Extension of a radical equals radical of the extension

Question

Let $R$ be a commutative ring with identity and let $f\colon R \to R[X]$ be the natural ring homomorphism. Let $I$ be an ideal of $R$. What I have shown is that $I$ is a prime ideal iff $I^e$ is a prime ideal. Now how can I prove $\sqrt{I^e}=\sqrt{I}^e$? Thanks in advance for your answer.

Answer 1

Fact 1: Consider the ring epimomoephis $\psi:R[x]\rightarrow (R/I)[x]$ by $\psi(a_0+a_1x+...+a_nx^n)=(a_0+I)+(a_1+I)x+...+(a_n+I)x^n$. Clearly, $ker(\psi)=I^e=I[x]$. Thus, $R[x]/I[x]\cong (R/I)[x]$.

Fact 2: Let $S$ be a ring. Then $S$ is an integral domain if and only if $S[x]$ is.

Form 1 and 2, we have:

$I$ is a prime ideal of $R$ if and only if $R/I $ is an integral domain if and only if $(R/I)[x]$ is an integral domain if and only if $R[x]/I[x]$ is an integral domain if and only if $I^e=I[x]$ is a prime ideal of $R[x]$.

For the last statment use the fact that $Nil(S[x])=Nil(S)[x]=Nil(S)^e$, where $Nil(A)$ is the set of all nilpotent element of a ring $A$( nilradical of $A$). Thus we have

$f\in \sqrt{I^e}$ if and only if $f^n\in I^e$ for some $n\in \mathbb{N}$ if and only if $f+I^e\in Nil(R[x]/I^e)$ if and only if $\psi(f)\in Nil(R[x]/I[x])$ if and only if $\psi(f)\in Nil(R/I)[x]$ if and only if ( the coefficients of $f$ are in $Nil(R/I)$) if and only if ( the coefficients of $f$ are in $\sqrt{I}$) if and only if $f\in \sqrt{I}^e.$

Answer 2

I guess that you mean that $f$ is the inclusion. My approach would be the following:

First, show that $I^e = I[X]$, the ideal of polynomials in $R[X]$ with coefficients in $I$.

Then, use what you proved about the primes and the fact that $\sqrt{I}$ is the intersection of all prime ideals containing $I$.