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Assume that $X$ is a non-singular projective curve in $P^n_k$, where $k=\bar{k}$ and $n \geq 3$. Prove that for every $m$ large enough, there always exists a non-singular hypersurface of deg $m$ containing $X$.

The hardest part of this problem is that you should properly choose conditions to restrict these hypersurfaces in order to make them non-singular. The existence of hypersurfaces containing $X$ can be shown if we consider the exact sequence $0 \rightarrow \mathcal{I}_X(m) \rightarrow \mathcal{O}_{P^n}(m) \rightarrow \mathcal{O}_X(m) \rightarrow 0$, take global sections and then estimate corresponding dimensions. But, you know, the greater m is, the more hypersurfaces containing $X$ there will be. So I need a good interpretation of non-singular hypersurfaces in the language of cohomology.

I think another possible approach is that we can construct a new model of this problem where hypersurfaces of deg $m$ containing $X$ form a nice variety. Then the non-singular ones are 'general' in this variety, like Bertini's theorem in Hartshorne's Algebraic Geometry.

Derek Allums
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user884626
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    Set $N = {n+m \choose m}$. There is a hypersurface $\Delta \subset P^N$ called the discriminant sitting in the $P^N$ which parametrizes all degree $m$ hypersurfaces in $P^n$, and $\Delta$ parametrizes the singular hypersurfaces. The basic idea should just be that containing $X$ is a codimension $2$ condition, i.e. the scheme $\mathcal H_X \subset P^N$ of hypersurfaces containing $X$ has dimension $N-2$, and hence should intersect $\Delta$ in a scheme of dimension $N-3$, i.e. the general point of $\mathcal H_X$ parametrizes a smooth hypersurface... – Tabes Bridges May 12 '23 at 05:57
  • ...My guess is that you should study the universal singular point $\Phi := \left{([Y],p) , | , Y \text{ is singular at } p\right} \subset P^N\times P^n$ and its natural projections. – Tabes Bridges May 12 '23 at 05:59
  • @TabesBridges do you have a short proof of $\mathcal{H}_X \not\subset\Delta$? – Levent May 12 '23 at 09:30
  • @Levent Otherwise every hypersurface containing $X$ would be singular, no? – Derek Allums May 12 '23 at 09:52
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    @DerekAllums , Yes. How do we know that this is not the case? (I know that it is impossible, I just wonder if there is a short proof) – Levent May 12 '23 at 10:55
  • @Levent You're right - and in fact my (implied) claim boils down to OP's question, since said another way, I am claiming there always exists a non-singular hypersurface containing $X$, which sounds awfully similar to OPs first sentence. I guess we'd need to follow the hint provided by Tabes Bridges above to prove it by construction. – Derek Allums May 12 '23 at 11:47

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