Radical of the Extension of Ideals and The Extension of an Radical Ideals

Question

First, here is the book that I would like to refer to this question. Any notation here directly from the context of this book.

The problem I have is doing Question 14e in pg. 198. Please take a look at the question as well as Question 13 for any of the following notations.

I managed to prove the first containment, however, I don't find any way to get the opposite containment with the condition $K\subseteq I$ and $\alpha _A$ being onto

Here is what I did: suppose $h\in \sqrt{\langle \alpha _A(I) \rangle}$, so that $h^m\in \langle \alpha _A(I) \rangle$ for some integer $m>0$

With the help of Hilbert Basis Theorem, I write $\sqrt{\langle \alpha _A(I) \rangle}=\langle h_1,\cdots ,h_s\rangle$ (with each $h_i$ not necessary in $\langle \alpha _A(I) \rangle$, obviously)

Then let $h=p_1h_1+\cdots +p_sh_s$ be some representation of $h$. By definition of radical ideal, some powers of $h_1,\cdots ,h_s$ must in $\langle \alpha _A(I) \rangle$. WLOG, we can assume $m$ can do this (for $m$ sufficiently large)

It is obvious that using the multinomial theorem, we have $h^M=p_1'h_1^m+\cdots +p_s'h_s^m$ for another large $M>m$, where $p_i'$ is just some other polynomials (not derivative)

Then my attempt is stuck here, so instead of going forward, I thought about what I have to prove 'backwardly'

To prove the containment, I have to show that $h\in \langle \alpha _A(\sqrt{I})\rangle$, so that is, finding some expression looks like $h=q_1f_1+\cdots +q_tf_t$ with each $f_i\in \langle \alpha _A(\sqrt{I}) \rangle$

I also know that I have to somehow use the condition $K\subseteq I$ and $\alpha _A$ is surjective, but so far I still don't see any clue using the first one. Meanwhile, the second one can be used almost everywhere (e.g. since $\alpha _A$ is onto, for each $i$, there exist $H_i$ such that $h_i=\alpha _A(H_i)$)

I appreciate any help and hint and answer, thanks

Answer 1

Please double check this answer, I am a noob, lol. So the point you are probably missing is a ring homomorphism $\alpha$ preserves an Ideal $I$ and its 'type', if it is onto and $\ker \alpha\subset I$:

Actually, the only thing we need from the following lemma is that under above conditions, (radical) ideals map to (radical) ideals, respectively. This can also be proven step-by-step easily, but I prefer to give this powerful Lemma aswell. [I hope it's true lol.]

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Lemma. Let $\phi:R\rightarrow S$ be a surjective homomorphism of rings. If $I$ is an ideal in $R$, then

1) $\phi(I)$ is an ideal in $S$
2) $J:=I+\ker\phi$ is an ideal in $R$
3) $R/J\cong S/\phi(I)$
4) If $\ker(\phi)\subset I$ then $R/I\cong S/\phi(I)$ $\quad\quad\quad$ (where $\cong$ denotes an isomorphism of rings.)

Proof: 1) One checks that the ideal properties pull through the surjective homomorphism $\phi$.

2) One easily confirms that the sum of ideals is an ideal. Note that for the special case $\phi=\alpha_A$ it is shown in 13c that $\ker\phi$ is an ideal. $\small{\textit{(In fact, }\ker\phi\textit{ is an ideal as it's the hom. preimage of the zero ideal.)}}$

3) [Copied from link above.] From 1) & 2), both sides are well defined. Claim: $\phi^{-1}(\phi(I))=I+\ker\phi.\:$ Proof of Claim:
'⊇': $\phi(I+\ker\phi)=\phi(I)+0=\phi(I).$
'⊆': Let $x\in R$ such that $\exists y \in I$: $\phi(x)=\phi(y) \Rightarrow x-y\in\ker\phi \Rightarrow x\in y+\ker\phi \subset I+\ker\phi.$

By applying the first isomorphism theorem to the composition: $R\twoheadrightarrow S\twoheadrightarrow S/\phi(I)$ we get: $$R/(I+\ker\phi)=R/\phi^{-1}(\phi(I))\cong S/\phi(I)$$

4) $\ker(\phi)\subset I\Rightarrow I+\ker\phi=I$ $$\tag*{$\square$}$$

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In your exercise 13b, it is shown that $\alpha_A$ is a ring homomorphism. (actually it is also a homomorphism of $k$-algebras since it fixes $k$.) Now given the onto & $\ker$ conditions, by 1) $\alpha_A$ preserves ideals and by 4) $\alpha_A$ maps radicals to radicals, since an ideal is radical if and only if it's quotient ring is reduced.

Thus, under the assumption that $\alpha_A$ is surjective and $\ker (\alpha_A)\subset I, $ we conclude that

$$\left\langle \alpha _A(I) \right\rangle=\alpha _A(I), \:\text{ analog: }\:\left\langle \alpha _A(\sqrt{I}) \right\rangle=\alpha _A(\sqrt{I}) \:\text{is radical, i.e.}\: \sqrt{ \alpha _A(\sqrt{I})}=\alpha _A(\sqrt{I}).$$

Finally, we have all we need to proof $\left\langle\alpha_{A}(\sqrt{I})\right\rangle \supseteq \sqrt{\left\langle\alpha_{A}(I)\right\rangle}:$

$$\sqrt{\left\langle\alpha_{A}(I)\right\rangle}=\sqrt{\alpha_{A}(I)}\subseteq \sqrt{ \alpha _A(\sqrt{I})}=\alpha _A(\sqrt{I})=\left\langle\alpha_{A}(\sqrt{I})\right\rangle$$