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| bio | website | |
|---|---|---|
| location | Pisa, Italy | |
| age | 24 | |
| visits | member for | 1 year, 9 months |
| seen | 18 hours ago | |
| stats | profile views | 502 |
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May 6 |
awarded | Necromancer |
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Apr 10 |
comment |
Associated Primes of Tensor Product I think that in general almost nothing can be said. See for example $M =A/I$ and $N = A/J$. |
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Apr 10 |
comment |
Does there exists a continuous surjection from $\mathbb{R}$ to $\mathbb{R}^2$? I don't know, but continuous is probably too weak as a condition. In fact there is the Peano curve, which is an example of continuous surjection $[0,1] \to [0,1] \times [0,1]$. |
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Mar 24 |
answered | When do two matrices have the same column space? |
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Jan 25 |
comment |
Similar Matrices I have just discovered that my proof is contained also in Problems 7 and 9 of Section 6.7 in Topics in algebra (2nd ed.) by I. N. Herstein. |
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Dec 6 |
comment |
Shrinking the base field of a scheme of finite type over a field It is true by the theory of limits of schemes of finite presentation: EGA IV §8. |
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Dec 6 |
answered | Elementary questions about regular rings and Zariski tangent spaces |
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Nov 19 |
accepted | Finite presentation of algebra of invariants |
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Nov 7 |
answered | Finite presentation of algebra of invariants |
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Oct 31 |
comment |
Limits of subrings and surjectivity Thanks! A simpler example is $A = \mathbb{Q}$, because every subring of $\mathbb{Q}$ of finite type over $\mathbb{Z}$ is of the form $\mathbb{Z}[1/d]$ and the map $\mathrm{Spec}(\mathbb{Q}) \to \mathrm{Spec}(\mathbb{Z}[1/d])$ cannot be surjective. |
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Oct 31 |
accepted | Limits of subrings and surjectivity |
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Oct 31 |
asked | Limits of subrings and surjectivity |
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Sep 22 |
awarded | Tumbleweed |
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Sep 15 |
comment |
Compute Hilbert function of a monomial ideal Thank you very much! |
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Sep 15 |
accepted | Compute Hilbert function of a monomial ideal |
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Sep 15 |
accepted | A consequence of Runge's theorem |
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Aug 15 |
comment |
Finite presentation of algebra of invariants Thanks! I had the suspect that the answer to (2) was no. But my real interest is in (1) and I hope that it has an adfirmative answer. |
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Aug 15 |
revised |
Finite presentation of algebra of invariants added 5 characters in body |
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Aug 13 |
asked | Finite presentation of algebra of invariants |
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Aug 12 |
comment |
Irreducible polynomial over algebraically closed field Great answer!!! |
