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Radicals and Prime Ideals

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading R.Y. Sharp: Steps in Commutative Algebra, Chapter 3: Prime Ideals and Maximal Ideals.

Exerise 3.47 on page 52 reads as follows:

=====================================================

Let P be a prime ideal of the commutative ring R.

Show that \(\displaystyle \sqrt (P^n) = P \)

======================================================

Can someone please help me get started on this problem.

Working from definitions of terms we have that

\(\displaystyle p_i \in P^n \Longrightarrow p_i = p_{11}p_{12} ... \ ... p_{1n} + p_{21}p_{22} ... \ ... p_{2n} + p_{i1}p_{i2} ... \ ... p_{in} \)

and then

\(\displaystyle (P^n) = \{ \sum_{i=1}^{n} r_i p_i , r_i \in R, p_i \in P^n \} \) ...

and

\(\displaystyle \sqrt (P^n) = \{ p \in R \ \ | \ \) there exists \(\displaystyle m \in \mathbb{N} \) with \(\displaystyle p^m \in (P^n) \} \)

But how to move beyond a restatement of the definitions ... ... ?

Can someone please help ...

Peter
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
One simplification of your statement should be noted:

$(P^n) = P^n$.

This comes from noting that $P^n$ is itself, an ideal.

Now, you have two sets, show they contain each other.

One containment should be obvious:

If $p \in P$, then $p^n \in P^n$ in which case there certainly is an $x \in R$ and an $m \in \Bbb N$ such that $x^m \in P^n$, namely: $x = p$ and $m = n$.

On the other hand, suppose $x^m \in P^n$. This means $x^m \in P$, since $P$ is closed under multiplication, and addition (so $P^n \subseteq P$).

But this means that $x(x^{m-1}) \in P$ and since $P$ is prime, either $x \in P$ or $x^{m-1} \in P$. If $x^{m-1} \in P$, rinse and repeat.

Let's look at a special example:

Suppose $R = \Bbb Z$ with $P = (2)$ and $n = 3$

Then $\displaystyle P^3 = \left\{\sum_{i = 1}^k (2r_i)(2s_i)(2t_i): r_i,s_i,t_i \in \Bbb Z\right\}$

It should be clear that $P^3 = (8)$.

Now suppose $x$ is such that:

$x^m = 8a$ for some integer $a$, and some $m \in \Bbb N$. Clearly $x$ must be even.

Now if $x$ is even, then either:

$x = 0 \text{( mod 8)}$ or:

$x = 2 \text{( mod 8)}$ or:

$x = 4 \text{( mod 8)}$ or:

$x = 6 \text{( mod 8)}$

In the first case, $x \in (8)$, in the second case $x^3 \in (8)$, in the third case $x^2 \in (8)$, and in the fourth case $x^3 \in (8)$.

Thus $\sqrt{(2)^3} = (2)$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
One simplification of your statement should be noted:

$(P^n) = P^n$.

This comes from noting that $P^n$ is itself, an ideal.

Now, you have two sets, show they contain each other.

One containment should be obvious:

If $p \in P$, then $p^n \in P^n$ in which case there certainly is an $x \in R$ and an $m \in \Bbb N$ such that $x^m \in P^n$, namely: $x = p$ and $m = n$.

On the other hand, suppose $x^m \in P^n$. This means $x^m \in P$, since $P$ is closed under multiplication, and addition (so $P^n \subseteq P$).

But this means that $x(x^{m-1}) \in P$ and since $P$ is prime, either $x \in P$ or $x^{m-1} \in P$. If $x^{m-1} \in P$, rinse and repeat.

Let's look at a special example:

Suppose $R = \Bbb Z$ with $P = (2)$ and $n = 3$

Then $\displaystyle P^3 = \left\{\sum_{i = 1}^k (2r_i)(2s_i)(2t_i): r_i,s_i,t_i \in \Bbb Z\right\}$

It should be clear that $P^3 = (8)$.

