 # 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. Brittney Lord 2020-11-22 Answered

Suppose that R and S are commutative rings with unites, Let $\varphi$ 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 ${\varphi }^{-1}\left(A\right)=\left\{x\in R\mid \varphi \left(x\right)\in A\right\}$ is prime $\in R$.
b. If A is maximal in S, show that ${\varphi }^{-1}\left(A\right)$ is maximal $\in R$.

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a. Suppose A i sprime ideal in S.
Consider $ab\in {\varphi }^{-1}\left(A\right)$. Then $\varphi \left(ab\right)=\varphi \left(a\right)\varphi \left(b\right)\in A.$
Since A is prime, either $\varphi \left(a\right)\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\varphi \left(b\right)$ belongs to A.
Hence either a or b belongs to ${\varphi }^{-1}\left(A\right)$
Thus, ${\varphi }^{-1}\left(A\right)$ is prime $\in R$
b. Consider the homomorphism phi: $R\to \frac{S}{A}$ defined by $\varphi \left(r\right)=\varphi \left(r\right)+A$
Then the kernel of $\varphi$ is ${\varphi }^{-1}\left(A\right).$
Then,$\frac{R}{ker\varphi }\approx \frac{S}{A}$
Therefore, ${\varphi }^{-1}\left(A\right)$ is maximal ideal.

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