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Let $V \subset {\mathbb{C}}^6$ be the set defined by $$ \operatorname{rank} \begin{bmatrix}z_1 & z_2 & z_3 \\\ z_4 & z_5 & z_6 \end{bmatrix} \leq 0 . $$ 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.

Let $V \subset {\mathbb{C}}^6$ be the set defined by $$ \operatorname{rank} \begin{bmatrix}z_1 & z_2 & z_3 \\\ z_4 & z_5 & z_6 \end{bmatrix} \leq 1 . $$ 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.

Let $V \subset {\mathbb{C}}^6$ be the set defined by $$ \det \begin{bmatrix}z_1 & z_2 & z_3 \\\ z_4 & z_5 & z_6 \end{bmatrix} = 0 . $$ 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.

Let $V \subset {\mathbb{C}}^6$ be the set defined by $$ \operatorname{rank} \begin{bmatrix}z_1 & z_2 & z_3 \\\ z_4 & z_5 & z_6 \end{bmatrix} \leq 0 . $$ 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.

Let $V \subset {\mathbb{C}}^6$ be the set defined by $$ \det \begin{bmatrix} z_1 & z_2 & z_3 \\ z_4 & z_5 & z_6 \end{bmatrix} = 0 $$ 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.

Let $V \subset {\mathbb{C}}^6$ be the set defined by $$ \det \begin{bmatrix}z_1 & z_2 & z_3 \\\ z_4 & z_5 & z_6 \end{bmatrix} = 0 . $$ 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.