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I came to realise my mistake... the $0$-chain is not the zero simplex (which doesn't make any sense anyway, right?) but the neutral element in the group $S_n(X)$. I first deleted the question, but now ...

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Note that for a polynomial $f=a_m x^m+\cdots +a_0$ being in $I$ is equivalent to say that the constant term is even and all coefficients add up to an even number. Now I claim that with $J:=(x^2+x,x^2-... View answer 1 answers 2 votes 37 views Accepted answer 1 votes First of all note that$P(z)$is irreducible in$\mathbb{Z}[z]$, because it is of degree$2$and does not have a root in$\mathbb{Z}$. Then it is also irreducible in$\mathbb{Q}[z]$, so the ideal ... View answer 2 answers 2 votes 212 views 1 votes Let‘s show that$R/(I+J)$satisfies the universal property of the tensor product; then we‘re done by the usual universal property argument. First, we need a bilinear map$\phi: R/I \times R/J\...
If one takes $R=K[x_1,x_2,\ldots]/(x_1^2,x_2^2,\ldots)$ to be the polynomial ring over a field in infinitely many variables modulo the ideal $(x_1^2,x_2^2,\ldots)$, then $R$ is not noetherian and $\... View answer 1 answers 0 votes 263 views Accepted answer 1 votes Consider an element f of the prime ideal and factor it as a product of prime elements in the UFD A. Since the ideal is prime, one factor must actually belong to the ideal and collecting these elements ... View answer 2 answers 2 votes 77 views 0 votes So I'm far from an expert, hence this is just a try. But maybe we can think for example of a two dimensional subspace in euclidean space, thus a plane, and a one dimensional subspace that is not ... View answer 1 answers 0 votes 49 views 0 votes I am trying to follow the hint from the comment. For a point$x\in X$, we can compute the stalk $$(i_*i^{-1}\mathcal{O}_X/\mathcal{I})_X)_x=0$$ if$x\notin Z$since$Z$is closed, and if$x\in Z$then ... View answer 1 answers 0 votes 55 views Accepted answer 0 votes I think I came up with the answer myself, at least for my second question. I tried to write out the details: the map$\Gamma(D(s),\mathcal{O}_Y)=A_s\rightarrow \Gamma(X_{\varphi(s)},\mathcal{O}_X)\$ is ...