Question about a proof: $U$ maximal among non-finitely generated ideals of $R$, then $U$ is a prime ideal.

Solution 1:

To answer your questions:

1. If $I$ and $J$ are ideals, then so is $I+J= \{x+y\mid x\in I, y\in J\}$. Indeed, it is nonempty since $I$ and $J$ are nonempty; if $x_1,x_2\in I$ and $y_1,y_2\in J$, then $(x_1+y_1)-(x_2+y_2) = (x_1-x_2) + (y_1-y_2)\in I+J$, so $I+J$ is an additive subgroup. And given $x\in I$, $y\in J$, and $r\in R$, we have $r(x+y) = (rx)+(ry) \in I+J$. This also works in a noncommutative ring (and for one sided ideals), as you also have $(x+y)r = (xr) + (yr)\in I+J$. For the case at hand, $U$ is an ideal and $(a)$ is an ideal, so $U+(a)$ is necessarily an ideal.
   
   In fact, $I+J$ is the smallest ideal that contains both $I$ and $J$.

2. This is, I suspect, a typo. Suppose that $ab\in U$. If $a\in U$, we are done. If $a\notin U$, then we look at $J$. If $J=R$, then $1\in J= U+(a)$, so there exists $u\in U$, $r\in R$ such that $1=u+ra$. Multiplying by $b$ we get $b = ub + r(ab)$. Since $u,ab\in U$, then $ub+r(ab)\in U$, and so we get that $b\in U$, and we are done. Thus, we are reduced to the case where $J\neq R$, $a\notin U$.

In this case, $J$ is a proper ideal that properly contains $U$. Now, we need to show that $J$ is non-finitely generated to reach a contradiction. I think this proof is a little too cavalier in the assertion that this is clear, though. It is clear that if $J$ is finitely generated, then you can find a finite set of elements $u_1,\ldots,u_n\in U$ such that $u_1,\ldots,u_n,a$ together generate $J$, but it is not clear to me that this implies that $U$ would then be finitely generated...

Solution 2:

There is a typo, that is $J$ is proper because $b\notin J$ (not $a$). However the “proof” is completely wrong.

Indeed, it follows from maximality of $U$ that $J=U+(a)$ is finitely generated, because it properly contains $U$. One might prove that it is not finitely generated in order to find a contradiction, but the given argument fails to address this.