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Another generalization of the theorem for vector spaces could be the following.

Assume that $X \subset Y$ is a closed subvariety, with $Y$ connected. If

$\dim X = \dim Y $ and

$\textrm{mult}_xX = \textrm{mult}_x Y$ for all $x \in X$,

then $X=Y$.

In fact, assume $X \neq Y$. Since $Y$ is connected there exists $x \in X \cap \overline{Y \setminus X}$, and for such a $x$ we have

$\textrm{mult}_xY \geq \textrm{mult}_xX + \textrm{mult}_x \overline{Y \setminus X} \geq \textrm{mult}_xX+1$,

contradiction. This can be useful, since it does not require the irreducibility of $X$ and $Y$. However, the assumption about the connectedness of $Y$ cannot be dropped: consider the case where $Y$ is the disjoint union of two copies of $X$.

EDITED

EDIT. In a first version of this answer I required the condition $\dim T_xX = \dim T_xY$ instead of $\textrm{mult}_xX = \textrm{mult}_x Y$. This actually did not work, see damiano's comment below.

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Another generalization of the theorem for vector spaces is could be the following.

Assume that $X \subset Y$ is a closed subvariety, with $Y$ connected. If

$\dim X = \dim Y $ and

$\dim T_xX \textrm{mult}_xX = \dim T_xY$textrm{mult}_x Y$ for all $x \in X$,

then $X=Y$.

In fact, assume $X \neq Y$. Since $Y$ is connected there exists $x \in X \cap \overline{Y \setminus X}$. Then , and for such a $x$ we must have

$\dim T_xY > \textrm{mult}_xY \dim T_xX$,geq \textrm{mult}_xX + \textrm{mult}_x \overline{Y \setminus X} \geq \textrm{mult}_xX+1$,

contradiction. This can be useful, since you don't need to check it does not require the irreducibility of $X$ and $Y$, but only the dimensions of their tangent spaces at each point. Y$. However, the assumption about the connectedness of $Y$ cannot be dropped: consider the case where $Y$ is the disjoint union of two copies of $X$.

EDITED. In a first version of this answer I required the condition $\dim T_xX = \dim T_xY$ instead of $\textrm{mult}_xX = \textrm{mult}_x Y$. This actually did not work, see damiano's comment below.

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Another generalization of the theorem for vector spaces is the following.

Assume that $X \subset Y$ is a closed subvariety, with $Y$ connected. If

$\dim X = \dim Y $ and $\dim T_xX = \dim T_xY$

for all $x \in X$, then $X=Y$.

In fact, assume $X \neq Y$. Since $Y$ is connected there exists $x \in X \cap \overline{Y \setminus X}$. Then for such a $x$ we must have

$\dim T_xY > \dim T_xX$,

contradiction. This can be useful, since you don't need to check the irreducibility of $X$ and $Y$, but only the dimensions of their tangent spaces (which can be computed locally)at each point. However, the assumption about the connectedness of $Y$ cannot be dropped: consider the case where $Y$ is the disjoint union of two copies of $X$.
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