Now suppose $x$ is such that:

$x^m = 8a$ for some integer $a$, and some $m \in \Bbb N$. Clearly $x$ must be even.

Now if $x$ is even, then either:

$x = 0 \text{( mod 8)}$ or:

$x = 2 \text{( mod 8)}$ or:

$x = 4 \text{( mod 8)}$ or:

$x = 6 \text{( mod 8)}$

In the first case, $x \in (8)$, in the second case $x^3 \in (8)$, in the third case $x^2 \in (8)$, and in the fourth case $x^3 \in (8)$.

Thus $\sqrt{(2)^3} = (2)$.
Thanks for the help Deveno

Just working through your post now.

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Thanks for the help Deveno

Just working through your post now.

Peter
Hi Deveno,

I hope you can clarify the following

You write:

"On the other hand, suppose \(\displaystyle x^m \in P^n \). This means \(\displaystyle x^m \in P \), since P is closed under multiplication, and addition (so \(\displaystyle P^n \subseteq P \))."

How exactly does it follow from \(\displaystyle x^m \in P^n \) that \(\displaystyle x^m \in P \)?

Certainly if you assume that \(\displaystyle x \in P \) then we have ...

\(\displaystyle x^m = x.x.x.... ... .x \) (m factors)

and then by closure of multiplication (don't need closure of addition) we have

\(\displaystyle x^m \in P \) but we are then assuming what we want to prove ...

Can you clarify this matter?

Peter
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
We are not assuming $x \in P$. We are assuming $x \in \sqrt{P^n}$.

All this tells us is some power of $x$ lies in $P^n$, say $x^m$.

But $P^n$ is contained within $P$...usually it is a much SMALLER ideal than $P$.

So we know that $x^m \in P$. Suppose that $x \not\in P$.

Since $x^m = x(x^{m-1})$ and $P$ is a prime ideal, it must be that:

$x^{m-1} \in P$. But now:

$x^{m-1} = x(x^{m-2})$ and since $x \not \in P$, $x^{m-2} \in P$.

Continuing in this way, we eventually wind up with:

$x^2 = xx \in P$. But this is impossible, since one of those factors must lie in $P$.

Our only assumption was $x \not \in P$, and this leads to a self-contradiction. SO that assumption must be wrong, which is to say: $x \in P$.

Let's pick an ideal in a ring that is not principal, to make this clearer. Suppose $R = \Bbb Z[x]$ and let $I = (2,x)$. Let's pick four elements of $I$ and multiply them together in pairs, and add them. So:

$r_1 = 2 + x$
$r_2 = 4 + 3x$
$r_3 = 4 + 2x + x^2 = 2(2 + x) + x(x)$
$r_4 = -2 + x^3$

Now $r_1r_2 = (2+x)(4 + 3x) = 8 + 10x + 3x^2 = 2(4 + 5x) + x(x)$ and

$r_3r_4 = (4 + 2x + x^2)(-2 + x^3) = -8 - 4x - 2x^2 + 2x^4 + x^5$

$= 2(-2 - 2x - x^2 + x^4) + x(x^4)$

I wrote them this way so you could see that $r_1r_2 + r_3r_4 \in I$.

But clearly $r_1r_2 + r_3r_4 \in I^2$. While this is not a "proof" that $I^2 \subseteq I$, I hope it lets you see how a proof could be constructed.

This is NOT to say $I^2 = I$, in fact, this is usually not true: in the above example, we have $x \in I$, but $x \not\in I^2$.

In my previous example in my previous post, we have $6 \in (2)$, but not $6 \not\in (2)^2 = (4)$. However, 6 IS in the radical of $(4)$ because $36 \in (4)$ (for example).

For another way to look at it, we have for any two ideals $I,J$ of a ring $R$:

$IJ \subseteq I\cap J$.

Letting $J = I$, this leads to:

$I^2 \subseteq I \cap I = I$.

