Hank Scorpio
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• Cypress Creek

It depends what you mean by "torus" - if you mean the manifold $S^1\times S^1$, the only possible choice is the Euclidean/analytic topology, because manifolds are Hausdorff and the Zariski ...

This is not as expansive as the other answer by Z Wu, but perhaps that will be a strength. First I claim that we can write every monomial in $R/I$ as $x^aw^d$, $x^aw^dy$, or $x^aw^dz$ by applying the ...

No, it is not the case that every homogeneous ideal is radical. Here is an easy counterexample: $(x^2)\subset k[x]$. This is homogeneous, being generated by a homogeneous element, but not radical, ...

A line bundle $\mathcal{L}$ on a scheme (or locally ringed space) $X$ is called trivial if $\mathcal{L}\cong\mathcal{O}_X$ as $\mathcal{O}_X$-modules. As $\operatorname{Spec} k$ is a one-point space, ...

Up to automorphisms of $\Bbb P^3$, we may assume the two lines are $\ell_1=[a:b:0:0]$ and $\ell_2=[0:0:c:d]$ with the isomorphism identifying $[a:b:0:0]$ and $[0:0:a:b]$ (why? the cones on the two ...

$f_0\mapsto f_0/f_0=1$, and any function in $\mathcal{O}_{X,x}/\mathfrak{m}_x^2$ is the sum of a constant function and a function in $\mathfrak{m}_x/\mathfrak{m}_x^2$. Yes, your exact sequence is ...

By the correspondence theorem, it suffices to analyze ideals $I$ of $\Bbb Z[x]$ containing $(x^2-1)$. Write such an ideal as $I=(x^2-1,f_1,\cdots,f_m)$ (if there are no $f_i$, then $I=(x^2-1)$). Then ...

Bezout. If three points on your conic were colinear, then that line is an irreducible component of your conic and therefore your conic is degenerate. Any transformation fixing $[0,0,1]$ keeps $b'=0$. ...

You don't have to worry so much about whether $a$ is zero or nonzero: you can just write down what the conditions "passing through $p$" mean. Explicitly, if $p=[p_0,p_1,p_2]$, then a conic $... View answer 2 votes This is not true. Purely inseparable maps of curves give fibers of cardinality one but are not isomorphisms. In characteristic$p$, consider the map$\Bbb P^1\to \Bbb P^1$by$[x:y]\mapsto [x^p:y^p]$... View answer Accepted answer 2 votes What you ask happens exactly when$K$is infinite. If$K=\Bbb F_q$is finite, just take$\prod_{a\in K} (x_1-a)$which vanishes on every element of$\Bbb F_q^n$but not on$(b,0,\cdots,0)$where$b\in\...

What you're doing isn't really the right way to go about this. Your expressions cannot build the element $\frac{1}{x^2+y^2+xy}$, for instance. (You're also claiming that a field is equal to an element,...

No and no, to the questions in the first line. $k[x,y]/(x,y)\cong k$, so we can define a map by sending $(p,0)\mapsto p\cdot(ax+by)$ and $(0,q)\mapsto q\cdot(cx+dy)$ for any $a,b,c,d,p,q\in k$. Each ...

I think you're a bit confused here. The Sylvester matrix $S$ is the matrix that represents the map $(P,Q)\mapsto AQ+BP$. That is, if you give me two polynomials $P,Q$ then I can determine $AQ+BP$ by ...

Consider the natural action of $GL_2$ on $\Bbb P^1$ by coordinate transformations. This induces an action on the global sections of any sheaf on $\Bbb P^1$. In particular, we get an induced action on $... View answer Accepted answer 2 votes Hint: look at$x^2$and$x^3$. Can you find a non-unit$f\in R$with$g,h\in R$so that$fg=x^2$and$fh=x^3$? There's a solution under the spoiler, but give yourself a chance before looking at it, ... View answer 2 votes Actually, there is something going on here, depending on your definition of a locally closed immersion. The definition of a locally closed immersion according to most sources (i.e. EGA) is a closed ... View answer Accepted answer 2 votes Let$f:X\to Y$be a morphism of schemes. The scheme-theoretic image$Z$is the smallest closed subscheme of$Y$through which$f$factors, and if$f$is quasi-compact or$X$is reduced, we have that ... View answer Accepted answer 2 votes Q1: projective transformations do indeed preserve intersection multiplicity. On one hand, this should be geometrically clear: the axioms are coordinate-independent, so why should our specific choice ... View answer 2 votes One low-tech way to go from here is to write any polynomial$f$in$k[x,y,z,w]$as a sum of$j\in J$and some other particularly nice$u$and then apply the homomorphism$\varphi$you've written down -... View answer Accepted answer 2 votes I pieced this together after the discussion with Mindlack in the comments. Define$B$to be the subset of$X\times\Bbb P^n$cut out by the minors of size$n-\dim X+1$of the matrix $$\begin{pmatrix} \... View answer Accepted answer 1 votes Brute force works without a ton of fuss, since it turns this in to a linear algebra problem: write$$\varphi = ax_0^2+bx_0x_1+cx_0x_2+dx_0x_3+ex_1^2+fx_1x_2+gx_1x_3+hx_2^2+ix_2x_3+jx_3^2$$and plug in.... View answer 1 votes Consider the sequence of ring maps \Bbb Z\to \Bbb Z[\sqrt{-3}] \to \Bbb Z[\frac{1+\sqrt{-3}}{2}]. This gives a corresponding sequence of maps of spectra \def\Spec{\operatorname{Spec}}\Spec \Bbb Z[\... View answer Accepted answer 1 votes To be clear, every time I say \dim here I'm talking about the dimension of a vector space over the base field k. Equality 1: \dim k[x,y]_{(x,y)} / (f,g) = \dim k[x,y]/(f,g). Take f=x-1, g=y-1... View answer 1 votes First, one should assume that l is not contained in f. Next, view l as a copy of \Bbb P^1: then the restriction of f to \Bbb P^1 is a homogeneous degree-d polynomial of two variables. ... View answer 1 votes Not unless you're working over \Bbb F_2. In any abelian category, there's a bijection between isomorphism classes of extensions of B by A and elements of \operatorname{Ext}^1(A,B) where the ... View answer Accepted answer 1 votes By the universal property of the tensor product, from the maps R\to A \to Frac(A/p) and R\to B\to Frac(B/q) we get a map$$A\otimes_R B\to Frac(A/p)\otimes_R Frac(B/q)\cong Frac(A/p)\otimes_{Frac(... View answer Accepted answer 1 votes The horseshoe lemma is not at play here - what Matsumura is saying is that if$P^\bullet\to m/xA\to 0$is a$B$-projective resolution of$m/xA$, then$P^\bullet\to B\to k\to 0$is a$B$-projective ... View answer Accepted answer 1 votes$\phi\times\psi$does indeed give an isomorphism of$U\times V$with$Z(f_1,\cdots,f_a)\times Z(g_1,\cdots,g_b) \subset \Bbb A^n\times\Bbb A^m$, which is exactly$Z(f_1,\ldots,f_a,g_1,\ldots,g_b)\...
If $f$ is not surjective, then $Rx$ is a proper nonzero submodule of $M$, contrary to our assumption that $M$ is simple. It could be better - it seems like you're using $1$ here instead of $2$. I ...