MHB Radicals of Ideal: Get Help Understanding D&F Prop. 11

[Math Amateur]

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

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

I need help with an aspect of the definition of a radical of a quotient ideal in order to proceed with understanding the proof of Proposition 11, D&F page 673, Section 15.2.

Proposition 11 plus preliminary definitions of radical, nilradical and radical ideal respectively read as follows:View attachment 4803I am trying to understand the proof that $$( \text{rad } I ) / I$$ is the nilradical of $$R/I$$ ... but first, I need to get the definition of rad $$R/I$$ clear, and also get the definition of the nilradical of $$R/I$$ clear ...Based on the definition of the text above from D&F, I assume the definition of rad $$R/I$$ is as follows:

$$ \text{rad } R/I = \{ a \in R \ | \ (a + I)^k \in R/I \text{ for some } k \ge 1 \}$$ ... ... ... ... (1)and the definition of the nilradical of R/I is as follows:

nilradical of $$ R/I = \{ a \in R \ | \ (a+I)^k = 0 + I \text{ for some } k \ge 1 \}$$ ... ... ... ... (2)Could someone please confirm that the above definitions are correct ... or point out where my definitions are in error ...

Help would be appreciated ...NOTE: I am concerned that my definitions above may be incorrect since D&F write:

"... ... By definition, the nilradical of $$R/I$$ consists of the elements in the quotient some power of which is 0 ... ... "

and this seems to imply that the elements of the nilradical of $$R/I$$ are elements in the quotient $$R/I$$ ... and not elements of $$R$$ as I have assumed in my point (2) above ...

 

[caffeinemachine]

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

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

I need help with an aspect of the definition of a radical of a quotient ideal in order to proceed with understanding the proof of Proposition 11, D&F page 673, Section 15.2.

Proposition 11 plus preliminary definitions of radical, nilradical and radical ideal respectively read as follows:I am trying to understand the proof that $$( \text{rad } I ) / I$$ is the nilradical of $$R/I$$ ... but first, I need to get the definition of rad $$R/I$$ clear, and also get the definition of the nilradical of $$R/I$$ clear ...Based on the definition of the text above from D&F, I assume the definition of rad $$R/I$$ is as follows:

$$ \text{rad } R/I = \{ a \in R \ | \ (a + I)^k \in R/I \text{ for some } k \ge 1 \}$$ ... ... ... ... (1)and the definition of the nilradical of R/I is as follows:

nilradical of $$ R/I = \{ a \in R \ | \ (a+I)^k = 0 + I \text{ for some } k \ge 1 \}$$ ... ... ... ... (2)Could someone please confirm that the above definitions are correct ... or point out where my definitions are in error ...

Help would be appreciated ...NOTE: I am concerned that my definitions above may be incorrect since D&F write:

"... ... By definition, the nilradical of $$R/I$$ consists of the elements in the quotient some power of which is 0 ... ... "

and this seems to imply that the elements of the nilradical of $$R/I$$ are elements in the quotient $$R/I$$ ... and not elements of $$R$$ as I have assumed in my point (2) above ...

(All rings are commutative with identity).

I am not sure why you are considering $\text{rad}(R/I)$. It would be $R/I$ itself. This is because radical of any ring is the ring itself (a trivial fact).

Further, your definition of the radical was wrong.

You have written:

$$ \text{rad } R/I = \{ a \in R \ | \ (a + I)^k \in R/I \text{ for some } k \ge 1 \}$$ ... ... ... ... (1)

suggesting that $\text{rad}(R/I)$ is not a subset of $R/I$.
The correct definition would be

$$ \text{rad } R/I = \{ a+I \in R/I \ | \ (a + I)^k \in R/I \text{ for some } k \ge 1 \}$$ ... ... ... ... (1)

which trivially gives you $\text{rad}(R/I)=R/I$.

You have made a similar error in defining the nilradical of $R/I$. The nilradical of any ring is the set of all the nilpotents in the ring. So in particular it should be a subset of the ring. Can you try an fix the error in your definition?

 

[Math Amateur]

(All rings are commutative with identity).

I am not sure why you are considering $\text{rad}(R/I)$. It would be $R/I$ itself. This is because radical of any ring is the ring itself (a trivial fact).

Further, your definition of the radical was wrong.

You have written:

$$ \text{rad } R/I = \{ a \in R \ | \ (a + I)^k \in R/I \text{ for some } k \ge 1 \}$$ ... ... ... ... (1)

suggesting that $\text{rad}(R/I)$ is not a subset of $R/I$.
The correct definition would be

$$ \text{rad } R/I = \{ a+I \in R/I \ | \ (a + I)^k \in R/I \text{ for some } k \ge 1 \}$$ ... ... ... ... (1)

which trivially gives you $\text{rad}(R/I)=R/I$.