Induction on $n$ leads to:

$I^n \subseteq I$, for any $n \in \Bbb Z^+$.

Still another approach presents itself in this way:

Suppose $P$ is prime. Pass to the quotient $R/P$. Suppose $x \in \sqrt{P^n}$.

Then the quotient mapping takes $x^m$ to a sum of powers of $0 = P \in R/P$ (which of course is STILL $0 = P$ in the quotient).

Since $P$ is prime, $R/P$ is an integral domain. this means that:

$[x^m] = [x]^m = [0]$ which implies $[x] = [0]$, which is to say $x \in P$

(writing $[x] = x + P$ for clarity).

This last approach is actually the easiest to understand, writing ideal elements as sums of indexed product terms gets cumbersome.

The whole idea of making a certain kind of ideal, is to get rid of "bad elements" the ideal characterizes. Prime ideals are maximal, and maximal ideals get rid of zero divisors.

As another example, nilpotent elements are frequently "undesirable". We can get rid of this by taking $R/N$, where $N = \sqrt{(0)}$, the nilradical.

But sometimes, we "throw out the baby with the bathwater", the ideal $I$ might be too big (if it's maximal, it's usually substantial). Other times, $I$ might not be "big enough", so we are always looking for ways to create "bigger" and smaller" ideals associated with $I$.

Now, let's take a look at where this will eventually be going:

It turns out that given a nice enough ring $R$ (typical a unital commutative ring), it makes more sense to study $R$-modules. The goal is to try to break a given $R$-module into "atomic pieces" that we can play with like tinker-toys (building up, and splitting down as we please). If we know enough about the individual "pieces" hopefully that gives us insight into the "bigger" things we make from them.

And the way we study how these $R$-modules interact is by studying $R$-linear maps ($R$-module homomorphisms), which allow us to transfer "arithmetic in $R$" from one $R$-module to another.

Even more exciting, is the fact that given $M$ as both an $R$-module and an $S$-module, we can leverage any homomorphic relationship $R \to S$ to examine the internal structure of $M$. For example, in linear algebra, we can learn a great deal about a particular vector space, by examining the relationship between the underlying field $F$, and the polynomial ring $F[x]$. This leads to deep results about $\text{Hom}_{\ F}(V,V)$ (which is equivalent to the study of matrices in the finite-dimensional case). And I don't think I would be understating the case to say that these kind of results are HUGE, with applications far beyond the algebraic context we might study them in.

As a historical aside: much of early ring theory came about in an attempt to understand various "sub-systems" of the complex numbers (for example: the Gaussian integers $\Bbb Z)$, or various complex extensions $\Bbb Z[\sqrt{-n}]$). The hope was, that in these number systems solutions to problems that "ordinary methods" failed to solve could be found (this approach turned out to be justified in solving a general cubic polynomial). It turned out the our normal concept of "prime number" wasn't quite GOOD enough, as it didn't transfer directly to these "larger systems". The question became: what kind of "grouping" of numbers could serve as a kind of "substitute" for "primality", and the numbers that worked as a kind of "pseudo-prime" were called "ideal numbers".

The goal then, was to try to get something like unique factorization of integers in these larger systems. And it turned out that you could GET that (in suitable rings), if you replaced "numbers" with "ideals" (a notion first worked out by Dedekind, I believe).

******

(Ahem!)

OK, that's a bit of a detour....but I wanted to point out that ring theory is, in a sense, much less cohesive than, say, group theory, and instead of broad strokes tying everything up into a neat little package, it developed much more incrementally (more nooks and crannies to explore). And one of the things that took much longer in ring theory, was to see that ideals were "kernel objects". Kernel objects are related to "zero objects" (which are some distinguished "trivial example" in an imprecise way). These make a theory "nice".

