# Polynomial with infinitely many zeros.

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?

• 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... Jan 25, 2013 at 2:48
• do you mean isolated roots? Jan 25, 2013 at 3:23
• Do you mean roots or zero coefficients? Jan 25, 2013 at 3:33

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 $\mathbb{C}^{n}$.

• 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 $\mathbb{C}^{n}$. Jan 25, 2013 at 7:02

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.

• A nice proof for the general case. Jan 25, 2013 at 4:07

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

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

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 $f(a,b)=0$.