# Polynomials in one variable with infinitely many roots.

Can a non-zero polynomial in one variable have infinitely many roots ?

Can a non-zero polynomial in one variable have uncountably many roots ?

Motivation : over $\mathbb Z/12\mathbb Z$, $X^2-4$ has 4 roots.

When it comes to polynomials with coefficents over an integral domain, the answer is clearly negative (in that case, a polynomial can’t have more roots than its degree).

What happens with a ring that has zero divisors ?

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Let $R = \prod_{i \in I} (\mathbb{Z}/4\mathbb{Z})$ with elementwise addition and multiplication. Then $t \mapsto (2,2,…)t$ is a non-zero polynomial over $R$ with $2^{|I|}$ many zeros.

Here’s a very simple construction: let $R$ be the ring

$$\mathbf{Z}[x_0, x_1, x_2, \ldots] / \langle x_0^2 + 1, x_1^2 + 1, x_2^2 + 1, \ldots \rangle$$

Then every $x_i$ is a root of the polynomial $t^2 + 1$ over $R$.

A more trimmed down example is the ring

$$\mathbf{Z}[x,y] / \langle x^2, xy, y^2 \rangle$$

This is the ring of all polynomials of the form $a + bx + cy$ with integer coefficients, subject to the relations $x^2 = xy = y^2 = 0$. So multiplication is

$$(a + bx + cy)(d + ex + fy) = ad + (ae+bd) x + (af+cd)y$$

Every number of the form $bx + cy$ is a root of the polynomial $t^2$.

Of course, this example only needed one variable, not two, but I think it’s more interesting with two.

Take $A = \prod_{n \geq 1} \mathbb{Z}/2^n\mathbb{Z}$ and consider the polynomial $f(x) = 2x$ over $A.$ It has infinitely many zeros.