhu
Nov 4, 2010, 12:34:49 PM11/4/10
to cocktai...@googlegroups.com
Hi Xudong,
If you only consider the base extension by field extensions, you question is easy to answer.
Let X be an affine variety over a field K and let L be a field extension of K. X_L=X \times_K L
Remember that the base change (by fiber product) means tensor product of the rings of regular functions K[X] and L.
i.e., everything (like formal variable x_i) remains same except the base field
if the variety is defined by an ideal I, the new ideal after the extension is I \times_K L, ie the "same" ideal.
Try x^2+y^2=1 over Q and extend over R and use the fiber product.
Best,
Yi
If you only consider the base extension by field extensions, you question is easy to answer.
Let X be an affine variety over a field K and let L be a field extension of K. X_L=X \times_K L
Remember that the base change (by fiber product) means tensor product of the rings of regular functions K[X] and L.
i.e., everything (like formal variable x_i) remains same except the base field
if the variety is defined by an ideal I, the new ideal after the extension is I \times_K L, ie the "same" ideal.
Try x^2+y^2=1 over Q and extend over R and use the fiber product.
Best,
Yi
zheng xudong
Nov 5, 2010, 2:21:06 AM11/5/10
to Prof. Shiu-chun Wong's Cocktail Seminar
Hi Ke,
I think I get it. Thank you. By the way, how is your working? How much
have you studied of Hartshorne?
best
Xudong
On Nov 4, 4:48 am, 1987xk47 <1987x...@163.com> wrote:
> Hi Xudong:
> I'm also working on Hartshorne now.
> About the "geometric something", my opinion is the following.
> $X \times_k k^\bar$ is a sheme of finite type over an algebraically closed field, if it also
> integral and separated then it comes from some variety (the funtor $t$ in section 2). So
> scheme over an algebraically closed field looks more like varieties than other shemes,
> i.e., they are more "geometric". So we say $X$ is "geometrically something" if $X \times_k k^\bar$
> is "something".
> First, if $X$ is affine, then any closed point of $X \times_k k^\bar$ maps to a closed point of $X$, i.e.,
> a maximal ideal $m$ of $k^\bar [...]$ contracts to a maximal ideal $n$ of $k[...]$. We have
> maps $k \longrightarrow k[...]/n \longrightarrow k^\bar [...]/m$. ($k^\bar [...]/m$ is just $k^\bar$).
> Since $k^\bar$ is algebraic closure, $k[...]/n$ is integral over $k$, hence a field itself, so $n$ is maximal.
> Conversely, for any maximal ideal $n$ of $k[...]$, if the extension of $n$ is not the whole ring (I believe so but I haven't found a rigorous proof), then it is the contraction of $m$ which contains $n^e$.
> So $X \times_k k^\bar$ has more closed points than $X$. General case follows from affine case.
> In general, I think the underlying set of product of two schemes $X$ and $Y$ is not easy to describe (it is definitely not the set-theoretic product of underlying sets of $X$ and $Y$), because generic points of
> the product do not look like $(x, y)$. In some cases, when $x$ and $y$ are closed points, we can say
> $(x, y)$ is closed point of $X \times Y$.
>
> Best
> Ke
>
> > -----原始邮件-----
>
> > 发件人: "zheng xudong" <hacken.zh...@gmail.com>
I think I get it. Thank you. By the way, how is your working? How much
have you studied of Hartshorne?
best
Xudong
On Nov 4, 4:48 am, 1987xk47 <1987x...@163.com> wrote:
> Hi Xudong:
> I'm also working on Hartshorne now.
> About the "geometric something", my opinion is the following.
> $X \times_k k^\bar$ is a sheme of finite type over an algebraically closed field, if it also
> integral and separated then it comes from some variety (the funtor $t$ in section 2). So
> scheme over an algebraically closed field looks more like varieties than other shemes,
> i.e., they are more "geometric". So we say $X$ is "geometrically something" if $X \times_k k^\bar$
> is "something".
> First, if $X$ is affine, then any closed point of $X \times_k k^\bar$ maps to a closed point of $X$, i.e.,
> a maximal ideal $m$ of $k^\bar [...]$ contracts to a maximal ideal $n$ of $k[...]$. We have
> maps $k \longrightarrow k[...]/n \longrightarrow k^\bar [...]/m$. ($k^\bar [...]/m$ is just $k^\bar$).
> Since $k^\bar$ is algebraic closure, $k[...]/n$ is integral over $k$, hence a field itself, so $n$ is maximal.
> Conversely, for any maximal ideal $n$ of $k[...]$, if the extension of $n$ is not the whole ring (I believe so but I haven't found a rigorous proof), then it is the contraction of $m$ which contains $n^e$.
