If any ideal in $R$ is principal, then any ideal in $R \times R$ is principal Let $R$ be a ring. We endow $R \times R$ with pointwise addition and pointwise multiplication. Then it's easy to verify that $R \times R$ is a ring under these operations. Then I conjecture that

If any ideal in $R$ is principal, then any ideal in $R \times R$ is principal.

Could you please verify if my conjecture and attempt to prove it are correct? Thank you so much for your help!


Lemma: Let $\sigma:R_1 \to R_2$ be a ring epimorphism and $I$ an ideal in $R_1$. Then $\sigma(I)$ is an ideal in $R_2$. Moreover, if $I$ is principal in $R_1$, then $\sigma(I)$ is principal in $R_2$.

Proof: It's straightforward to verify this lemma.
Let $I$ be an ideal in $R \times R$. Let $I_1,I_2$ be the sets of first and second coordinates of $I$ respectively. Consider the maps $\sigma_1: R \times R \to R, (r_1,r_2) \mapsto r_1$ and $\sigma_2: R \times R \to R, (r_1,r_2) \mapsto r_2$.
It's easy to verify that $\sigma_1$ and $\sigma_2$ are ring epimorphism. By Lemma, $\sigma_1(I_1)$ and $\sigma_2(I_2)$ are ideals in $R$. By hypothesis, $\sigma_1(I_1)$ and $\sigma_2(I_2)$ are principal. Then $\sigma_1(I_1) = \langle a_1 \rangle$ and $\sigma_2(I_2) = \langle a_2 \rangle$ for some $a_1,a_2 \in R$. It's straightforward to verify that $\langle (a_1,a_2) \rangle = I$.

Update: Throughout my question, $R$ is a commutative ring with unity. Here I add the proof of $\langle (a_1,a_2) \rangle = I$.

Then there are $(a_1,y), (x,a_2) \in I$ such that $\sigma_1(a_1,y) = a_1$ and $\sigma_2(x,a_2) = a_2$. It follows from $(1,0) \cdot(a_1,y) + (0,1) \cdot(x,a_2) = (a_1,a_2)$ that $(a_1,a_2) \in I$.
Next we prove $\langle (a_1, a_2) \rangle = I$. For $(a,b) \in I$, we have $a = xa_1$ and $b=ya_2$ for some $x,y \in R$. As such, $(a,b) = (x,y) \cdot (a_1,a_2)$. This completes the proof.
 A: More generally, the following is true. I want to emphasize that this is remarkable, because the corresponding claim about subgroups or normal subgroups of a product of groups is not true.

Proposition: Let $R, S$ be two rings. Every left ideal $K$ of the product ring $R \times S$ is of the form $I \times J$ where $I$ is a left ideal of $R$ and $J$ is a left ideal of $S$. Furthermore $K = (r, s)$ is principal if and only if $I = (r)$ and $J = (s)$ are.

Proof. Let $e = (1, 0)$. Every element of $R \times S$ has a canonical decomposition $(r, s) = e(r, s) + (1 - e)(r, s)$ (this is just a fancy way of saying $(r, s) = (r, 0) + (0, s)$). If $K$ is a left ideal of $R \times S$, then (since $K$ is by hypothesis closed under left multiplication) this decomposition applies to every element of $K$ and occurs entirely in $K$, so we have
$$K = eK + (1 - e) K.$$
Now $eK$ is a left ideal $I$ of $R$ and $(1 - e) K$ is a left ideal $J$ of $S$ and we have $K = I \times J$ as desired (the notation is a little confusing here because we could equally well say $K = I \oplus J$). And this argument gives straightforwardly that if $K$ is the principal ideal generated by some $(r, s)$ then $eK$ is the principal ideal generated by $(r, 0)$ and $(1 - e) K$ is the principal ideal generated by $(0, s)$. $\Box$
In the commutative case, what this says is that every quotient ring of a product $R \times S$ of commutative rings is a product of quotients $R/I \times S/J$, and again, I want to emphasize that this is remarkable because the corresponding claim about quotients of a product of groups is not true.
(Further comments, feel free to ignore: geometrically the upshot is that the affine scheme $\text{Spec } R \times S$ is a disjoint union $\text{Spec } R \sqcup \text{Spec } S$; the disjointness here is a stronger condition than just the condition that we have a coproduct, it really means that any affine scheme mapping in disconnects into an affine scheme mapping into $\text{Spec } R$ and an affine scheme mapping into $\text{Spec } S$. See extensive category for more about this.)
