Let $f: R \rightarrow R[X]$ be the natural map, and let $I$ be an ideal of $R$. Show that $I \in {\rm Spec}(R) \iff I^{e} \in {\rm Spec}(R[X])$.

Question

This is exercise in Commutative Algebra:

Let $R$ be a commutative ring and let $X$ be an indeterminate; use the extension and contraction notation of 2.41 in conjunction with the natural ring homomorphism $f: R \rightarrow R[X]$ , and let $I$ be an ideal of $R$. Show that $I \in \operatorname{Spec}(R) \Leftrightarrow I^{e} \in \operatorname{Spec}(R[X])$.

I proof the following:

Let a homomorphism ring $$ \begin{aligned} \psi: & R[X] & \longrightarrow &(R / I)[X] \\ & \sum_{i=0}^{n} r_{i} X^{i} & \longmapsto & \sum_{i=0}^{n} \bar{r}_{i} X^{i} \end{aligned} $$ And $$ \operatorname{ker} \psi=\left\{\sum_{i=0}^{n} r_{i} X^{i}: r_{i} \in I, \forall i=0, \ldots, n\right\}=I[X]=I R[X]=f(I) R[X]=I^{e} $$ Use isomorphism theorem, we have $$ (R / I)[X] \cong R[X] / I[X]=R[X] / I^{e} $$ Thus $I \in \operatorname{Spec}(R) \Leftrightarrow R / I$ is an integral domain $\Leftrightarrow(R / I)[X]$ is an integral domain $\Leftrightarrow R[X] / I^{e}$ is an integral domain $ \Leftrightarrow I^{e} \in \operatorname{Spec}(R[X]) $

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I have a lot of problem:

i) Why $\operatorname{ker} \psi=I[X]$, $I[X]=I R[X]$, $I R[X]=f(I) R[X]$

ii) And why $(R / I)[X] \cong R[X] / I[X]$

iii) $R / I$ is an integral domain $\Leftrightarrow(R / I)[X]$ is an integral domain

Thank you very much.

Answer 1

For the first question:

The map $f:R\rightarrow R[X]$ is the inclusion, meaning $a\in R$ is mapped to $a\in R[X]$. Therefore, $I=f(I)$. Now, since $I\subseteq R$, the ideal $I$ is the same as the ideal $IR$. First of all since $I$ is closed under multiplication of elements from $R$ and addition, $IR\subseteq I$. Now let $i\in I$. Since $1\in R, 1i=i\in IR$, and so $I=IR$. Therefore $I[X]=IR[X]=f(I)R[X]$.

For the second question:

We take the following map $\varphi:R[X]\rightarrow (R/I)[X]$ given by $\varphi(\sum_{i=1}^na_ix^i)=\sum_{i=1}^n\bar{a_i}x^i$. Obviosly it's surjective. Remember that a polynomial is zero iff all coefficients are zero, so:

$\operatorname{Ker}(\varphi)=\{f(x)\in R[X]:\varphi(f)=\bar{0}\}=\{\sum_{i=1}^na_ix^i:\bar{a_i}=\bar{0}\}={\sum_{i=1}^na_ix^i:\forall i, a_i\in I\}=I[X]}$.

So we have a surjective map $\varphi:R[X]\rightarrow (R/I)[X]$ with kernel $I[X]$, so the first isomorphism theorem gives us $R[X]/I[X]\simeq (R/I)[X]$ - the idea is that we have a surjective map, and we took all the polynomials that are mapped to zero, and by taking the quotient you now say "all those polynomials are equal to zero in the new ring" which now makes your original map also injective and therefore isomorphism of rings.