You have made a similar error in defining the nilradical of $R/I$. The nilradical of any ring is the set of all the nilpotents in the ring. So in particular it should be a subset of the ring. Can you try an fix the error in your definition?

Hi caffeinemachine,

Sorry to be slow in responding ... been traveling in regional Victoria ... back to Tasmania on Saturday ...

Thanks for your help ...

... given your guidance I now regard the nilradical of $$R/I$$ is the radical of the trivial ideal $$(0 +I)$$ in $$R/I$$ ... ...

So ... nilradical of $$(0 + I)$$ in $$R/I$$ $$= \{ (a + I) \in R/I \ | \ (a +I)^k \in (0 + I) \text{ for some } k \ge 1 \}
$$
... is that correct ...?
Will now (when I get a moment) move on to the proof of Proposition 11 ...

Thanks again for your help ...

Peter

 

[caffeinemachine]

Hi caffeinemachine,

Sorry to be slow in responding ... been traveling in regional Victoria ... back to Tasmania on Saturday ...

Thanks for your help ...

... given your guidance I now regard the nilradical of $$R/I$$ is the radical of the trivial ideal $$(0 +I)$$ in $$R/I$$ ... ...

So ... nilradical of $$(0 + I)$$ in $$R/I$$ $$= \{ (a + I) \in R/I \ | \ (a +I)^k \in (0 + I) \text{ for some } k \ge 1 \}
$$
Peter

It is correct. You made a typo in the last statement saying the nilradical of $(0+I)$. What you mean is the radical of $(0+I)$.

 

[Deveno]

I thought I would just give a simple example (it's not a proof).

Let's use a ring we know and understand well, $R = \Bbb Z$.

For $I$ we will use the ideal $I = (8) = 8\Bbb Z$.

Now $\text{rad }I = \{k \in \Bbb Z: k^n \in I\}$.

Let's think about this a moment: if $k^n = 8t$, it is clear that $k$ is even (because $2$ is prime).

On the other hand, if $k = 2s$, then $k^3 = 8s^3 \in I$.

Hopefully, this convinces you that $\text{rad }I = 2\Bbb Z$.

Now $\text{rad }I/I$ is just $2\Bbb Z/8\Bbb Z$, which is the following:

$\{8\Bbb Z, 2+8\Bbb Z, 4+8\Bbb Z, 6+8\Bbb Z\}$.

We turn our attention now to $R/I$, which is just the integers modulo $8$, $\Bbb Z_8$.

What is its nilradical? Well, let's just examine the powers of *everything* in $\Bbb Z_8$.

$[0]^1 = [0]$, so there is one element in the nilradical.

$[1]^k = [1]$, for any power $k$, so this is a no-go.

$[2]^3 = [0]$, so $[2]$ is in the nilradical.

$[3]^2 = [1]$, so the only possible powers of $[3]$ are $[1]$ and $[3]$. So $[3]$ is not in the nilradical.

$[4]^2 = [0]$ so $[4]$ is in the nilradical.

$[5]^2 = [1]$, so this is just the same as $[3]$, and so not in the nilradical.

$[6]^2 = [4]$, and $[6]^3 = [4][6] = [0]$, so $[6]$ is in the nilradical.

$[7]^2 = [1]$ (same situation as $[3]$ and $[5]$, not in the nilradical.

So the nilradical of $\Bbb Z_8$ is $\{[0],[2],[4],[6]\}$, and it should be clear that:

$[2k] \mapsto 2k + 8\Bbb Z$ is an isomorphism between the nilradical of $\Bbb Z_8$ and $2\Bbb Z/8\Bbb Z$.

 

[Math Amateur]

I thought I would just give a simple example (it's not a proof).

Let's use a ring we know and understand well, $R = \Bbb Z$.

For $I$ we will use the ideal $I = (8) = 8\Bbb Z$.

Now $\text{rad }I = \{k \in \Bbb Z: k^n \in I\}$.

Let's think about this a moment: if $k^n = 8t$, it is clear that $k$ is even (because $2$ is prime).

On the other hand, if $k = 2s$, then $k^3 = 8s^3 \in I$.

Hopefully, this convinces you that $\text{rad }I = 2\Bbb Z$.

Now $\text{rad }I/I$ is just $2\Bbb Z/8\Bbb Z$, which is the following:

$\{8\Bbb Z, 2+8\Bbb Z, 4+8\Bbb Z, 6+8\Bbb Z\}$.

