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 coefficients 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 ?
 A: 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.
A: 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.
A: 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.
