# Tag Info

7

Here's an example showing that $S$ is not always a finite union of algebraic sets. Let $Z$ be the zero locus of the single polynomial $x_1x_2 - 1$. Then $S = \mathbb{A}^1\setminus \{0\}$. What is true is that $S$ is always a finite union of sets defined by finitely many polynomial equations (basic Zariski closed sets) and negated equations (basic Zariski ...

5

The operation is called projection. Tbe first-order theory of the complex field (which is the same as the first-order theory of algebraically closed fields of characteristic $0$) admits quantifier elimination. This means that $\exists x_n (x_1, \ldots, x_n) \in Z$ is equivalent to a propositional combination of primitive formulas of the form $p_j(x_1, ... 4 The module of Kahler differentials is spanned by$dx$and$dy$, and differentiating the relation$y^2 = x^3$gives the relation$2y \, dy = 3x^2 \, dx$. Multiplying by$y$gives $$2y^2 \, dy = 2x^3 \, dy = 3 x^2 y \, dx$$ from which it follows that$2x \, dy - 3y \, dx$is torsion, since it is annihilated by multiplication by$x^2$. (Of course then we need ... 4 Many of these are proved using the universal hypersurface. Let$P$the projective space of all degree$d$forms (since the equation and any non-zero constant multiple give the same variety) and consider$Z\subset P\times\mathbb{P}^n$the universal hypersurface defined in the obvious way - these are pairs$(f,p)$with$f(p)=0$. Consider$T\subset Z$, defined ... 3 The answer is no. Why did you expect that? For example, consider, say$A$is a domain and$B=A[x]$,$I=(x-a)B$for some$0\neq a\in A$. Then the natural map$A\to B/I$is an isomorphism and in particular flat. If$J=I+xB$, clearly$A\to B/J=A/aA$is not flat. 3 The forgetful functor$\mathsf{LRS} \to \mathsf{RS}$has a right adjoint. The right adjoint "$\mathrm{Spec}$" is a rather direct generalization of the spectrum of a commutative ring. You can find the construction in W. D. Gillam's Localization of ringed spaces, for instance. The underlying set of$\mathrm{Spec}(X,\mathcal{O}_X)$consists of all pairs ... 3 It depends upon your definition of variety. If you, as Hartshorne, think of a variety as a quasi-projective variety, then yes - just take the set of smooth points contained in$X$(this is a variety in the sense of Harthorne - because the set of singular points is a closed subset). In that case$\dim X_{smooth} = \dim X$. If however, you think of as a ... 3 You have to use Newton Identities. See https://en.wikipedia.org/wiki/Newton%27s_identities In general if you have$n$variables$x_1\ldots.x_n, define the polynomials $$p_k(x_1,\ldots,x_n)=\sum_{i=1}^nx_i^k = x_1^k+\cdots+x_n^k,$$ and \begin{align} e_0(x_1, \ldots, x_n) &= 1,\\ e_1(x_1, \ldots, x_n) &= x_1 + x_2 + \cdots + x_n,\\ e_2(x_1, ... 3 The story is rather complicated in characteristicp$, but I think the answer of Drike takes care of it. On the other hand, Gregory Grant makes a claim that seems to me to be not right. We’re working with the algebraic group$(K^3,+)$, which I think would usually be called$\mathbf G_{\mathrm a}^3$. Consider now the subgroup of$(K^3,+)$given ... 3 You can also use the Jacobian criterion directly for the homogenous polynomial and notice that the only point, where all partials vanish, is$(0,0,0)$, which is not a point of the quadric (it is not even a point of the projective space). You can check Hartshorne, Exercise$I.5.8$for details. A key ingredient is Euler's Lemma. 2 If$U\subseteq X$is an open set and$V\subseteq Y$is any open set containing$f(U)$, then the restriction of$f^*\mathcal{G}$to$U$depends only on the restriction of$\mathcal{G}$to$V$(more precisely,$(f^*\mathcal{G})|_U=g^*(\mathcal{G}|_V)$, where$g:U\to V$is the restriction of the morphism$f$to a morphism$U\to V$). To show$f^*\mathcal{G}$is ... 2 Unless I missed something,$R[x]/\mathscr{P}$is generated as an algebra over$R/\mathfrak{m}$by the class$\bar x$of$x$(because any element of$R[x]$is a polynomial in$x$with coefficients in$R$, and its class mod$\mathscr{P}$is therefore a polynomial in$\bar x$with coefficients in$R/\mathfrak{m}$). But a field extension which is finitely ... 2 Section 2.4 of Waterhouse's Introduction to Affine Group Schemes (he doesn't number theorems for some reason). You forgot "commutative." 2 Let$\DeclareMathOperator{Spec}{\operatorname{Spec}}\mathfrak p \in \Spec A$. We first show that the map$f^\#_\mathfrak p : \mathcal O_{\Spec A, \mathfrak p} \to f_* \mathcal O_{\Spec B, \mathfrak p}$is the homomorphism$\varphi_\mathfrak p : A_\mathfrak p \to B_\mathfrak p$. Since the basic open subsets$D(g)$for$g \in A$form a basis for$\Spec A$, we ... 2 Strictly speaking the association of$D_f$to$R_f$only defines the sections on a basis; to construct the regular functions on an arbitrary open$U$, we have to take the inverse limit of the$D_f$contained in$U$. The fact that this is a contravariant functor is pretty straightforward from the universal property of inverse limits. Separatedness is also ... 2 The definition is stating that the map$f \mapsto \{f(s)\}_{s \in S}$is injective, meaning that a function is completely determined by its values on$S$. This is just the scheme theoretic version of the statement that a standard continuous function in topology (e.g.$\mathbf{R} \to \mathbf{R}$) is determined by its values on a dense subset. 2 Let$C$be an algebraic curve. A collection of points$P_1,…,P_n$in$C$with assigned integer multiplicities$k_1,…,k_n$is called a divisor on$C$and it is denoted $$D=k_1P_1+...+k_nP_n.$$ And that is it, it is a formal sum, so as it is defined it does not have any immediate meaning. For instance, on your example we can define the divisors$D_1=(i,1), ...

