I know that to show a ring is isomorphic to another ring, I have to find a bijective ring homomorphism between the two rings. Or I could use the F.H.T. but I would also need a function to make that happen (If I'm thinking about it correctly). Do I need to just make up my own function and prove that it is a bijective ring homomorphism? If so, where do I start? Thanks!
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Try to find a surjective map $p: R[x] \to R$ such that $\ker p = (x)$. Then you can use homomorphism theorem, which states that for any map $T : A\to B$ between commutative rings $A/\ker T \cong \operatorname{Im} T$. – Boris Bilich May 15 '19 at 16:17
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Where to start: look for your question first. – rschwieb May 15 '19 at 17:14
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Define $f: R[x] \to R$ by $f(p(x))=p(0)$. Prove this is a surjective homomorphism with kernel $(x)$ and use first isomorphism theorem
Consequence: If in addtion $R[x]$ is a PID, then the ideal $(x)$ is maximal and so $R$ is a field
Chinnapparaj R
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