This is exercise 12.2.I in Vakil's notes:     Suppose $X$ is a finite type scheme of pure dimension $d$ over an   algebraically closed field $k$. Show that $X$ is regular at its closed points if and   only if it is smooth.   For the forward direction, for a closed point $p$, $\operatorname{dim}\mathcal{O}_{X,p}=d$. From Exercise 12.1.G, one has that the Zasiki cotangent space at a closed point is given by the cokernel of the Jacobian matrix. If $X$ is regular at closed points, then corank of the Jacobian matrix is $d$ at all closed points, hence by Exercise 12.2.H, the Jacobian matrix has corank $d$ at all points. This shows the smoothness. Conversely, smoothness of $X$ implies the Jacobian has corank $d$ at all points, in particular closed points.   Is the result true for $k$ not algebraically closed? I don't see why it's required that $k$ is algebraically closed.
123361
Take $X=\operatorname{Spec}(k[x,y]/(y^2-x^p+t))$ with $k=\mathbb{F}_p(t)$ and $p\neq2$.  The Jacobian ideal is $J=(y,y^2-x^p+t)=(y,x^p-t)$ which is maximal ie a closed point of $X$ so Jacobian criterion work to say that $\mathfrak{p}=(y,x^p-t)$ is singular.   Let's $A=k[x,y]/(y^2-x^p+t)$ then in $A_\mathfrak{p}$ one have $\mathfrak{p}A_\mathfrak{p}=(y)$ because $x^p-t=y^2\in(y)$ so $\mathfrak{p}A_\mathfrak{p}$ has so many generator as $\dim A_\mathfrak{p}$ (which is 1 because $\mathfrak{p}$ is maximal in $A$ so $\dim A=\dim A_\mathfrak{p}$). So $A_\mathfrak{p}$ is regular ie $\mathfrak{p}$ is regular.  Where is my stupid mistake?
111350
Let $V$ be a vector space and $\alpha \in \operatorname{End}(V)$   (i) If $V$ is finite dimensional, then $\alpha$ is injective iff $\alpha$ is surjective.  (ii) Give example showing (i) is false if $V$ is not finite dimensional.  So on (i), since $V$ is finite dimensional, then $V$ has a basis with finite cardinality and hence $\dim(V)=n$. Also, the following holds, $\dim(V)=\dim(\ker(\alpha) + \dim(\operatorname{im}(\alpha))$ Since $\alpha$ is injective, then the $\ker(\alpha)$ is $0$ which implies that the $\dim(\ker(\alpha)$ is $0$ which implies that $\dim(\operatorname{im}(\alpha))=\dim(V)$ and that shows that $\alpha$ is surjective. (Is that right?)   On the second one, (ii), I just have no idea. I mean shouldn't Axiom of Choice be able to work here?
98625
Let $K$ be an algebraic extension field (not necessarily finite) of $\mathbb{Q}$. Let $\mathscr{O}_K$ be the integral closure of $\mathbb{Z}$ in $K$.  Then, is $\mathscr{O}_K$ Noetherian?  If $K$ is a finite extension of $\mathbb{Q}$, then $\mathscr{O}_K$ can be shown to be Noetherian. However, if we do not assume that finiteness condition, is it still true?  (It can be shown that $\mathscr{O}_K$ is an integrally closed domain of Krull-dimension $1$. Hence, if it is Noetherian, it is a Dedekind domain.)
115311
The following exercise comes from Fulton's Algebraic Curves:     Let $k$ be an infinite field and $V = \{(t,t^2,t^3) \, ; \, t \in k \} \subset \mathbb{A}^3(k)$. Show that $V$ is algebraic and compute $I(V)$ by giving a finite set of generators.   Now, showing $V$ is algebraic is easy by noting that $V = V(X^2-Y, X^3-Z)$, but I can't find a way to describe $I(V)$ explicitly. Obviously $(X^2-Y)(X^3-Z) \in V$, but how can I find the actual generators of this ideal?
