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 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 $$\displaystyle\in{R}$$. 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.