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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 characteristic $p$, 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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