Let $Y$ be an affine variety of $k^n$, defined by polynomials $f_1,\dots,f_s \subset A = k[y_1,\dots,y_n]$, i.e., $Y=V(f_1,\dots,f_s)$ where $k$ is algebraically closed. Now let $g \in A$ such that $f_1,\dots,f_s,g$ are algebraically dependent over $k$. Is it then possible that the complement of the variety $W=V(f_1,\dots,f_s,g)$ in $Y$ is of infinite cardinality, i.e., the set $Y \setminus W$ to have infinite cardinality?
complement of a subvariety by an algebraic element
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algebraic-geometry
2 Answers
1
Yes, it's possible.
For example, let $k=\mathbb{C}$, $A = \mathbb{C}[x,y]$, $f = (x - y)(x - y - 1)$, $g = x - y$.
Then \begin{align*} &\bullet\;\;f,g\text{ are algebraically dependent, since }f = g^2 - g\\[6pt] &\bullet\;\;Y=V(f) = \{(z,z) \mid z \in \mathbb{C}\} \cup \{(z,z-1) \mid z \in \mathbb{C}\}\\[6pt] &\bullet\;\;W = V(f,g) = V(g) = \{(z,z) \mid z \in \mathbb{C}\}\\[6pt] &\bullet\;\;Y \setminus W = \{(z,z-1) \mid z \in \mathbb{C}\}\\[6pt] \end{align*}
so $Y \setminus W$ is infinite.
2
Yes. Let $n=2$, $f_1=x$, $g=x-1$. Then as $W\cap Y$ is empty, $Y\setminus W= Y\cong \mathbb{A}^1_k$ which is infinite.