85442
There's a question in my complex analysis book:     Let $G$ be a region and let $\gamma_0$ and $\gamma_1$ be two closed smooth curves in $G$. Suppose $\gamma_0\sim\gamma_1$ and $\Gamma$ is a homotopy between them. Also suppose that $\gamma_t(s)=\Gamma(s,t)$ is smooth for each $t$. If $w\in \mathbb{C}-G$, define $h(t)=n(\gamma_t;w)$ and show that $h:[0,1]\rightarrow\mathbb{Z}$ is continuous.    This is an exercise right after the section that proves the following version of Cauchy's theorem:     If $\gamma_0$ and $\gamma_1$ are two closed rectifiable curves in $G$ and $\gamma_0 \sim \gamma_1$ then $$\int_{\gamma_{0}}f=\int_{\gamma_1} f$$ for every function $f$ analytic on $G.$   I must be missing something. It seems like you can just say the following:      'Proof:' Since $\frac{1}{z-w}$ is analytic in $G$ ($w$ isn't in $G$), and $\gamma_0\sim\gamma_t\ \forall t\in[0,1]$ (consider a rescaled version of $\Gamma$), then by Cauchy's theorem, $n(\gamma_t;w)=n(\gamma_0;w)=n(\gamma_1;w)$. Thus, the winding number is constant along $t$ and thus certainly continuous.   My problem with that is that this doesn't require the assumption that $\Gamma$ is a smooth homotopy. I think the author had meant for the reader to work harder for this result. What is the proof 'supposed' to look like?
64970
In my recent reading of various books and notes on algebraic geometry and scheme theory, I have come across three definitions of affine $n$-space over a field $k$:   $\mathbb{A}_k^n$ is $k^n$ 'without an origin'; $\mathbb{A}_k^n$ is simply $k^n$ with the Zariski topology (i.e. zero loci of sets of polynomials are closed); $\mathbb{A}_k^n=\operatorname{Spec}k[x_1,\ldots,x_n]$.   I know that things like the Nullstellensatz give us a nice way of going from one point of view to another when it comes to polynomial rings and affine varieties, but here I'm having some confusion. For $k=\mathbb{C}$ (which is algebraically closed, so things should all act reasonably nicely) we have that the points of $\operatorname{Spec}\mathbb{C}[x]$ are in bijective correspondence with points of $\mathbb{C}$, apart from the general point $[(0)]$. So it seems to me that these definitions can't all be the same, but searching for affine space on Google or in books leads to any one of the above definitions, depending on the level of the text.  Am I just making a silly mistake, or is there a difference in these definitions? If so, which one is the 'right' one?
56500
Motivation for the question: in Neukirch - Algebraic Number Theory (Chapter II - Section 6 - Henselian Fields) the author discusses the context of a nonarchimedean valued field $K$ and its completion $\widehat{K}$, and considers the separable closure $K_v$ of $K$ in $\widehat{K}$, so that $K \subseteq K_v \subseteq \widehat{K}$.  Then he writes the following:     When $K_v$ is algebraically closed in $\widehat K$ - hence in particular $\text{char}(K) = 0$ - this is immediately obvious [...]   (emphasis mine).  My question is about the implicit claim that if $K_v$ is algebraically closed,  then $\text{char}(K) = 0$.  Now we know that $K_v$ is algebraically closed if and only if $K$ is perfect (since every $K_v$ will be the algebraic closure of $K$ and so every algebraic extension of $K$ is separable).  So we're asking whether it is true that if $K$ is perfect, then it must have characteristic $0$.  I know that all fields of characteristic $0$ are perfect. I also know that we have the example of finite fields which are also perfect fields but have prime characteristic. This does not contradict the claim above because the finite fields have only the trivial valuation.  So is there something about having a nonarchimedean valuation on a perfect field that forces its characteristic to be zero?
91328
$p : X \to Y$ is continuous, closed and surjective, and $X$ is a normal space. Show $Y$ is normal.   There is a hint, which I'm trying to prove: show that if $U$ is open in $X$ and $p^{-1}(\{y\}) \subset U$, $y \in Y$, then there is a neighbourhood $W$ of $y$ such that $p^{-1}(W) \subset U$.  I have a candidate for $W$, namely $W=Y\setminus p(X \setminus U)$. I did prove that this $W$ is open, and that $p^{-1}(W) \subset U$, but I don't see how $y \in W$. I think this would require injectivity of $p$.  I have also shown that $y \in p(U)$ and that $W \subset p(U)$, so if also $W \supset p(U)$, then $y \in W$.   Can anyone help me?
57055
Let $K$ be an algebraically closed field ($\operatorname{char}K=p$). Denote $${\mathbb F}_{p^n}=\{x\in K\mid x^{p^n}-x=0\}.$$ It's easy to prove that ${\mathbb F}_{p^n}$ consists of exactly $p^n$ elements.  But if $|K|&lt;p^n$, we have collision with previous statement (because ${\mathbb F}_{p^n}$ is subfield of $K$).  So, are there any finite algebraically closed fields? And if they exist, where have I made a mistake?  Thanks.
