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Let $I$ be an ideal of the polynomial ring $P=K[x_{1},...,x_{n}]$ that is generated by degree two polynomials ${f_1,...,f_k}$. The zero set $\mathcal{Z}(I)$ is isomorphic to an affine space of dimension $m,$ where $m<n$. Let $\mathfrak{m}$ be the maximal ideal generated by $\{\bar{x}_1,...,\bar{x}_n\}$ in $P/I.$

Let $Y=\{y_1,...,y_m\}\subset \{x_1,...,x_n\}$ where the equivalence classes of $y_1,...,y_m$ form a basis of $\mathfrak{m}/\mathfrak{m}^2.$ Consider the isomorphism below,

$$P/I \rightarrow K[Y].$$

Is it straightforward to say that $I$ can be generated by $n-m$ polynomials?

To rephrase my question, I have a smooth, connected and irreducible variety. Is it straightforward to say it is an ideal theoretic complete intersection?

I am sorry if i am sloppy in my description.

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  • $\begingroup$ I would like to note that the polynomials are not homogenous, they are just quadratic. $\endgroup$
    – Bil
    Commented Apr 29, 2015 at 22:40
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    $\begingroup$ In general smooth subvarieties are NOT complete intersections. The least you need is that the module of Kahler differentials is stably free and in this case if the codimension is large enough, it is. I do not think one knows whether $\mathbb{A}^2\subset\mathbb{A}^5$ is in general a complete intersection. Instead of 5, for any $n>5$ it is. $\endgroup$
    – Mohan
    Commented Apr 30, 2015 at 2:45
  • $\begingroup$ There seems to be a hypothesis built into the way you pose the problem. Just to make this explicit: are you assuming that there exist a subset $\{y_1,\dots,y_m\}$ of the set of coordinates $\{x_1,\dots,x_n\}$ such that the images $\{\overline{y}_1,\dots,\overline{y}_m\}$ in $P/I$ are generators? If so, then certainly $I$ is a complete intersection generated by $n-m$ polynomials of the form $x_j - g_j(y_1,\dots,y_m)$ for $x_j\not\in \{y_1,\dots,y_m\}$. $\endgroup$
    – Jason Starr
    Commented Apr 30, 2015 at 9:19
  • $\begingroup$ @Mohan thanks for the comment. Would you please write the reference for dimension n>5? $\endgroup$
    – Bil
    Commented Apr 30, 2015 at 18:27
  • $\begingroup$ @JasonStarr thanks for your comment, as well. Actually those $y_1,...,y_m$ are not random. Their equivalence classes form a base of the cotangent space $m/m^2,$ where $m\subset P/I.$ I believe (not sure) that $\{\bar{y}_1,...,\bar{y}_m\}$ form a generating set for $P/I.$ $\endgroup$
    – Bil
    Commented Apr 30, 2015 at 18:59

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