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

In summary: Well,the fact that $\alpha^{k-n}\in R$ follows from the fact that $\alpha^{k}=\underbrace{\alpha^{k-n}}_{\in R}$. Well,the fact that $\alpha^{k-n}\in R$ follows from the fact that $\alpha^{k}=\underbrace{\alpha^{k-n}}_{\in R}$.
  • #1
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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
 
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  • #2
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.
 
  • #3
Fallen Angel said:
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
 
  • #4
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 $\alpha^{k}\in I$
 
  • #5
Fallen Angel said:
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 \(\displaystyle \alpha^{k-n}\) actually belongs to \(\displaystyle R\) ...

Can you help further?

Peter
 
  • #6
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 $\alpha^{0}=1$).
 
  • #7
Fallen Angel said:
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 \(\displaystyle \alpha \in I \) ... then how do we know \(\displaystyle \alpha \in R\)? ... but that was really silly since \(\displaystyle I \subseteq R\) so that \(\displaystyle \alpha \in I \rightarrow \alpha \in R\) ...

Thanks for your assistance ... was very helpful!

Peter
 

FAQ: Radicals and Radical Ideals, Dummit and Foote Proposition 11, Section 15.2

What are radicals and radical ideals?

Radicals are elements in a ring that cannot be expressed as a product of other elements in the ring. Radical ideals are ideals in a ring that contain all the radicals in the ring.

What is Dummit and Foote Proposition 11?

Dummit and Foote Proposition 11 is a theorem in abstract algebra that states that if R is a commutative ring with unity and I is an ideal in R, then the radical of I (denoted as rad(I)) is equal to the intersection of all prime ideals containing I.

Why is radical ideal important?

Radical ideals are important because they give us a way to generalize the notion of prime ideals and irreducible elements in a ring. They also help us understand the structure of a ring by identifying the elements that cannot be factored into smaller pieces.

How can we use radical ideals to simplify computations?

Radical ideals can be used to simplify computations by reducing the number of elements we need to consider. Instead of working with all the elements in a ring, we can focus on the radicals and their corresponding radical ideals, which may have simpler properties.

Can you give an example of a radical ideal?

One example of a radical ideal is the zero ideal in a ring R. This is because every element in the zero ideal is a radical element, and the intersection of all prime ideals containing the zero ideal is also the zero ideal. Therefore, the radical of the zero ideal is equal to the zero ideal, rad(0) = 0.

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