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Consider the following question from my ring theory assignment:

Let $A$ be a commutative ring, $\mathfrak a_1,...,\mathfrak a_m $ be pairwise comaximal ideals in $A$ and $\mathfrak a=\mathfrak a_1\cdots\mathfrak a_m$. If $f\in A[X]$ and if $V_{A/\mathfrak a_i}(f)\neq\emptyset$ for every $i=1,...,m$, then show that $V_{A/\mathfrak a}(f) \neq \emptyset$. (Set $V_A(f)$ means the solutions set of $f=0$ in $A$.)

So, for all $i=1,..., m$, $V_{A/\mathfrak a_i}(f)\neq \emptyset$ means that there exists an element $x_i$ in $A/\mathfrak a_i$ such that $f(x_i) =0$. Now, I have to construct an element $y$ in $A/\mathfrak a$ such that $f(y)=0$.

$\mathfrak a_1\cdots\mathfrak a_n =\mathfrak a_i$ for any $i =1,...,m$ as $\mathfrak a_i$'s are ideals. So, $x_1$ is such an element.

Is my proof fine?

user26857
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  • Hint: use the ring-theoretic version CRT - just like in the classical case in $\Bbb Z$ in the Remark here – Bill Dubuque Dec 05 '21 at 06:57
  • I have no idea how you arrived at the conclusion that $a_1...a_n =a_i$, which is definitely wrong. – user26857 Jan 21 '22 at 19:21
  • @user26857 As all of $a_1,...,a_m$ are ideals so , using definition of ideals $aa_i \in a_i$ for all $a\in A$ and using it on $a_1 ...a_m$ along with commutativity I get $a_1 ... a_n =a_i$ for all i . Let me know your thoughts on this. –  Jan 22 '22 at 07:22
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    Actually you get $a_1\cdots a_m\subseteq a_i$ for all $i$. For instance, in $\mathbb Z$ we have $(2)(3)\subsetneq(2)$. – user26857 Jan 22 '22 at 07:24

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We know that there is $x_i\in A$ such that $f(x_i)\in\mathfrak a_i$. By CRT there is $x\in A$ such that $x-x_i\in\mathfrak a_i$ for all $i$. Write $x=x_i+a_i$ with $a_i\in\mathfrak a_i$. Now notice that $f(x)=f(x_i)+a_ir_i$ with $r_i\in A$, and therefore $f(x)\in\mathfrak a_i$ for all $i$. Since $\bigcap\mathfrak a_i=\prod\mathfrak a_i$ we are done.

user26857
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  • How did you deduced that there is $x_i \in A$ such that $f(x_i) \in a_i$ . You must have used $f(x)\in A[x]$ and $V_{A/a_i}(f)\neq \phi$ but I think that means that there exists an element $a /a_i$ in $A/a_i$ such that $f(a/a_i)=0$. Kindly explain how you wrote the 1st line of your answer. –  Jan 22 '22 at 07:38
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    $f(a/a_i)=0$ where? Answer: in $A/a_i$. This means that $f(a)\in a_i$, right? – user26857 Jan 22 '22 at 07:56
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    Btw, $f(a/a_i)=f(a)/a_i$. – user26857 Jan 22 '22 at 07:57
  • How does CRT implies exiestence of $x\in A$ in line 1 ? Do you mind showing the step? I am not able to understand how you used CRT here. –  Jan 22 '22 at 08:09
  • CRT says the following: the system of congruences $x\equiv x_i\bmod\mathfrak a_i$, $i=1,\dots,m$, has a solution. – user26857 Jan 22 '22 at 08:11
  • How did you deduced that $\cap a_i =\prod a_i$? –  Jan 22 '22 at 11:19
  • This is a well known property of comaximal ideals. – user26857 Jan 22 '22 at 11:51
  • Link to a proof: https://math.stackexchange.com/questions/4356916. – user26857 Jan 22 '22 at 14:58