4 <= 1 of course :)

Let $V \subset {\mathbb{C}}^6$ be the set defined by $$\operatorname{rank} \begin{bmatrix}z_1 & z_2 & z_3 \\ z_4 & z_5 & z_6 \end{bmatrix} \leq 0 1 .$$ Then dimension of $V$ is 4, but near the origin you need at least 3 holomorphic functions to define $V$ (the three $2\times 2$ subdeterminants being zero). That is, $V$ is not a set-theoretic complete intersection.

There is an extra hickup in this. The minimal number of germs of holomorphic functions necessary to define the set (the germ of the set) need not be the same as the number of germs of holomorphic functions necessary to define the ideal. I do not have an example offhand for this.

3 I meant of course that rank <= 1, or that the 3 2x2 subdeterminants are zero

Let $V \subset {\mathbb{C}}^6$ be the set defined by $$\det operatorname{rank} \begin{bmatrix}z_1 & z_2 & z_3 \\ z_4 & z_5 & z_6 \end{bmatrix} = \leq 0 .$$ Then dimension of $V$ is 4, but near the origin you need at least 3 holomorphic functions to define $V$. V$(the three$2\times 2$subdeterminants being zero). That is,$V$is not a set-theoretic complete intersection. There is an extra hickup in this. The minimal number of germs of holomorphic functions necessary to define the set (the germ of the set) need not be the same as the number of germs of holomorphic functions necessary to define the ideal. I do not have an example offhand for this. 2 fix the matrix Let$V \subset {\mathbb{C}}^6$be the set defined by $$\det \begin{bmatrix} z_1 begin{bmatrix}z_1 & z_2 & z_3 \\ z_4 & z_5 & z_6 \end{bmatrix} = 0 .$$ Then dimension of$V$is 4, but near the origin you need at least 3 holomorphic functions to define$V$. That is,$V\$ is not a set-theoretic complete intersection.

There is an extra hickup in this. The minimal number of germs of holomorphic functions necessary to define the set (the germ of the set) need not be the same as the number of germs of holomorphic functions necessary to define the ideal. I do not have an example offhand for this.

1