The fact I want to prove is given $A$ and $B$ are rings, then $$J(A\oplus B) = J(A)\oplus J(B).$$
I am not happy with my current proof, which I feel like I just patched things from different sources together and I think it barely works. I want to check if it's correct and I am looking for a (simpler) proof that perhaps using a similar kind of argument both ways.
My current proof:
$J(A)\oplus J(B)$ is a (right) quasi-regular ideal since both $J(A)$ and $J(B)$ are, there exists $x$ and $y$ such that $(1-a)x=1$ and $(1-b)y=1$ for all $a\in J(A)$, $b\in J(B)$, $(1,1)-(a,b)=(1-a,1-b)\times (x,y)=(1,1)$.
I know that the Jacobson radical contains all right quasi-regular ideals, so $J(A)\oplus J(B)\subset J(A\oplus B)$.
False Reasoning, correction in answers. Conversely, every maximal right ideal of $A\oplus B$ is of the form $M_A\oplus M_B$ where $M_A$ is a maximal ideal of A etc.
If $(a,b)\in J(A\oplus B)$, then $(a,b)\in M_A\oplus M_B$ for all maximal right ideals $M_A$, $M_B$ in $A$ and $B$ respectively. Therefore $(a,b)\in J(A)\oplus J(B)$, so $J(A\oplus B) \subset J(A)\oplus J(B)$.
Whoa whoa whoa. It doesn't work that way. If $M_A$ and $M_B$ are maximal right ideals of $A$ and $B$ respectively, $M_A\oplus M_B$ is obviously properly contained in $M_A\oplus B$.
But you are close. You can compute $J(A\oplus B)$ exactly via an intersection of maximal right ideals, if you spot them all.
Every maximal right ideal has the form $M_A\oplus B$ or $A\oplus M_B$ where $M_A$ is a maximal right ideal of $A$ and $M_B$ is a maximal right ideal of $B$. Intersecting the members of the former collection gets you $J(A)\oplus B$, intersecting the members of the latter collection gets you $A\oplus J(B)$, and intersecting all of them gets you $J(A)\oplus J(B)$.
If you haven't already done so, it would be good for you to prove that every right ideal of $A\oplus B$ is of the form $T_A\oplus T_B$ where $T_A$ is a right ideal of $A$ and $T_B$ is a right ideal of $B$. Then confirm via quotients that my claim above about the maximal right ideals captures all maximal right ideals.
Finally, you could work out that the same thing works for two-sided ideals, and then prove that the prime ideals of $A\oplus B$ are of the form $P_A\oplus B$ or $A\oplus P_B$ where $P_A$ is a prime ideal of $A$ and $P_B$ is a prime ideal of $B$.