We turn our attention now to $R/I$, which is just the integers modulo $8$, $\Bbb Z_8$.

What is its nilradical? Well, let's just examine the powers of *everything* in $\Bbb Z_8$.

$[0]^1 = [0]$, so there is one element in the nilradical.

$[1]^k = [1]$, for any power $k$, so this is a no-go.

$[2]^3 = [0]$, so $[2]$ is in the nilradical.

$[3]^2 = [1]$, so the only possible powers of $[3]$ are $[1]$ and $[3]$. So $[3]$ is not in the nilradical.

$[4]^2 = [0]$ so $[4]$ is in the nilradical.

$[5]^2 = [1]$, so this is just the same as $[3]$, and so not in the nilradical.

$[6]^2 = [4]$, and $[6]^3 = [4][6] = [0]$, so $[6]$ is in the nilradical.

$[7]^2 = [1]$ (same situation as $[3]$ and $[5]$, not in the nilradical.

So the nilradical of $\Bbb Z_8$ is $\{[0],[2],[4],[6]\}$, and it should be clear that:

$[2k] \mapsto 2k + 8\Bbb Z$ is an isomorphism between the nilradical of $\Bbb Z_8$ and $2\Bbb Z/8\Bbb Z$.

Thanks Deveno ... I find such examples more helpful than proofs in helping one to understand the nature and the functioning of concepts/constructs ...

Will now work carefully through your example ...

Thanks again ... examples are such great teachers ...

Peter

 

[Math Amateur]

I thought I would just give a simple example (it's not a proof).Let's use a ring we know and understand well, $R = \Bbb Z$.For $I$ we will use the ideal $I = (8) = 8\Bbb Z$.Now $\text{rad }I = \{k \in \Bbb Z: k^n \in I\}$.Let's think about this a moment: if $k^n = 8t$, it is clear that $k$ is even (because $2$ is prime).On the other hand, if $k = 2s$, then $k^3 = 8s^3 \in I$.Hopefully, this convinces you that $\text{rad }I = 2\Bbb Z$.Now $\text{rad }I/I$ is just $2\Bbb Z/8\Bbb Z$, which is the following:$\{8\Bbb Z, 2+8\Bbb Z, 4+8\Bbb Z, 6+8\Bbb Z\}$.We turn our attention now to $R/I$, which is just the integers modulo $8$, $\Bbb Z_8$.What is its nilradical? Well, let's just examine the powers of *everything* in $\Bbb Z_8$.$[0]^1 = [0]$, so there is one element in the nilradical.$[1]^k = [1]$, for any power $k$, so this is a no-go.$[2]^3 = [0]$, so $[2]$ is in the nilradical.$[3]^2 = [1]$, so the only possible powers of $[3]$ are $[1]$ and $[3]$. So $[3]$ is not in the nilradical.$[4]^2 = [0]$ so $[4]$ is in the nilradical.$[5]^2 = [1]$, so this is just the same as $[3]$, and so not in the nilradical.$[6]^2 = [4]$, and $[6]^3 = [4][6] = [0]$, so $[6]$ is in the nilradical.$[7]^2 = [1]$ (same situation as $[3]$ and $[5]$, not in the nilradical.So the nilradical of $\Bbb Z_8$ is $\{[0],[2],[4],[6]\}$, and it should be clear that:$[2k] \mapsto 2k + 8\Bbb Z$ is an isomorphism between the nilradical of $\Bbb Z_8$ and $2\Bbb Z/8\Bbb Z$.

Hi Deveno,Thanks again for the example ... it is just great to have an example to illustrate new ideas ...

Just reflecting on your example ... and I need some help to fully understand a part of your argument that $$\text{ rad I } = 2 \mathbb{Z}$$ ...In your post above, you write the following:

" ... ... For $I$ we will use the ideal $I = (8) = 8\Bbb Z$.Now $\text{rad }I = \{k \in \Bbb Z: k^n \in I\}$.Let's think about this a moment: if $k^n = 8t$, it is clear that $k$ is even (because $2$ is prime).On the other hand, if $k = 2s$, then $k^3 = 8s^3 \in I$.Hopefully, this convinces you that $\text{rad }I = 2\Bbb Z$. ... ... "As I mentioned above, I do not quite follow your argument that $$\text{ rad I } = 2 \mathbb{Z}$$ ...

I can see that $$k$$ must be even ... so then it seems that $$\text{ rad I } = 2 \mathbb{Z}$$ or $$4 \mathbb{Z}$$ or $$ 6 \mathbb{Z}$$ or ... ... etc (at least I think this is true!)

