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Ideal proof

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Let R be a commutative ring with 1. Prove that the principal ideal generated by x in the polynomial ring R[x] is a prime ideal if and only if R is an integral domain. Prove that (x) is a maximal ideal if and only f R is a field.

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Q1: (x) is a prime ideal if and only if R is an integral domain.
Since R is a commutative ring with 1, we only need to show that R does not have non-trivial zero divisor.
"=>": Suppose (x) is a prime ideal in R[x]. If R is not an integral domain, we can find two non-zero element r and s in R, such that rs = 0. Now r and s are also elements in R[x] with degree 0, then rs = 0 is in (x). So r is in (x) or s is in (x). Then we must have r = 0 or s = 0. We get a contradiction. Therefore, R must be an integral domain.
"<=": Suppose R is an integral domain. We consider f(x), g(x) in R[x] and f(x)g(x) is in (x). We can rewrite f(x), g(x) as
f(x) = xp(x) + r and g(x) = xq(x) + s, where r, s are constant ...

Solution Summary

This provides an example of proving a principal ideal is a prime ideal.

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