Polynomials in one variable with infinitely many roots.

Question

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 ?

Answer 1

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.

Answer 2

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 $R$.

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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.

Answer 3

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.

Answer 4

Here is a very simple example: Every abelian group $R$ admits a unique commutative ring structure such that if $f\in R[x]$ is a polynomial whose constant term is zero, then every element in $R$ is a root of $f$.

Let $(R,+)$ be an abelian group with identity element $0$. Define multiplication $*$ on $R$ by $$ a*b=0 \text{ for all }a,b\in R.$$ It is easy to verify that $(R,+,*)$ is a commutative ring. Every polynomial $f$ in $R[x]$ with a zero constant term has every element of $R$ as its root. So, to obtain a polynomial whose roots form a set of a given cardinality, it suffices to have an abelian group of that cardinality.

To prove the uniqueness of this structure, suppose $(R,+,*)$ is a commutative ring such that if $f\in R[x]$ is a polynomial whose constant term is zero, then every element in $R$ is a root of $f$. Let $a,b$ be elements of $R$. Since $ax\in R[x]$ is a polynomial whose constant term is zero, $b$ is a root of $ax$, that is, $a*b=0$ .