But then you write:

" ... ... On the other hand, if $k = 2s$, then $k^3 = 8s^3 \in I$. ... ... "

Can you please explain how this guarantees that $$\text{ rad I } = 2 \mathbb{Z}$$?

Hope you can clarify exactly why $$\text{ rad I } = 2 \mathbb{Z}$$ ...

Peter***EDIT***

After a little more reflection I can now follow your argument that $$\text{ rad I } = 2 \mathbb{Z}$$ ... ...

In fact, I have to say your argument was quite clear ... no idea why I did not see it straight away ... ...

Now will continue with your example ...

Thanks again,

Peter

 

[Math Amateur]

I thought I would just give a simple example (it's not a proof).

Let's use a ring we know and understand well, $R = \Bbb Z$.

For $I$ we will use the ideal $I = (8) = 8\Bbb Z$.

Now $\text{rad }I = \{k \in \Bbb Z: k^n \in I\}$.

Let's think about this a moment: if $k^n = 8t$, it is clear that $k$ is even (because $2$ is prime).

On the other hand, if $k = 2s$, then $k^3 = 8s^3 \in I$.

Hopefully, this convinces you that $\text{rad }I = 2\Bbb Z$.

Now $\text{rad }I/I$ is just $2\Bbb Z/8\Bbb Z$, which is the following:

$\{8\Bbb Z, 2+8\Bbb Z, 4+8\Bbb Z, 6+8\Bbb Z\}$.

We turn our attention now to $R/I$, which is just the integers modulo $8$, $\Bbb Z_8$.

What is its nilradical? Well, let's just examine the powers of *everything* in $\Bbb Z_8$.

$[0]^1 = [0]$, so there is one element in the nilradical.

$[1]^k = [1]$, for any power $k$, so this is a no-go.

$[2]^3 = [0]$, so $[2]$ is in the nilradical.

$[3]^2 = [1]$, so the only possible powers of $[3]$ are $[1]$ and $[3]$. So $[3]$ is not in the nilradical.

$[4]^2 = [0]$ so $[4]$ is in the nilradical.

$[5]^2 = [1]$, so this is just the same as $[3]$, and so not in the nilradical.

$[6]^2 = [4]$, and $[6]^3 = [4][6] = [0]$, so $[6]$ is in the nilradical.

$[7]^2 = [1]$ (same situation as $[3]$ and $[5]$, not in the nilradical.

So the nilradical of $\Bbb Z_8$ is $\{[0],[2],[4],[6]\}$, and it should be clear that:

$[2k] \mapsto 2k + 8\Bbb Z$ is an isomorphism between the nilradical of $\Bbb Z_8$ and $2\Bbb Z/8\Bbb Z$.

I have now worked through your example ...

What an extremely helpful example to illustrate Dummit & Foote, Proposition 11 of Chapter 15: Commutative Rings and Algebraic Geometry ... ...

I note that Proposition 11 states that ... ... :

" ... ... $$( \text{ rad } I )/ I$$ is the nilradical of $$R/I$$ ... ... " (my emphasis)

whereas you have shown that $$( \text{ rad } I )/ I \cong R/I$$ ... ...

hmm ... reflecting ...

I know that you have said previously that if things in mathematics are isomorphic then they are essentially the same ... well, I think you have said that

... but it bothers me a bit ... being isomorphic seems like things having the same structure and this seems to me a bit different from being identically the same thing as ...

But anyway, your example certainly illustrated the theorem in an excellent way ... thanks ..

Peter

 

[Deveno]

No the niradical of $\Bbb Z_8$ is $\{[0],[2],[4],[6],[8]\}$, whereas $R/I = \Bbb Z_8 = \{[0],[1],[2],[3],[4],[5],[6],[7],[8]\}$

(in this case, the nilradical of $\Bbb Z_8$ is exactly "half" of the quotient ring).

I used "isomorphism" in this case, just to keep the construction of nilradical$(R/I)$ distinct from $\text{rad }I/I$-in other words two paths to the same place.

The isomorphism $\Bbb Z_8 \to \Bbb Z/8\Bbb Z$ is, in this case, actually the identity mapping! In fact, all an isomorphism essentially amounts to is a "re-naming" of the elements of a group/ring/algebra, etc. It is another example of the tendency, prevalent *everywhere* in mathematics to replace "equal" to "equivalent" (egual-valued).

For example, the NATURAL equivalence relationship with respect to a function:

$f: A \to B$

is to set $a \sim a' \iff f(a) = f(a')$.

For the function $f: \Bbb R \to \Bbb R$ given by $f(x) = x^2$, this equivalence sends $a \mapsto |a|$.

