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1 vote

Relation between free resolutions and minimal free resolutions

The proof is based on the following observation: If $A \in \mathrm{M}_{n,m}(R)$ is an $n \times m$ matrix with coefficients $a_{ij}$ such that there exists indices $i,j$ where $a_{ij}$ is a unit, then ...
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2 votes
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Faithful module cancellation with maximal ideal

Consider first the case where $\mathfrak{a}\subseteq \mathfrak{m}^2$. In this case we have $\mathfrak{a}\cdot M\subseteq \mathfrak{m}^2\cdot M\subseteq \mathfrak{m}\cdot M=\mathfrak{a}\cdot M$. In ...
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4 votes
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Rational even polynomials maximally tangent to the unit circle

Denote $\varepsilon=\frac1{2m-1}$ (I guess it should be minus in the denominator, not plus, please recheck). We want to find a polynomial $h_{m-1}(t)$ of degree $m-1$ and a polynomial $p_m(t)$ of ...
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4 votes
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Is this a true weakening of the quasi-coherence property?

Any submodule of a quasicoherent $O_X$-module satisfies (#): this is clear via reduction to principal open sets, and the fact that localization is exact. More generally, as Neil observes, if $F$ ...
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3 votes
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On the Artin-Rees Lemma for non-commutative rings

There is some discussion of this in Rowen's "Ring Theory", volume I, Section 3.5, with additional references therein. Exercise 19 on p. 462 in op. cit. states that a polycentral ideal $I$ of ...
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0 votes

An exercise in fuzzy logics built from a t-norm

My teacher has provided a solution: Take a $[0, 1]_*$-interpretation with $I(\phi \rightarrow \phi * \phi ) = 1$, and say $a:=I(\phi)$. Define the function \begin{align*} h: [0, 1] & \rightarrow [...
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  • 301
1 vote
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Prove that $\overline{a}_{11}$ is a prime element in $R$

Now that I have thought about this further, I realize that you need much less than the Samuel Conjecture to solve this problem. By a dimension count, the ring $R$ is a complete intersection ring, ...
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A confusion about power series and p-adic measures

You are asking about substitution of one power series into another. This can be interpreted in two ways, which ultimately amount to the same thing (like two different ways of thinking about anything) ...
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0 votes

Cohen-Macaulyness of Milnor algebra

I am just posting my comment as one answer. Using the local flatness criterion, the ring $R/\langle \partial_1(f),\dots,\partial_n(f)\rangle$ is Cohen-Macaulay if and only if it is flat as a module ...
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On presentations of universal rings of deformations

Doesn't this kind of prove itself? Pick some elements $\alpha_1, \dots, \alpha_n \in \mathfrak{m}$ which represent $\mathfrak{m} / (p, \mathfrak{m}^2)$. Clearly sending $t_i$ to $\alpha_i$ defines a ...
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12 votes

Is the ring of power series with coefficients in a field free as a module over the polynomials subring?

No. Take some element $f$ of $K[x]$ which is not in the ideal $(x)$ and is not invertible. Then $$K[[x]] \otimes_{K[x]} K[x]/(f) \cong K[[x]]/(f) = 0$$ where the latter follows from the fact that ...
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5 votes
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Is $\mathbb{Z}[i,\varphi]$ a Euclidean domain?

It turns out that $K=\mathbb{Q}[i,\varphi]$ is norm-Euclidean. The proof of this fact appears as Appendix A in https://arxiv.org/abs/2205.03007. I'll explain the heft of the argument here. Let $R=\...
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  • 345
1 vote

Completion reducing to localization on Noetherian rings

No. If $A$ is a complete noetherian local ring, then $A$ is complete for the adic topology defined by any other prime ideal.
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  • 473
1 vote

Cellular and primary binomial ideals

Consider the ideal $I=\left\langle x_{1}^{3} x_{3}-x_{1}^{3}, x_{1}^{4}, x_{1}^{2} x_{2} x_{4}-x_{1}^{2} x_{2}, x_{2}^{2}, x_{4}^{3}-1\right\rangle \subseteq \mathbb{k}\left[x_{1}, x_{2}, x_{3}, x_{4}\...
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