Question

Suppose that R and S are commutative rings with unites, Let PSJphi be a ring homomorphism from R onto S and let A be an ideal of S. a. If A is prime in S, show that phi^-1(A)={x in R| phi(x) in A} is prime in R. b. If A is maximal in S, show that phi^-1(A) is maximal in R.

Commutative Algebra
ANSWERED
asked 2020-11-22
Suppose that R and S are commutative rings with unites, Let PSJphiZSK be a ring homomorphism from R onto S and let A be an ideal of S.
a. If A is prime in S, show that \(\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}={\left\lbrace{x}\in{R}{\mid}\phi{\left({x}\right)}\in{A}\right\rbrace}\) is prime \(\displaystyle\in{R}\).
b. If A is maximal in S, show that \(\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}\) is maximal \(\displaystyle\in{R}\).

Answers (1)

2020-11-23

a. Suppose A i sprime ideal in S.
Consider \(\displaystyle{a}{b}\in\phi^{{-{{1}}}}{\left({A}\right)}\). Then \(\displaystyle\phi{\left({a}{b}\right)}=\phi{\left({a}\right)}\phi{\left({b}\right)}\in{A}.\)
Since A is prime, either \(\displaystyle\phi{\left({a}\right)}{\quad\text{or}\quad}\phi{\left({b}\right)}\) belongs to A.
Hence either a or b belongs to \(\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}\)
Thus, \(\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}\) is prime \(\displaystyle\in{R}\)
b. Consider the homomorphism phi: \(\displaystyle{R}\to\frac{{S}}{{A}}\) defined by \(\displaystyle\phi{\left({r}\right)}=\phi{\left({r}\right)}+{A}\)
Then the kernel of \(\displaystyle\phi\) is \(\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}.\)
Then,\(\displaystyle\frac{{R}}{{{k}{e}{r}\phi}}\approx\frac{{S}}{{A}}\)
Therefore, \(\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}\) is maximal ideal.

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