In other words, we IDENTIFY $a$ by it's VALUE under $f$. Since $f$ may not be one-to-one, this is NOT equality, but $a,a'$ are "equivalent", that is to say "equal-valued" under $f$ ($f$ treats them the same, takes them both to the same image).

This "identification" let's us make a 1-1 correspondence $A/f$ with $f(A)$. If $f$ is injective, then "as sets", $A$ and $f(A)$ may as well "be the same", if we have, for example:

$A = \{a_1,a_2,a_3\}$ and $B = \{x,y,z,w\}$ with:

$f(a_1) = x$
$f(a_2) = y$
$f(a_3) = z$

Then any argument we make with $A$, purely as a SET (without regard to the PARTICULAR definition of what $a_1,a_2,a_3$ ARE), can ALSO be made for $f(A) = \{x,y,z\}$ by substituting $x$ for $a_1, y$ for $a_2$ and $z$ for $a_3$.

It should be clear from D&F's definition that $\text{rad }I/I$ is a subset of $R/I$. The proof that it is an IDEAL of $R/I$ is exactly the same as the proof that the nilradical (perhaps we should call this $\text{rad_0}(R)$?) is an ideal of $R$.

The property that $\{a+I: a^k + I \in I\}$ *is*, for $a+I \in R/I$ the definition of the nilradical in $R/I$ (since $I$ is the "$0$" of $R/I$). It should be almost transparent that this happens if and only if $a^k \in I$, that is $a \in \text{rad } I$.

My example was not chosen "off the cuff". Prime ideals are particularly important in ring theory, and ideals which represent "powers" of prime ideals are likewise important: this concepts become "front-and-center" when considering localizations of rings.

 

[Math Amateur]

I have now worked through your example ...

What an extremely helpful example to illustrate Dummit & Foote, Proposition 11 of Chapter 15: Commutative Rings and Algebraic Geometry ... ...

I note that Proposition 11 states that ... ... :

" ... ... $$( \text{ rad } I )/ I$$ is the nilradical of $$R/I$$ ... ... " (my emphasis)

whereas you have shown that $$( \text{ rad } I )/ I \cong R/I$$ ... ...

hmm ... reflecting ...

I know that you have said previously that if things in mathematics are isomorphic then they are essentially the same ... well, I think you have said that

... but it bothers me a bit ... being isomorphic seems like things having the same structure and this seems to me a bit different from being identically the same thing as ...

But anyway, your example certainly illustrated the theorem in an excellent way ... thanks ..

Peter

Oh my goodness ... Deveno, your last post has alerted me to a typo ...

I wrote in the above post ...I note that Proposition 11 states that ... ... :

" ... ... $$( \text{ rad } I )/ I$$ is the nilradical of $$R/I$$ ... ... " (my emphasis)

whereas you have shown that $$( \text{ rad } I )/ I \cong R/I$$ ... ..."BUT ... ... I meant to write:I note that Proposition 11 states that ... ... :

" ... ... $$( \text{ rad } I )/ I$$ is the nilradical of $$R/I$$ ... ... " (my emphasis)

whereas you have shown that $$( \text{ rad } I )/ I \cong \text{ the nilradical of } R/I$$ ... ...

Sorry for the silly error ...My issue was with the difference between "is" and "is isomorphic to" ... ...

Peter

 

[Math Amateur]

No the niradical of $\Bbb Z_8$ is $\{[0],[2],[4],[6],[8]\}$, whereas $R/I = \Bbb Z_8 = \{[0],[1],[2],[3],[4],[5],[6],[7],[8]\}$

(in this case, the nilradical of $\Bbb Z_8$ is exactly "half" of the quotient ring).

Thanks for the help ... yes, I made a typo ... see other post ...

Just a point regarding the above post

In $$\mathbb{Z}_8$$ isn't $$[0] = [8]$$ and hence $R/I = \Bbb Z_8 = \{[0],[1],[2],[3],[4],[5],[6],[7]\}$

and similarly for the nilradical of $$\mathbb{Z}_8$$ ...

Let me know if I am wrong in this matter ...

Peter

 

[Deveno]

Thanks for the help ... yes, I made a typo ... see other post ...

Just a point regarding the above post

In $$\mathbb{Z}_8$$ isn't $$[0] = [8]$$ and hence $R/I = \Bbb Z_8 = \{[0],[1],[2],[3],[4],[5],[6],[7]\}$

and similarly for the nilradical of $$\mathbb{Z}_8$$ ...

Let me know if I am wrong in this matter ...

Peter

Yup-I was clearly in a hurry when I typed that out this morning-the $[8]$ doesn't belong in either set.