# Bertini's Theorem

Let $p_1,...,p_n\in\mathbb{P}^{N}$ be general points. Consider the linear system $|L|$ of hypersurfaces of degree $d$ in $\mathbb{P}^{N}$ with prescribed multiplicities $m_1,...,m_n$ at $p_1,...,p_n$. I would like to know if the following variation of Bertini's theorem holds.

Suppose $|L|\neq\emptyset$. Is it true that a general element $H\in|L|$ is an hypersurface of degree $d$ such that $Sing(H) = \{p_1,...,p_n\}$ and $mult_{p_i}H = m_i$ for $i = 1,...,n$ ?

With multiplicity $m_i$ I mean that $p_i$ is an ordinary $m_i$-uple singular point.

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$|L|$ may consist of a single surface that is not what you want! – Alex Degtyarev Mar 15 '14 at 23:40

As pointed out by Alex in his comment, this is in general not true.

For instance, consider the case $N=d=n=m=2$. Then $|L|$ is the linear system of plane curves of degree $2$ passing through $p_1$ and $p_2$ with multiplicity $2$. Of course there is only one such a curve, namely the line through $p_1$ and $p_2$ counted with multiplicity $2$. In particular, $|L|$ is not empty but no element in $|L|$ has ordinary singularities.

However, what you want is true under some additional hypotheses. For instance, the following result is proven in Shustin's paper Lower deformation of isolated hypersurface singularities, Proposition 10 p. 242.

For any pair $N$, $k$ of positive integers set $$M(N, \,k):=2 \cdot \binom{k+N}{N}, \quad M(N, \,2k-1):=\binom{k+N}{N} + \binom{k+N-1}{N}.$$ Then the following holds.

Proposition. Assume $$\sum_{i=1}^n \binom{m_i+N-1}{N} < M(N, \, d).$$ Then if $p_1, \ldots, p_n$ are general points in $\mathbb{P}^N$ there exists a hypersurface $L$ of degree $d$ having ordinary singularities of multiplicity $m_1, \ldots, m_n$ at the points $p_1, \ldots, p_n$ and which is smooth elsewhere. Consequently, the general element of $|L|$ has the same properties.

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Thank you. It seems to me that this condition means that the number on the left is the greatest number of independent conditions imposed by the singular points. So, this means that the dimension of the space of hypersurfaces must be strictly greater than the number of expected independent conditions. In the example $d = N = m_1 = m_2 = 2$ this fails because the number of expected conditions is 6 and even if these equations are not independent we have $M(2,2) = 6$. Is this correct? Why did you define the number $M(N,2k-1)$? It seems that you do not need it in the proposition. – user47036 Mar 16 '14 at 14:33
Dear Francesco, I guess there is a little mistake in your definition of $M(N,k)$. Did you mean $M(N,2k) = 2\cdot\binom{k+N}{N}$ ? – S_Z_S May 14 '14 at 11:42
Right, thank you. Now it is fixed. – Francesco Polizzi May 15 '14 at 14:53