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

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

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.]

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

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 $$\left\langle\alpha_{A}(\sqrt{I})\right\rangle \supseteq \sqrt{\left\langle\alpha_{A}(I)\right\rangle}:$$

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