70 lines
3.7 KiB
Markdown
70 lines
3.7 KiB
Markdown
# Things that we can do
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Given $f := (\tilde f,h) : (\mathbb{C}^n,0) \to (\mathbb{C}^p\times \mathbb{C}^q,0)$ such that
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$\tilde f : (\mathbb{C}^n,0) \to (\mathbb{C}^p,0)$ is finite.
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1. Set $X_i = f_i(x_1,\ldots,x_n)$ for $i = 1,\ldots, p$, and $Y_i = h_(x)$ for $i=1,\ldots, q$. **done**
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1. Find a $\mathbb{C}$-basis $g_1,\ldots, g_m$ of $\mathcal{O}[x_1,\ldots,x_n]/(f_1,\ldots,f_p)$. **done**
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1. Given $\alpha \in \mathbb{C}[x_1,\ldots,x_n]^{p + q}$ such that there exists a unit $u\in \mathcal{O}[x_1,\ldots,x_n]$, and $A^1_1\ldots,A^{p+q}_m \in \mathbb{C}[X]$ with $u\cdot \alpha = g_1\cdot (A_1^1,\ldots, A_1^{p+q})+\cdots+g_m\cdot (A_m^1,\ldots, A_m^{p+q})$ modulo some submodule, find the $A_i^j$ and tell us if $u$ lives in $\mathcal{O}[X]$ or not.
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- If there exists a solution with $u\in \mathbb{C}$ **can be done with syzygies, or reduction modulo de right ideal**.
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- If there is a solution but not with $u$ being a constant, **to-do**.
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- Even if in theory we are interested only in vectors of length $p+q$, in some cases it will be usefull to do it for length $1$.
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1. Apply the previous step to the vectors $(\frac{\partial f_1}{\partial x_j},\ldots, \frac{\partial f_{p+q}}{\partial x_j})$ to obtain $n$ vectors of length $m\times (p+q)$. Call them $(F_1^1,\ldots, F_{m\times (p+q)}^1), \ldots, (F_1^n,\ldots, F_{m\times (p+q)}^n)$.
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1. For each $(e_1,\ldots,e_q)$ with $0\leq e_i<m$, apply algorithm 3 to $h_1^{e_1}\cdots h_q^{e_q}$ to get vectors $H_1,\ldots, H_{(p+q)m^q}$. If in some point algorithm 3 said that it used a unit not in $\mathcal{O}[X]$, raise a nonimplemented error (we don't know how to handle this case).
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1. Compute a standard basis of the module $M$ generated by the $F$'s and the $H$'s.
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1. Use the standard basis to compute a $\mathbb{C}$-basis of the quotient $\mathcal{O}[X]^{m(p+q)}/M$. Its size is the Ae-codimension.
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# Pending generalizations:
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- Change the source by a Cohen-Macaulay subvariety.
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# Theoretical remarks:
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- The reason for taking $e_i<m$ in step 5 is because $Y_i^m$ can be expressed as a polynomial in $X,Y$ with degree in the $Y$ smaller than $m$. **check the theory**
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# Relevant examples:
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If $f(x)=x + x^3$, there is no solution for (3) with polynomial unit.
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# Comments
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- At some point we thought that solving (3) when $u$ is a polynomial could be done with Syzygies, but we found a problem with that approach.
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# To do
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- Check if Pfistetr's and Hiroshi's algorithms can handle examples that we can. - Done
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- List of mappings vs algorithms and check if they work
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- Failing examples
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- Failing examples that can be converted into working examples
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- Other nice working examples
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- Positive characteristic
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- Find finite determinacy criteria
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- Algorithm modulo a submodule of O_n^p (usually the one generated by a power of the maximal ideal)
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- Submodule in source or in target??
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- Let this sumbodule to be the zero module (this is just the usual algorithm)
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- It always converges up to a maximal ideal power, can we find which?
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- If it is not too expensive: search and use a power of the maximal ideal for which it converges
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- If it is too expensive: let the user choose if they want to search for this maximal ideal power
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- Finite determinacy criteria (modulo a submodule) -- Guillermo
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- Implement complete transversal algorithm
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- Stopping criteria for the algorithm
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- Verify that if the local-global basis is the same then the algorithm:
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1. Finishes
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2. It finishes correctly
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3. If it does not, it gives an upper bound
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- In all rings generated: change ordering to (dp/ds,c/C)
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- Change the use of the maximal ideal so that it works modulo a power in the target already.
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- Do a baby example of why the implementation of the modulo power in the target is not working (try crosscap or Hiroshi example). |