One of the difficulties in set theory, for example, is that it lacks a suitable "zero object". It turns out that the empty set is a good "default sub-thingy", but a bad thing to "map onto" (so not very useful for "quotient maps" that is to say, equivalence classes). Singleton sets make good things to map TO, but one is hampered by the fact that one cannot choose a map:

$f: \{\ast\} \to A$

in a unique way: if $A$ has several elements, we get a different map for each element (we lose UNIQUENESS). This makes the theory of sets "lop-sided" the directionality of mappings keeps us from exploiting symmetry.

This situation is somewhat better in groups, rings and similar structures: kernels and images occur in pairs, with a nice canonical (similar for every example) relationship between them. This provides some very nice "machinery" for making difficult computations easier (my personal theory is that mathematics moves in a direction of increasing laziness :p).
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
We are not assuming $x \in P$. We are assuming $x \in \sqrt{P^n}$.

All this tells us is some power of $x$ lies in $P^n$, say $x^m$.

But $P^n$ is contained within $P$...usually it is a much SMALLER ideal than $P$.

So we know that $x^m \in P$. Suppose that $x \not\in P$.

Since $x^m = x(x^{m-1})$ and $P$ is a prime ideal, it must be that:

$x^{m-1} \in P$. But now:

$x^{m-1} = x(x^{m-2})$ and since $x \not \in P$, $x^{m-2} \in P$.

Continuing in this way, we eventually wind up with:

$x^2 = xx \in P$. But this is impossible, since one of those factors must lie in $P$.

Our only assumption was $x \not \in P$, and this leads to a self-contradiction. SO that assumption must be wrong, which is to say: $x \in P$.

Let's pick an ideal in a ring that is not principal, to make this clearer. Suppose $R = \Bbb Z[x]$ and let $I = (2,x)$. Let's pick four elements of $I$ and multiply them together in pairs, and add them. So:

$r_1 = 2 + x$
$r_2 = 4 + 3x$
$r_3 = 4 + 2x + x^2 = 2(2 + x) + x(x)$
$r_4 = -2 + x^3$

Now $r_1r_2 = (2+x)(4 + 3x) = 8 + 10x + 3x^2 = 2(4 + 5x) + x(x)$ and

$r_3r_4 = (4 + 2x + x^2)(-2 + x^3) = -8 - 4x - 2x^2 + 2x^4 + x^5$

$= 2(-2 - 2x - x^2 + x^4) + x(x^4)$

I wrote them this way so you could see that $r_1r_2 + r_3r_4 \in I$.

But clearly $r_1r_2 + r_3r_4 \in I^2$. While this is not a "proof" that $I^2 \subseteq I$, I hope it lets you see how a proof could be constructed.

This is NOT to say $I^2 = I$, in fact, this is usually not true: in the above example, we have $x \in I$, but $x \not\in I^2$.

In my previous example in my previous post, we have $6 \in (2)$, but not $6 \not\in (2)^2 = (4)$. However, 6 IS in the radical of $(4)$ because $36 \in (4)$ (for example).

For another way to look at it, we have for any two ideals $I,J$ of a ring $R$:

$IJ \subseteq I\cap J$.

Letting $J = I$, this leads to:

$I^2 \subseteq I \cap I = I$.

Induction on $n$ leads to:

$I^n \subseteq I$, for any $n \in \Bbb Z^+$.

Still another approach presents itself in this way:

Suppose $P$ is prime. Pass to the quotient $R/P$. Suppose $x \in \sqrt{P^n}$.

Then the quotient mapping takes $x^m$ to a sum of powers of $0 = P \in R/P$ (which of course is STILL $0 = P$ in the quotient).

Since $P$ is prime, $R/P$ is an integral domain. this means that:

$[x^m] = [x]^m = [0]$ which implies $[x] = [0]$, which is to say $x \in P$

(writing $[x] = x + P$ for clarity).

This last approach is actually the easiest to understand, writing ideal elements as sums of indexed product terms gets cumbersome.

The whole idea of making a certain kind of ideal, is to get rid of "bad elements" the ideal characterizes. Prime ideals are maximal, and maximal ideals get rid of zero divisors.