2

$A$ is quasi-compact, since it is a closed subset of the quasi-compact set $K^n$. Any quasi-compact discrete set is finite, this is a very easy exercise in basic topology.

2

In general, $k'$-points of $X$ need not be determined by their image (and I'm not sure where you're seeing Milne treat them as if they were, except in the case $k'=k$). For instance, if you take $X=Y$, then $X(k')$ is in bijection with the set of automorphisms of $k'$ over $k$ (of which there can be many), but all of these maps send the unique point of $Y$ ...

2

This is actually fairly easy. Indeed, if $f \colon A \to B$ is of finite presentation, then so is $B \otimes_A B \to B$ by Tag 00F4 (4). Moreover, the latter is flat by assumption, and it is always a surjective ring map (i.e., closed immersion). Thus, by Tag 0819, it is an open immersion onto a clopen subset. Thus, the diagonal is an open immersion, which ...

2

Have a look at the formulation of the exercise again. He explicitly states that a regular curve is implicitly assumed to be locally noetherian and in part $b)$ (your exercise) he assumes the curve to be quasi-compact. We have quasi-compact + locally noetherian = noetherian. So you are done, aren't you?

2

Sketch of the solution: Consider ellipse $\cal E$ with foci $B$ and $C$ which is tangent to the graph of $f$. Observe that the point of tangency is the point $P$. Indeed, each point $Q(s, f(s))$ lies inside the ellipse $\cal E$ which means that $QB+QC\le PB+PC$. The tangent $\ell$ to the graph of $f$ at point $P$ is tangent to the ellipse $\cal E$. But ...

1

Yes, we need the field to be algebraically closed. Then it is clear. $B(x,y,0,0)$ is a homogenous polynomial in $x,y$. Such a polynomial certainly admits a root other than $(x=0,y=0)$.

1

Let $A$ be a commutative ring and $S$ a multiplicative set. Then the family of rings $\left\{A_s \right\}_{s \in S}$ forms a directed family. To see this, first we define a partial order on $S$ by $s \le t$ if $t = u s$ for some $u \in S$. Next for $s \le t$ with $t = u s$, there exists a ring homomorphism $f_{s,t}: A_s \rightarrow A_t$, which is defined by ...

1

Show $$I(V):I(W) \supset I(V-W)$$ This is the easy part. By the definition of the ideal quotient, we have to show $I(V-W)I(W) \subset I(V)$ and this immediate: If $f$ vanishes on $V-W$ and $g$ vanishes on $W$, then $fg$ vanishes on all of $V$. Show $$I(V):I(W) \subset I(V-W)$$ For this we need a crucial ingredient: Lemma. If $W \subset \mathbb ... 1 This is kind of useless at this point, but since I was able to get a copy of one of Olivier's original articles on weakly étale/absolutely flat morphisms, I thought I'd share what I found. The article Ferrand, Daniel. "Epimorphismes d'anneaux et algèbres séparables." C. R. Acad. Sci. Paris Sér. A-B 265 1967 A411–A414. MR0244313 (39 #5628) is cited as ... 1$m_p$is the maximal (irrelevant) ideal of$K[X_1,\dots,X_n]/I(V)$, that is, the ideal generated by the images of$X_1,\dots,X_n$hence$m_p=M_P/I(V)$. Then$m_p^2=(M_P^2+I(V))/I(V)$. Now it's clear why $$m_p/m_p^2=M_P/(M_P^2+I(V)).$$ 1 By the genus-degree formula (or Riemann-Hurwitz), the projective closure of this curve has genus$3$, so looks like a three-holed torus. Next you need to figure out how many points taking the projective closure added, and as in the comments it's not hard to see that there are four. 1 If$F=f^{-1}(p)$, then the normal bundle is the pull back of the normal bundle of$p\in B$and this of course is trivial. The second sequence doesn't look right. You should have$\omega_{F_i}$s on the right. You always have the natural exact sequence,$0\to O_X\to O_X(F_1+\cdots+F_r)\to \oplus O_{F_i}\to 0$, since$O_{F}(F)=O_F$. Tensoring with$\omega_X$... 1 First of all, "equivariant version of$\mathcal{O}(1)$" sounds a bit confusing to me. What we actually do, we endow$\mathcal{O}(1)$with equivariant structure.$U(1)$acts as follows. Point of total space is a pair$(x, \xi)$, where$x \in \mathbb{P}^1$i.e.$x$is line in$\mathbb{C}^2$.$\xi$is a linear function on this line$l$.$U(1)\$ acts on hole ...

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