Ring Homomorphisms and Radical Ideals/ Solution

a) Prove that f(I–√)⊂f(I)−−−−√$f(\sqrt{I})\subset \sqrt{f(I)}$.

Let x∈f(I–√)$x\in f(\sqrt{I})$ be an arbitrary element. Then there is a∈I–√$a\in \sqrt{I}$ such that f(a)=x$f(a)=x$. As a∈I–√$a\in \sqrt{I}$, there exists a positive integer n$n$ such that an∈I$a^n\in I$.

It follows that we have

xn=f(a)n=f(an)∈f(I).$$\begin{array}{r}{x}^n=f(a{)}^n=f(a^n)\in f(I).\end{array}$$

This implies that x∈f(I)−−−−√$x\in \sqrt{f(I)}$.
Hence we have f(I–√)⊂f(I)−−−−√$f(\sqrt{I})\subset \sqrt{f(I)}$.

Prove that f−1(I′)−−−−−−√=f−1(I′−−√)$\sqrt{f^{-1}(I^{\prime })}=f^{-1}(\sqrt{I^{\prime }})$

(⊂)$(\subset )$ Let x∈f−1(I′)−−−−−−√$x\in \sqrt{f^{-1}(I^{\prime })}$. Then there is a positive integer n$n$ such that xn∈f−1(I′)$x^n\in f^{-1}(I^{\prime })$ and thus f(xn)∈I′$f(x^n)\in I^{\prime }$.
As f$f$ is a ring homomorphism, it follows that f(x)n=f(xn)∈I′$f(x{)}^n=f(x^n)\in I^{\prime }$.

Hence f(x)∈I′−−√$f(x)\in \sqrt{I^{\prime }}$, and then x∈f−1(I′−−√)$x\in f^{-1}(\sqrt{I^{\prime }})$.
This proves that f−1(I′)−−−−−−√⊂f−1(I′−−√)$\sqrt{f^{-1}(I^{\prime })}\subset f^{-1}(\sqrt{I^{\prime }})$.

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(⊃)$(\supset )$ Let x∈f−1(I′−−√)$x\in f^{-1}(\sqrt{I^{\prime }})$. Then f(x)∈I′−−√$f(x)\in \sqrt{I^{\prime }}$. It follows that there exists a positive integer n$n$ such that f(xn)=f(x)n∈I′$f(x^n)=f(x{)}^n\in I^{\prime }$.
Hence xn∈f−1(I′)$x^n\in f^{-1}(I^{\prime })$, and we deduce that x∈f−1(I′)−−−−−−√$x\in \sqrt{f^{-1}(I^{\prime })}$.

This proves that f−1(I′−−√)⊂f−1(I′)−−−−−−√$f^{-1}(\sqrt{I^{\prime }})\subset \sqrt{f^{-1}(I^{\prime })}$.
Combining this with the previous inclusion yields that f−1(I′)−−−−−−√=f−1(I′−−√)$\sqrt{f^{-1}(I^{\prime })}=f^{-1}(\sqrt{I^{\prime }})$