> So $X \times_k k^\bar$ has more closed points than $X$. General case follows from affine case.
> In general, I think the underlying set of product of two schemes $X$ and $Y$ is not easy to describe (it is definitely not the set-theoretic product of underlying sets of $X$ and $Y$), because generic points of
> the product do not look like $(x, y)$. In some cases, when $x$ and $y$ are closed points, we can say
> $(x, y)$ is closed point of $X \times Y$.
>
> Best
> Ke
>
> > -----原始邮件-----
>
zheng xudong
Nov 5, 2010, 2:22:45 AM11/5/10
to Prof. Shiu-chun Wong's Cocktail Seminar
Hi Yi,
Thank you. I was obstructed in the topological fiber product.
best
Xudong
On Nov 4, 11:34 am, Yi Zhu <y...@math.sunysb.edu> wrote:
> Hi Xudong,
>
> If you only consider the base extension by field extensions, you question is
> easy to answer.
>
> Let X be an affine variety over a field K and let L be a field extension of
> K. X_L=X \times_K L
>
> Remember that the base change (by fiber product) means tensor product of the
> rings of regular functions K[X] and L.
>
> i.e., everything (like formal variable x_i) remains same except the base
> field
>
> if the variety is defined by an ideal I, the new ideal after the extension
> is I \times_K L, ie the "same" ideal.
>
> Try x^2+y^2=1 over Q and extend over R and use the fiber product.
>
> Best,
>
> Yi
>
> > .
Thank you. I was obstructed in the topological fiber product.
best
Xudong
On Nov 4, 11:34 am, Yi Zhu <y...@math.sunysb.edu> wrote:
> Hi Xudong,
>
> If you only consider the base extension by field extensions, you question is
> easy to answer.
>
> Let X be an affine variety over a field K and let L be a field extension of
> K. X_L=X \times_K L
>
> Remember that the base change (by fiber product) means tensor product of the
> rings of regular functions K[X] and L.
>
> i.e., everything (like formal variable x_i) remains same except the base
> field
>
> if the variety is defined by an ideal I, the new ideal after the extension
> is I \times_K L, ie the "same" ideal.
>
> Try x^2+y^2=1 over Q and extend over R and use the fiber product.
>
> Best,
>
> Yi
>
> On Thu, Nov 4, 2010 at 1:39 AM, zheng xudong <hacken.zh...@gmail.com> wrote:
> > Dear seniors,
>
> > While I am reading Hartshorne, a notion, namely 'geometrically
> > something' as in the title, comes about which I am not quite clear.
> > Basically, statements are contained in Ch.II, Ex. 3.15.
>
> > Intuitively, say in the affine case, consider a variety over a field $k
> > $, if we work over an extension of the base field, we will get more
> > points, since it is saying we are looking for zero loci for the same
> > ideal in a larger field. But how can this be seen from the definition
> > of the product notion?
>
> > What topological information can we get from base extension?
>
> > Thanks a lot.
> > Xudong
>
> > --
> > You received this message because you are subscribed to the Google Groups
> > "Prof. Shiu-chun Wong's Cocktail Seminar" group.
> > To post to this group, send email to cocktai...@googlegroups.com.
> > To unsubscribe from this group, send email to
> > cocktailsemin...@googlegroups.com<cocktailseminar%2Bunsu...@googlegroups.com>
> > Dear seniors,
>
> > While I am reading Hartshorne, a notion, namely 'geometrically
> > something' as in the title, comes about which I am not quite clear.
> > Basically, statements are contained in Ch.II, Ex. 3.15.
>
> > Intuitively, say in the affine case, consider a variety over a field $k
> > $, if we work over an extension of the base field, we will get more
> > points, since it is saying we are looking for zero loci for the same
> > ideal in a larger field. But how can this be seen from the definition
> > of the product notion?
>
> > What topological information can we get from base extension?
>
> > Thanks a lot.
> > Xudong
>
> > --
> > You received this message because you are subscribed to the Google Groups
> > "Prof. Shiu-chun Wong's Cocktail Seminar" group.
> > To post to this group, send email to cocktai...@googlegroups.com.
> > To unsubscribe from this group, send email to
> > .
Pan Xuanyu
Nov 5, 2010, 2:40:05 PM11/5/10
to cocktai...@googlegroups.com
| A good reference for those geometrical "things" is Mumford's Red book,namely , Chapter 2 Section 4 the field of definition.....It has a clear "picture" Best Xuanyu --- 10年11月5日，周五, Yi Zhu <yz...@math.sunysb.edu> 写道：
|
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