An equation of form $x^{2}+ax+b=0$ might have infinite amount of solutions in a ring $(R,+,\cdot)$

Question

An equation of form $x^{2}+ax+b=0$ might have infinite amount of solutions in a ring $(R,+,\cdot)$.

Now I am a bit lost here.

The definition for ring is that $(R,+)$ is Abel and $(R,\cdot)$ is a monoid.

I just wonder what in earth they are after in this exercise?

I should find a equation of that form and show that it has infinite amount of solutions. But it feels just a bit absurd.

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After receiving these very good answers. I feel that I should write an example.

Let's look at matrix ring $(M_{2}(\mathbb{Z}/4\mathbb{Z}),+,\cdot)$, which has the usual matrix addition and multiplication. Now, when $n>1$, the $n \times n$ matrix is not commutative. Now we can calculate that

$$ \begin{bmatrix} 2 & 0\\ 0 & 0 \end{bmatrix}\cdot \begin{bmatrix} 2 & 0\\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 4 & 0\\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} $$ so we have solution to a equation $$ X^{2}=0. $$

We can find an example of infinite ring that has infinitely many solutions to the equation above.

For example such is matrix ring $(M_{2}(\mathbb{R}),+,\cdot)$ where infinitely many solutions can be found using matrix of form $$ \begin{bmatrix} 0 & a\\ 0 & 0 \end{bmatrix} $$ where $a\in\mathbb{R}$.

Answer 1

Example

$$x^2+1=0\;,\;\;x\in\Bbb H=\text{Hamilton's Quaternions}$$

Answer 2

HINT: Let $R$ be the ring of infinite sequences of zeroes and ones with coordinatewise addition and multiplication modulo $2$.

Answer 3

Remember that a ring may have zero divisors.

For example, if $M_2(R)$ is a $2\times 2$ matrices ring, over any non-zero ring $R$ including a nilpotent element of the form $a^2=0$, then, even the equation $$ x^2=\mathbf{0}= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} $$ has an infinite number of solutions. In fact, any matrix $$ x= c\begin{bmatrix} a & 0 \\ 0 & 0 \end{bmatrix}= \begin{bmatrix} ca & 0 \\ 0 & 0 \end{bmatrix} $$ for any $c\in R$, is a solution of the above.

P.S.: If you want a concrete example of the above idea, take any infinite ring $R$ with zero divisors $b,d$, such that $bd=0$ and $db\neq 0$ (matrix rings provide a host of such examples). Then the element $a=db\neq 0$ is nilpotent, since $a^2=(db)^2=d(bd)b=0$. Now, your required ring will be $M_2(R)$ and inside it, the equation $x^2=\mathbf{0}$, will have an infinite number of solutions of the above form.

An even simpler example has already been mentioned in an update of OP: Take $R=\mathbb{R}$. Then the equation $x^2=\mathbf{0}$ will have an infinity of solutions inside the ring $M_2(\mathbb{R})$, of the form: $$ \begin{bmatrix} 0 & a\\ 0 & 0 \end{bmatrix} $$ for any $a\in\mathbb{R}$.

Answer 4

Consider the ring $R=\mathbb{Z}\times \mathbb{Z}$ and the equation $(a,0)x=(0,0)$ with $a \in \mathbb{Z}$ over $R$. It has infinitely many solutions of the form $\{(0,n):n \in \mathbb{Z}\}$. Try to use this idea.

Answer 5

Summarizing a bit all the answers here: If $R$ is an integral domain (i.e. it is commutative and does not have divisors of $0$), then you may consider its field of fractions $Q(R)$ and then embed this one in its algebraic closure $\overline {Q(R)}$. Since $\overline {Q(R)}$ is commutative and algebraically closed, any $2$nd degree equation will have at most $2$ distinct roots. Since $R \subseteq \overline {Q(R)}$, it follows that any equation of the form $x^2 + ax + b = 0$ with $a,b \in R$ will also have at most $2$ distinct roots in $R$, clearly not what you want.

In order to avoid the conclusion obtained above, it follows that you must look for a ring which is not an integral domain. There are two possible approaches: either look for a non-commutative ring (rings of matrices, for instance), or for a ring with divisors of $0$ (products of integral domains, rings of matrices).