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I'm trying to prove a lemma, so I wanted to use the next claim (which I do not know if it is true or if there is a counter example): every smooth irreducible affine variety is set theoretical complete intersection. Or if this is not true I would like to know if there is a counter example to it.

I have tried to find a counter example or find the result in the literature, but I had no luck. I suspected that this fact is true, however, I was unable to prove it.

The lemma I want to prove using this plausible claim is: Given a smooth irreducible variety $V$ of dimension $d$ with ideal $I(V) = (f_1,…,f_r)⊆k[x_1,…,x_n]$, we can construct the ideal $I ' (V) ⊆k[x_1,…,x_n,y_1,…,y_n]$ given by $I ′ (V) =(∇f_1⋅(y_1,…,y_n),…,∇f_r⋅(y_1,…y_n))$. Then every irreducible component $W$ of $Z(I ′ (V))$ such that $π(W)∩V≠∅$ has $dim(W)≥n+d$ where $π$ is the projection to the first $n$ coordinates

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    $\begingroup$ Welcome new contributor. Your conclusion can be proved “Zariski locally.” Every smooth subvariety is Zariski locally a complete intersection. $\endgroup$
    – Jason Starr
    Commented Dec 25, 2022 at 0:28
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    $\begingroup$ All smooth curves in affine spaces are set- theoretic complete intersections. Higher dimensional varieties, this is an open question. The only general case known is when the conormal bundle is trivial, for example affine spaces embedded in higher dimensional affine spaces. $\endgroup$
    – Mohan
    Commented Dec 25, 2022 at 20:31
  • $\begingroup$ @Mohan : do you have a reference or a sketch of proof for your claim about affine smooth curves? $\endgroup$
    – Libli
    Commented Dec 27, 2022 at 8:41
  • $\begingroup$ I think its a Theorem from Ferrand @Libli $\endgroup$
    – Leonardo Lanciano
    Commented Dec 27, 2022 at 13:54
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    $\begingroup$ @Libli For a curve in three space, this is a theorem of Ferrand ( and Szpiro). Higher dimensional case is due to me, Inv. Math., late seventies. $\endgroup$
    – Mohan
    Commented Dec 27, 2022 at 14:19

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