Inverse Image of Ideal in R is an Ideal of S

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The discussion centers on proving that the inverse image of an ideal in a ring S, under a ring homomorphism f: R → S, is indeed an ideal in R. Participants emphasize the importance of adhering to forum guidelines by demonstrating an attempt at solving the problem before seeking help. The proof involves showing that the preimage of an ideal satisfies the necessary properties of an ideal in R. Engaging with the problem directly is encouraged to facilitate better assistance. Overall, the focus is on the mathematical proof and the necessity of following community standards for effective collaboration.
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Let f:R...s be a ring homomorphism .Prove that the inverse image of an ideal of S is an ideal of R






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You aren't getting help for your questions because you are aren't following the forum guidelines and showing us at least an attempt at doing the problem.
 
Show some attempt.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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