Context: Let $R$ a commutative ring, and $\mathfrak{a}$, $\mathfrak{b}$, $\mathfrak{b^{'}}$ are ideals.
Problem statement: Let $\mathfrak{a}$ be comaximal to both $\mathfrak{b}$ and $\mathfrak{b^{'}}$. Prove that $\mathfrak{a}$ is also comaximal to $\mathfrak{b}\mathfrak{b^{'}}$.
My solution
First, $\mathfrak{b}\mathfrak{b^{'}} = \mathfrak{b} \cap \mathfrak{b^{'}}$. We can see this because all sums of the form $\sum a_i b_i$ with $a \in \mathfrak{b}, b \in \mathfrak{b^{'}}$ must be in both ideals (ideals by definition absorb multiplication and each term in this sum is the product of an element from each ideal, plus of course ideals are closed under summation). We'll need this in a second.
So, if $\mathfrak{a}$, $\mathfrak{b}$ comaximal, then for all $r \in R$ and $b \in \mathfrak{b}$, there exists some $a \in \mathfrak{a}$ such that $a + b = r$.
Same thing applies for $\mathfrak{a}$ and $\mathfrak{b^{'}}$.
So consider some $b$ in $\mathfrak{b}\mathfrak{b^{'}}$. By our statement above, this element must be in both $\mathfrak{b}$ and $\mathfrak{b^{'}}$. By the above, for any $r \in R$, there then must exist some $a \in \mathfrak{a}$ such that $a + b = r$. So $\mathfrak{b}$ and $\mathfrak{b^{'}}$ are both comaximal to $\mathfrak{a}$. ◻
The solution in my text book uses a different argument and I was just curious if mine was correct as well?
Thanks in advance!
PS - not a student, long-graduated hobbyist mathematician working through a commutative algebra text during quarantine
Hint for a shorter proof, if you're assuming that your ring has $1$: if $a+b=1$ and $a+b'=1$, consider $(a+b)(a+b') = a(a+b+b')+bb'$.