Hank Scorpio
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In what topology, elliptic curve is homeomorphic to torus?
5 votes

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 ...

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How to prove a specific quotient of polynomial ring is a free module?
5 votes

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 ...

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Is a homogeneous ideal radical?
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5 votes

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, ...

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Any line bundle (invertible sheaf) on $\operatorname{Spec}k$ is trivial?
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4 votes

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, ...

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Rational Normal Scroll $S(1,1)$
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3 votes

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 ...

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Let $X \subseteq \mathbb{P}_k^n$ be a smooth projective variety over $k$ algebraically closed. What is $\text{dim}_k(\mathcal{O}_{X, x}/m_x^2)$?
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3 votes

$f_0\mapsto f_0/f_0=1$, 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 ...

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How to find ideals of $\mathbb{Z}[x]/(x^2-1)$?
3 votes

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 ...

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Non-degenerate conic is projectively equivalent to smooth conic
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3 votes

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 $b'=0$. ...

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Showing that pencil of conics through four points in general position in $\mathbb{P}^2$ forms a line when it's considered in $\mathbb{P}^5$
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3 votes

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 $...

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Let $C$ and $C'$ be algebraic (smooth) curves.If morphism $φ:C→C'$ satisfies the condition that $#φ^{-1}(Q)=1$ for all $Q∈C'$, why $φ$ is isomorphism?
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 $[x:y]\mapsto [x^p:y^p]$ ...

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Nilradical in an algebra over a field
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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\...

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Understanding that $\mathbb{R}(X^2 + Y^2, XY)(x) \supset \mathbb{R}(Y)$?
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2 votes

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,...

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$R$ module homomorphisms between two modules
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2 votes

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 ...

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Why does this linear map from a definition of two polynomials' resultant involve two other polynomials?
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2 votes

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 ...

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Action of $PGL_2$ on $\mathbb{P}^n$
2 votes

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 $...

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How to prove that ideal is not principal
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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, ...

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Question on Smooth Completion of curves
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 ...

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Factorization as morphism between schemes vs. as morphism between underlying sets
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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 ...

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Proving uniqueness of intersection multiplicities
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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 ...

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Showing an ideal is prime in polynomial ring
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 -...

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Why can you check closedness on the fibers? (Hartshorne's proof of Bertini's theorem)
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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} \...

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Finding a non-degenerated quadric containing three projective lines in $\mathbb{P}^3(\mathbb{C})$
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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....

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Fiber of a map between spectra
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[\...

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Dimension of quotients and localizations
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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 $f=x-1$, $g=y-1$...

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degree of intersections in $\mathbb{P}^n_k$
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. ...

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Decision problem for a bundle in a line bundle extension of $\dim Ext^1=1$.
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 ...

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How to find the lifting in the spectra of tensor product?
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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(...

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Horseshoe lemma for modules in left or right.
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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 ...

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Product of smooth varities is smooth
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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)\...

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Proving that $M$ is a simple $R-$module.
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1 votes

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 ...

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