MHB Radicals and Radical Ideals, Dummit and Foote Proposition 11, Section 15.2

[Math Amateur]

I am trying to gain an understanding of the basics of elementary algebraic geometry and am reading Dummit and Foote Chapter 15: Commutative Rings and Algebraic Geometry ...

At present I am focused on Section 15.2 Radicals and Affine Varieties ... ...

I need help on an apparently simple aspect of the proof of Proposition 11

Proposition 11 and its proof reads as follows:View attachment 4781Now in the above text the first task is to show that rad I is an ideal ... but ... D&F do not bother to do this ... presumably they think it is obvious ... but i have not been able to formulate a formal and rigorous proof of this ... can someone help by providing a rigorous and formal proof that rad I is an ideal ...

Help will be appreciated ...

Peter

 

[Fallen Angel]

Hi Peter,

I'm assuming your problem is proving that the sum of two elements in the radical is also in the radical, since it seems to be the harder part.

Let $I$ be an ideal of $R$ and $rad(I)=\{a\in R \ : \ \exists \ n\in \mathbb{N} \ \mbox{with} \ a^{n}\in I\}$ it's radical.

If $\alpha, \beta \in rad(I)$, then there exists $n,m\in \mathbb{N}$ s.t. $\alpha^{n},\beta^{m}\in I$. Now apply the binomial theorem to $(\alpha + \beta)^{n+m}$, and you will see that $(\alpha + \beta)\in rad(I)$. (remember that $\alpha^{n}\in I$ implies $\alpha^{k}\in I$ for any $k\geq n$).

Try to complete the proof and tell us if you encounter some other problem.

 

[Math Amateur]

Hi Peter,

I'm assuming your problem is proving that the sum of two elements in the radical is also in the radical, since it seems to be the harder part.

Let $I$ be an ideal of $R$ and $rad(I)=\{a\in R \ : \ \exists \ n\in \mathbb{N} \ \mbox{with} \ a^{n}\in I\}$ it's radical.

If $\alpha, \beta \in rad(I)$, then there exists $n,m\in \mathbb{N}$ s.t. $\alpha^{n},\beta^{m}\in I$. Now apply the binomial theorem to $(\alpha + \beta)^{n+m}$, and you will see that $(\alpha + \beta)\in rad(I)$. (remember that $\alpha^{n}\in I$ implies $\alpha^{k}\in I$ for any $k\geq n$).

Try to complete the proof and tell us if you encounter some other problem.

Thanks Fallen Angel, appreciate your help ...

Working on this now ... but as a preliminary issue ... can you demonstrate exactly why $\alpha^{n}\in I$ implies $\alpha^{k}\in I$ for any $k\geq n$)?

Hope you can help with this matter ...

Peter

 

[Fallen Angel]

Hi again,

$\alpha \in R$ and $I$ is an ideal, and we know $\alpha^{k}=\underbrace{\alpha^{k-n}}_{\in R} \ \underbrace{\alpha^{n}}_{\in I}$, so $\alpha^{k}\in I$

 

[Math Amateur]

Hi again,

$\alpha \in R$ and $I$ is an ideal, and we know $\alpha^{k}=\underbrace{\alpha^{k-n}}_{\in R} \ \underbrace{\alpha^{n}}_{\in I}$, so $\alpha^{k}\in I$

Thanks again for helping ...

I was thinking along the lines of your argument/demonstration ... BUT ... I could not see why or how we can be sure that $$\alpha^{k-n}$$ actually belongs to $$R$$ ...

Can you help further?

Peter

 

[Fallen Angel]

Well,

$\alpha \in R$ and $k\geq n$ so $\underbrace{\alpha \cdot \alpha \cdots \alpha}_{k-n \ times}\in R$ because $R$ is closed under multiplication (as always, I'm taking $\alpha^{0}=1$).

 

[Math Amateur]

Well,

$\alpha \in R$ and $k\geq n$ so $\underbrace{\alpha \cdot \alpha \cdots \alpha}_{k-n \ times}\in R$ because $R$ is closed under multiplication (as always, I'm taking $\alpha^{0}=1$).

Yes, thanks ... I was thinking that $$\alpha \in I $$ ... then how do we know $$\alpha \in R$$? ... but that was really silly since $$I \subseteq R$$ so that $$\alpha \in I \rightarrow \alpha \in R$$ ...

Thanks for your assistance ... was very helpful!

Peter