2036
Consider the topological subspace $\mathbb{Q} \cap [0,1]$ endowed with the usual topology of $[0,1]$, since $[0,1]$ is Hausdorff and that $\mathbb{Q} \cap [0,1]$ is not closed, we conclude that $\mathbb{Q} \cap [0,1]$ is not compact, i.e., there exist an open covering $\{O_i\}_{i \in I}$ of $\mathbb{Q} \cap [0,1]$ such that for any finite index set $J \subseteq I$, $\displaystyle \bigcup_{j \in J} O_j \subset \mathbb{Q} \cap [0,1]$, where the inclusion is strict.  My question is if there is an explicite example of such open covering?
117065
A proposition from Fulton's book 'Algebraic Curves': Let $I$ be an ideal in $k[X_1, ..., X_n]$ ($k$ algebraically closed), and suppose $V(I) = \{ P_1, \ldots, P_N \}$ is finite. Let $\mathcal{O}_i = \mathcal{O}_{P_i}(\mathbb{A}^n)$. Then there is a natural isomorphism of $k[X_1, \ldots, X_n] \mathbin{/} I$ with $\prod_{i = 1}^{N} \mathcal{O}_i \mathbin{/} I\mathcal{O}_i$. But if $k$ is not algebraically closed, is it still true that, if $V(I)$ is finite, $k[X_1, \ldots, X_n] \mathbin{/} I$ is a direct product of local rings? And if so, then some factors are most likely given by localizations at the points of $V(I)$, whereas other factors are not; how to describe these latter factors? The reason for this question is the observation that field extensions of $k$ are fields and thus local rings. For instance, $\mathbb{Q}[X] \mathbin{/} (X^2 + 1)$ is a field, but $V(I)$ is empty and $\mathbb{Q}$ is not algebraically closed. Hence, by the Chinese remainder theorem, it seems to me that, for any $f$ in $k[X]$, $k[X] \mathbin{/} (f)$ is the direct product of local rings anyway. Is it at least that correct?
64894
I am reading  that finite-dimensional subspaces of normed vector spaces have closed direct sum complements. This is the proof:      Let $\{e_1, ..., e_n\}$ be a basis for $\mathcal M$. Every $x \in   \mathcal M$ has then a unique representation $$x = \alpha_1(x)e_1 +...+  \alpha_n(x)e_n.$$ Each $\alpha_i$ is a continuous linear functional on    $\mathcal M$ (a linear map from  finite dimensional space is always    continuous) which extends to a member of $\mathcal X^*$,  by the    Hahn-Banach theorem ($\mathcal X^*$ is the dual of $\mathcal X$). Let    $\mathcal N$ be the intersection of the null spaces of these exten­sions. Then $\mathcal X = \mathcal M\oplus \mathcal N$.   I follow it until the last sentence. Clearly $\mathcal N$ is closed, as it is the intersection of closed sets (the inverse of $0$ under any continuous function, and in particular any linear functional on a vector space, is closed). Also it is easy to see that $\mathcal M \cap \mathcal N= \{0\}$, since if some $m\in\mathcal M$ is in every null space, it must have the zero coefficient for every basis vector, and so it must be zero.  My question: Why do we have $\mathcal X =\mathcal M + \mathcal N$? This seems to only hold if the extension given by Hahn-Banach is zero off of $\mathcal M$, but I see no reason why this should be the case.
43360
Atiyah Macdonald, Exercise 8.3. Let $k$ be a field and $A$ a finitely generated $k$-algebra. Prove that the following are equivalent:   (1) $A$ is Artinian.   (2) $A$ is a finite $k$-algebra.   I have a question in the proof of (1$\Rightarrow$2): By using the structure theorem, we may assume that $(A,m)$ is an Artin local ring. Then $A/m$ is a finite algebraic extension of $k$ by . Since $A$ is Artinian, $m$ is the nilradical of $A$ and thus $m^n=0$ for some $n$. Thus we have a chain $A \supseteq m \supseteq m^2 \supseteq \cdots \supseteq m^n=0$. Since $A$ is Noetherian, $m$ is finitely generated and hence each $m^i/m^{i+1}$ is a finite dimensional $A/m$-vector space, hence a finite dimensional $k$-vector space.  But now how can I deduce that $A$ is a finite dimensional $k$-vector space?
