Let $R$ be the ring $\frac{\mathbb{Q}[x]}I$ , where $I=\langle x^2-x\rangle$.Then
$(1) R$ has infinitely many unit elements
$(2)R$ has exactly $4$ idempotent elements
$(3)R $ has infinitely many prime ideals.
$(4) R$ is UFD.
My thoughts:-
$I=\langle x^2-x\rangle=\langle x(x-1)\rangle$
Let $A=\langle x\rangle$ and $B=\langle x-1\rangle$
A and B are comaximal i.e $A+B=\mathbb{Q}[x]$. This is trivial as $x-(x-1)=1$
Thus by Chinese Remainder Theorem $\langle x^2-x\rangle=A\cap B=AB$ and $\frac{\mathbb{Q}[x]}{\langle x^2-x \rangle}\simeq \frac{\mathbb{Q}[x]}{\langle x \rangle } ×\frac{\mathbb{Q}[x]}{\langle x-1 \rangle } $
Now $\frac{\mathbb{Q}[x]}{\langle x \rangle } \simeq \mathbb{Q}$ and $\frac{\mathbb{Q}[x]}{\langle x-1 \rangle }\simeq \mathbb{Q}$ via the ring homomorphism $\phi :\mathbb{Q}[x] \to \mathbb{Q}$ given by $\phi (p(x))=p(1),$ where $p(x) \in \mathbb{Q}[x]$.
So $R \simeq \mathbb{Q}× \mathbb{Q}$
Now every element $(a,b) \neq (0,0)$ from $\mathbb{Q}× \mathbb{Q}$ is a unit , so infinitely many unit elements
If $(a,b)^2=(a,b)$ then $a^2=a,b^2=b$ so the idempotent elements are $(1,0),(0,1),(1,1),(0,0)$
I claim $\mathbb{Q}×\{0\}$ and $\{0\}×\mathbb{Q}$ are the only proper non trivial ideals .
Proof: If not let $J$ be such an ideal other than the ideals mentioned above. Then there is at least one $(a,b)\neq (0,0) \in J $ but then it is a unit , so $(1,1)\in J$ and $J= \mathbb{Q}×\mathbb{Q}$ .Thus the claim is proved.
So there are no infinitely many prime ideals.
Again $(a,0)(0,b)=(0,0) $ where $a\ne 0, b\neq 0$ implies that $R$ is not an Integral domain, so it's not a UFD.
Have I missed something or done it wrong? Please go through it and suggest improvements or better ideas.
Thanks for your valuable time.
(1.) Unfortunately, it is not true that every nonzero element of $\mathbb Q \times \mathbb Q$ is a unit. (Can you find the inverse of $(1, 0),$ for instance?) Be careful about what your identity element is. In a unital ring $R \times S$ with pointwise multiplication, the identity element is always $(1_R, 1_S),$ where $1_R$ and $1_S$ are the respective identity elements of $R$ and $S.$ Ultimately, though, it is true that there are infinitely many units in $\mathbb Q \times \mathbb Q.$ (Use the aforementioned hint about the identity element.)
(2.) Correct.
(3.) Correct intuition; incorrect proof. One proof of your correct observation is the following.
Proof. Consider a prime ideal $P$ of $R = \mathbb Q \times \mathbb Q.$ Given any two elements $(a, b)$ and $(c, d)$ of $R$ such that $(a, b)(c, d)$ is in $P,$ we must have (by definition) that $(a, b) \in P$ or $(c, d) \in P.$ Considering that $(0, 0) = (a, 0)(0, a) \in P$ for all nonzero elements $a \in \mathbb Q,$ it follows that $(a, 0) \in P$ or $(0, a) \in P$ for all nonzero $a \in \mathbb Q.$ We will assume first that $(a, 0) \in P.$ We have therefore that $(1/a, 0) (a, 0) = (1, 0)$ is in $P$ so that $P = \mathbb Q \times \{0\}.$ On the other hand, if we have that $(0, a) \in P,$ then the analogous argument will show that $P = \{0\} \times \mathbb Q.$ QED.
(d.) Correct.