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Can a polynomial in $ \mathbb{C}[x,y] $ have infinitely many zeros? This is clearly not true in the one-variable case, but what happens in two or more variables?

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    $\begingroup$ I would add that the in the one-variable case it's true that a non-zero polynomial has only a finite number of zeroes. A zero polynomial on the other hand... $\endgroup$
    – JSchlather
    Jan 25, 2013 at 2:48
  • $\begingroup$ do you mean isolated roots? $\endgroup$
    – Maesumi
    Jan 25, 2013 at 3:23
  • $\begingroup$ Do you mean roots or zero coefficients? $\endgroup$
    – Joe Z.
    Jan 25, 2013 at 3:33

5 Answers 5

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Examples

The polynomials $ XY $ and $ X + Y $ have infinitely many zeros in $ \mathbb{C}^{2} $.

The zero-set of $ XY $ is $ (\{ 0 \} \times \mathbb{C}) \cup (\mathbb{C} \times \{ 0 \}) $, while the zero-set of $ X + Y $ is $ \{ (x,-x) ~|~ x \in \mathbb{C} \} $. Both sets are clearly uncountable.


General Theory

Suppose that $ p(X,Y) \in \mathbb{C}[X,Y] $ has a non-vanishing $ X $-degree $ n $. We can thus find polynomials $ {q_{0}}(Y),\ldots,{q_{n}}(Y) \in \mathbb{C}[Y] $, where $ {q_{n}}(Y) \neq 0 $, such that $$ p(X,Y) = \sum_{i=0}^{n} {q_{i}}(Y) \cdot X^{i}. $$ Let $ \Delta $ denote the zero-set of $ {q_{n}}(Y) $ in $ \mathbb{C} $, and for each $ \lambda \in \mathbb{C} $, let $ \mathcal{Z}_{\lambda} $ denote the zero-set of $ p(X,\lambda) \in \mathbb{C}[X] $ in $ \mathbb{C} $. It follows from $ {q_{n}}(Y) \neq 0 $ that $ \Delta $ is finite, which implies that $ \mathbb{C} \setminus \Delta $ is uncountable.

For each $ \lambda \in \mathbb{C} \setminus \Delta $, the polynomial $ p(X,\lambda) \in \mathbb{C}[X] $ has non-vanishing $ X $-degree $ n $ precisely because $ {q_{n}}(\lambda) \neq 0 $. Therefore, by the Fundamental Theorem of Algebra, $ \mathcal{Z}_{\lambda} \neq \varnothing $, which implies that $ \displaystyle \bigcup_{\lambda \in \mathbb{C} \setminus \Delta} (\mathcal{Z}_{\lambda} \times \{ \lambda \}) $ is an uncountable set of zeros of $ p(X,Y) $ in $ \mathbb{C}^{2} $.

Similarly, if $ p(X,Y) $ has a non-vanishing $ Y $-degree, then $ p(X,Y) $ has uncountably many zeros in $ \mathbb{C}^{2} $.


Conclusion: If $ p(X,Y) \in \mathbb{C}[X,Y] $ is not of the form $ p(X,Y) = c $, where $ c \in \mathbb{C}^{\times} $, then $ p(X,Y) $ has uncountably many zeros in $ \mathbb{C}^{2} $.

In general, for $ n \in \mathbb{N}_{\geq 2} $, if $ p(X_{1},\ldots,X_{n}) \in \mathbb{C}[X_{1},\ldots,X_{n}] $ is not of the form $ p(X_{1},\ldots,X_{n}) = c $, where $ c \in \mathbb{C}^{\times} $, then $ p(X_{1},\ldots,X_{n}) $ has uncountably many zeros in $ \mathbb{C}^{n} $.

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    $\begingroup$ Basically, if $ n \in \mathbb{N}_{\geq 2} $, then only the non-constant polynomials and the zero polynomial in $ \mathbb{C}[X_{1},\ldots,X_{n}] $ have uncountably many zeros in $ \mathbb{C}^{n} $. $\endgroup$ Jan 25, 2013 at 7:02
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Any nonconstant polynomial $p(x,y)\in\mathbb{C}[x,y]$ will always have infinitely many zeros.

If the polynomial is only a function of $x$, we may pick any value for $y$ and find a solution (since $\mathbb{C}$ is algebraically closed).

If the polynomial uses both variables, let $d$ be the greatest power of $x$ appearing in the polynomial. Writing $p(x,y)=q_d(y)x^d+q_{d-1}(y)x^{d-1}+\cdots+q_0(y)$, let $\hat y$ be any complex number other than the finitely many roots of $q_d$. Then $p(x,\hat y)$ is a polynomial in $\mathbb{C}[x]$, which has a root.

EDIT: I neglected to mention that this argument generalizes to any number of variables.

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  • $\begingroup$ A nice proof for the general case. $\endgroup$ Jan 25, 2013 at 4:07
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Any non constant polynomial in one variable has at least one root in $\mathbb C$.

If $P(x,y)\in\mathbb C[x,y]$ is nonconstant and $a$ is a root of the polynomial $P(x,0)$ then $(a,0)$ is a root of $P(x,y)$...

Consider the polynomials $P(x,0),P(x,1),P(x,2),P(x,3),\ldots$

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The function $x$ has zeros $\{(0,y):y\in \mathbb C\}$.

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For variety, only the zero polynomial has infinitely many zeroes for $y$.

That is, if there are infinitely many values $b$ such that $f(x,b) = 0$, then $f$ is the zero polynomial.

But this is a very different question than asking about the zeroes in $x$ and $y$ together: that is, pais $(a,b)$ such that $f(a,b)=0$.

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