18023
In chapter 2 section 7 (pg 151) of Hartshorne's algebraic geometry there is an example given that talks about automorphisms of $\mathbb{P}_k^n$. In that example Hartshorne states that $\mathcal{O}(-1)$ has no global sections. However, we know that $\mathcal{O}(1)$ is generated by global sections. This is stated at the first of the section that if $\mathbb{P}_k^n = Proj k[x_0,...,x_n]$ then the $x_0,...,x_n$ give rise to global sections $x_0,...,x_n\in\Gamma(\mathbb{P}_k^n,\mathcal{O}(1))$.  I guess I don't understand this twisted structure sheaf very well, or to be honest I don't think I understand structure sheaves in general as well as I would like. The twisting part seems simple at first- you shift the grading of Proj over and then take the structure sheaf.  Admittedly I don't feel comfortable using the structure sheaf other than using the basic facts about it that Hartshorne gives when it is introduced. If anyone could give some insight as to what's going on here or how I might be able to understand this better it would be much appreciated. Thanks.
80906
Can anyone give an explicit basis of the $k$-vector space $k(X) = \operatorname{Quot}(k[X])$ of rational functions over $k$?   The dimension is given by $$\dim_k k(X) = \max(|k|, |\mathbb N|).$$  If $k$ is infinite, this follows from $|k| \leqslant |k(X)| \leqslant |k[X] \times k[X]| = |k \times k| = |k|$ and the linear independence of $\lbrace \frac{1}{X - \alpha} \mid \alpha \in k\rbrace.$ If $k$ is finite, the result can be obtained similarly from $|k(X)|=|\mathbb N|$ and the linear independence of the monomials $X^n$, $n \geqslant 0$.
117564
This is an exercise from Miles Reid, Undergraduate Algebraic Geometry. The proof has two parts, one of which I can do and one of which I can't. I could have some misunderstandings about notations here as well so anything people can point out will be helpful.  a) Let $k$ be an infinite field and $f\in k[X_1,...,X_n]$ be nonconstant. Prove  $$V(f)=\{P\in \mathbb{A}^n_k\mid f(P)=0\}\neq \mathbb{A}^n_k.$$  $A^n_k$ here is the point set $k^n$.   My proof (which follows from a hint of his) is write $f=\sum a_i(X_1,...,X_{n-1})X_n^i$. Then $V(f)$ consists of two kinds of points   $P=(p_1,...,p_{n-1},p_n=0)$ $P=(p_1,...,p_{n-1},p_n\neq 0 )$ with $\sum a_i(p_1,...,p_{n-1})p_{n}^i=0$   In the first case we have $V(f)\neq \mathbb{A}^n_k$. Assume we have $P$ like in the second case and now write $f=\sum b_i(X_1,...,\hat{X}_{n-1},X_n)X_{n-1}^i$. Here again, either $P=(p_1,...,p_{n-1}=0,p_n\neq 0)$ and $V(f)\neq \mathbb{A}^n_k$ or $p_{n-1}\neq 0$, and by induction we should get to $V(f)\neq \mathbb{A}^n_k$.  b) Now let $k$ be algebraically closed. Let $a_m(X_1,...,X_{n-1})X_n^m$ be the leading term of $f$. Show that when $a_m\neq 0$, there is a finite set of points of $V(f)$ corresponding to every value $(X_1,...,X_{n-1})$. Thus, $V(f)$ infinite for $n \geq 2$.  So I have kinda no idea what to do. A given value of $(X_1,...,X_{n-1})$ corresponds to points in $V(f)$ $P=(p_1,...,p_{n-1},p_n)$ where $f(P)=0$, but I don't know anything about the value of $p_n$; if it vanishes then clearly there is a correspondence $(X_1,...,X_n)\to (p_1,...,p_{n-1},0)$ but I guess I need to show there are a finite number of $p_n$ where this is true and I don't know how.
34799
I've read this exercise on my book, but I don't know how to prove it... Can someone help me?  Let X $\subseteq \mathbb{A}^n$ be an algebraic set, and $f, g:X\rightarrow k$ two regular functions.  a) If we see $f$ as an application $X \rightarrow \mathbb{A}^1(k)$, prove that $f$ is continuous in the Zariski topology.  b) If $X$ is irreducible and an open $U \subseteq X$ exists such that $f_{|U} = g_{|U}$, prove that $f=g$.  c) Give an example of $X, f$ and $g$ such that $f \neq g$ but $f_{|U} = g_{|U}$ in an open $U \subseteq X$.  Thank you!  For b) I have been able to prove that:  Let E be the subset where $f = g$ so E is closed (Why? I know that $E = (f,g)^{-1}(\Delta_{k})$ but... why is $(\Delta_{k})$ closed?). So $U \subseteq E \Rightarrow \bar{U} \subseteq \bar{E}=E$ (beacuse E is closed) $\Rightarrow X \subseteq E$, because $\bar{U}=X \Rightarrow E = X$ and $f=g$ in X.