As another example, nilpotent elements are frequently "undesirable". We can get rid of this by taking $R/N$, where $N = \sqrt{(0)}$, the nilradical.

But sometimes, we "throw out the baby with the bathwater", the ideal $I$ might be too big (if it's maximal, it's usually substantial). Other times, $I$ might not be "big enough", so we are always looking for ways to create "bigger" and smaller" ideals associated with $I$.

Now, let's take a look at where this will eventually be going:

It turns out that given a nice enough ring $R$ (typical a unital commutative ring), it makes more sense to study $R$-modules. The goal is to try to break a given $R$-module into "atomic pieces" that we can play with like tinker-toys (building up, and splitting down as we please). If we know enough about the individual "pieces" hopefully that gives us insight into the "bigger" things we make from them.

And the way we study how these $R$-modules interact is by studying $R$-linear maps ($R$-module homomorphisms), which allow us to transfer "arithmetic in $R$" from one $R$-module to another.

Even more exciting, is the fact that given $M$ as both an $R$-module and an $S$-module, we can leverage any homomorphic relationship $R \to S$ to examine the internal structure of $M$. For example, in linear algebra, we can learn a great deal about a particular vector space, by examining the relationship between the underlying field $F$, and the polynomial ring $F[x]$. This leads to deep results about $\text{Hom}_{\ F}(V,V)$ (which is equivalent to the study of matrices in the finite-dimensional case). And I don't think I would be understating the case to say that these kind of results are HUGE, with applications far beyond the algebraic context we might study them in.

As a historical aside: much of early ring theory came about in an attempt to understand various "sub-systems" of the complex numbers (for example: the Gaussian integers $\Bbb Z)$, or various complex extensions $\Bbb Z[\sqrt{-n}]$). The hope was, that in these number systems solutions to problems that "ordinary methods" failed to solve could be found (this approach turned out to be justified in solving a general cubic polynomial). It turned out the our normal concept of "prime number" wasn't quite GOOD enough, as it didn't transfer directly to these "larger systems". The question became: what kind of "grouping" of numbers could serve as a kind of "substitute" for "primality", and the numbers that worked as a kind of "pseudo-prime" were called "ideal numbers".

The goal then, was to try to get something like unique factorization of integers in these larger systems. And it turned out that you could GET that (in suitable rings), if you replaced "numbers" with "ideals" (a notion first worked out by Dedekind, I believe).

******

(Ahem!)

OK, that's a bit of a detour....but I wanted to point out that ring theory is, in a sense, much less cohesive than, say, group theory, and instead of broad strokes tying everything up into a neat little package, it developed much more incrementally (more nooks and crannies to explore). And one of the things that took much longer in ring theory, was to see that ideals were "kernel objects". Kernel objects are related to "zero objects" (which are some distinguished "trivial example" in an imprecise way). These make a theory "nice".

One of the difficulties in set theory, for example, is that it lacks a suitable "zero object". It turns out that the empty set is a good "default sub-thingy", but a bad thing to "map onto" (so not very useful for "quotient maps" that is to say, equivalence classes). Singleton sets make good things to map TO, but one is hampered by the fact that one cannot choose a map:

$f: \{\ast\} \to A$

in a unique way: if $A$ has several elements, we get a different map for each element (we lose UNIQUENESS). This makes the theory of sets "lop-sided" the directionality of mappings keeps us from exploiting symmetry.

This situation is somewhat better in groups, rings and similar structures: kernels and images occur in pairs, with a nice canonical (similar for every example) relationship between them. This provides some very nice "machinery" for making difficult computations easier (my personal theory is that mathematics moves in a direction of increasing laziness :p).


Hi Deveno,

Thanks so much for the above post!!!

I am now working through it carefully.

I have only been working on it a short time now, yet already I am finding it exceptionally helpful!!!

Your examples are always a great learning experience.

Thanks again,

Peter