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Let $A$ and $B$ be Noetherian rings and $f: A \rightarrow C$ and $g: B \rightarrow C$ ring homomorphisms. If both $f$ and $g$ are surjective show $\{(a,b) \in A \times B: f(a)=g(b)\}$ is a Noetherian ring.

Here's a hint: show first (I already did this) that if $I_{1},..I_{n}$ are ideals of a ring $A$ such that $I_{1} \cap ... \cap I_{n} = \{0\}$ if each $A/I_{i}$ is Noetherian then so is $A$.

How to apply the above result?

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Hint2: let's call the fibre $D$, the maps from $D$ to $A$ denote $\alpha$(resp. to B $\beta$), show that $\alpha$ and $\beta$ are both surjective.

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    $\alpha$ maps $(a,b)$ to $a$.2011-07-16