# The Ideal (x1, x2, ... ... xn)^m

#### Peter

##### Well-known member
MHB Site Helper
in the book, Sharp: Steps in Commutative Algebra, in Chapter 2 on ideals on page 29 we find Exercise 2.29 which reads as follows: (see attachment)

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Let $$\displaystyle R$$ be a commutative ring and let $$\displaystyle m \in \mathbb{N}$$.

Describe the ideal $$\displaystyle (x_1, x_2, ... \ ... ,x_n)^m$$ of the ring $$\displaystyle R[x_1, x_2, ... \ ... ,x_n]$$ of polynomials over R in indeterminates $$\displaystyle x_1, x_2, ... \ ... ,x_n$$.

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Peter

Note: On pages 28 and 29 of Sharp we have the following relevant information: (see attachment)

"we can unambiguously define the product $$\displaystyle {\prod}_{i=1}^{n} I_i$$ of ideals $$\displaystyle I_1, I_2, ... \ ... ,I_n$$ of $$\displaystyle R$$: we have

$$\displaystyle {\prod}_{i=1}^{n} I_i = I_1I_2 ... \ ... I_n = RL$$ ... ... (1)

where

$$\displaystyle L = \{a_1, a_2, ... \ ... , a_n \ | \ a_1 \in I_1, a_2 \in I_2, ... \ ... a_n \in I_n \}$$

We therefore see that a typical element of $$\displaystyle I_1I_2 ... \ ... I_n$$ is a sum of finitely many elements of L."

Note however that I have some trouble with reconciling the last statement: "a typical element of $$\displaystyle I_1I_2 ... \ ... I_n$$ is a sum of finitely many elements of L." with the equation (1) above.

BUT ... from one it appears to me that an element of RL would be of the form $$\displaystyle r (a_1, a_2, ... \ ... , a_n)$$ where $$\displaystyle r \in R$$ - however this is not a finite sum ...

Can someone please clarify this issue for me ... as well as help get a start on the problem ...

Last edited:

#### Peter

##### Well-known member
MHB Site Helper
in the book, Sharp: Steps in Commutative Algebra, in Chapter 2 on ideals on page 29 we find Exercise 2.29 which reads as follows:

----------------------------------------------------------------------------

Let $$\displaystyle R$$ be a commutative ring and let $$\displaystyle m \in \mathbb{N}$$.

Describe the ideal $$\displaystyle (x_1, x_2, ... \ ... ,x_n)^m$$ of the ring $$\displaystyle R[x_1, x_2, ... \ ... ,x_n]$$ of polynomials over R in indeterminates $$\displaystyle x_1, x_2, ... \ ... ,x_n$$.

-----------------------------------------------------------------------------

Peter

Note: On pages 28 and 29 of Sharp we have the following relevant information:

"we can unambiguously define the product $$\displaystyle {\prod}_{i=1}^{n} I_i$$ of ideals $$\displaystyle I_1, I_2, ... \ ... ,I_n$$ of $$\displaystyle R$$: we have

$$\displaystyle {\prod}_{i=1}^{n} I_i = I_1I_2 ... \ ... I_n = RL$$ ... ... (1)

where

L = \{a_1, a_2, ... \ ... , a_n \ | \ a_1 \in I_1, a_2 \in I_2, ... \ ... a_n \in I_n \}

We therefore see that a typical element of $$\displaystyle I_1I_2 ... \ ... I_n$$ is a sum of finitely many elements of L."

Note however that I have some trouble with reconciling the last statement: "a typical element of $$\displaystyle I_1I_2 ... \ ... I_n$$ is a sum of finitely many elements of L." with the equation (1) above.

BUT ... from one it appears to me that an element of RL would be of the form $$\displaystyle r (a_1, a_2, ... \ ... , a_n)$$ where $$\displaystyle r \in R$$ - however this is not a finite sum ...

Can someone please clarify this issue for me ... as well as help get a start on the problem ...
SOLVED ???

Sharp gives the following information on page 29.

"Note in particular that the powers $$\displaystyle I^m$$ for $$\displaystyle m \in \mathbb{N}$$, of I are defined; we adopt the convention that $$\displaystyle I^0 = R$$.

Note that a general element of $$\displaystyle I^m$$ (for positive m) has the form:

$$\displaystyle a_{11}a_{12} ... a_{1m} + a_{21}a_{22} ... a_{2m} + ... \ ... + a_{n1}a_{n2} ... a_{nm}$$ ... ... (1)

where

$$\displaystyle n \in \mathbb{N}$$ and $$\displaystyle a_{ij} \in I$$ for all $$\displaystyle i = 1,2, ...,n$$ and $$\displaystyle j = 1,2, ...,m$$ "

Given this I suspect that for $$\displaystyle (x_1, x_2, ... \ ... ,x_n)^m$$ we have (following (1) above) that each $$\displaystyle a_{ij}$$ is as follows:

$$\displaystyle a_{ij} = (x_1, x_2. ... \ ... x_n)$$

so that

$$\displaystyle a_{11}a_{12} ... a_{1m} = (x_1, x_2. ... \ ... x_n)^m$$

So the general element of $$\displaystyle (x_1, x_2, ... \ ... ,x_n)^m$$ is as follows:

MATH] a_{11}a_{12} ... a_{1m} + a_{21}a_{22} ... a_{2m} + ... \ ... + a_{n1}a_{n2} ... a_{nm} [/MATH]

$$\displaystyle = (x_1, x_2. ... \ ... x_n)^m + ... \ ... + (x_1, x_2. ... \ ... x_n)^m$$ (n terms)

$$\displaystyle = n(x_1, x_2. ... \ ... x_n)^m$$ when $$\displaystyle n \in \mathbb{N}$$

Can someone please confirm that the above is correct ... or not ...?

Peter

Note:

I was a bit surprised when Sharp defined $$\displaystyle I^0$$ to be equal to the whole ring R. Can someone please comment on the reasonableness of this definition.

#### Deveno

##### Well-known member
MHB Math Scholar

Typically, the "closer to 1" an element of an ideal is, the larger an ideal it generates:

For example, the ideal generated by $x^2$ in $R[x]$ is smaller than the ideal generated by $x$.

It is "harder" (given an element $a \in R$, and an ideal $I$) for $a$ to be in $I^2$ than it is to be in $I$, for example, every element of $I^2$ is of course in $I$, but not vice-versa. Adopting the convention $I^0 = (1) = R$, says that basically, no matter what ideal $I$ we're building the series of ideals:

$I,I^2,I^3,\dots$

out of, our "starting spot" is the ideal EVERY element ALWAYS belongs to: the entire ring. This is analogous (in my polynomial example) to building the series of ideals:

$(x^0) = R, (x),(x^2),(x^3),\dots$

*********

It might help, for your other question, to consider which polynomials AREN'T in:

$(x,y)^2 \subseteq R[x,y]$

For example, do we have $a + bx + cy \in (x,y)^2$?

#### Peter

##### Well-known member
MHB Site Helper

Typically, the "closer to 1" an element of an ideal is, the larger an ideal it generates:

For example, the ideal generated by $x^2$ in $R[x]$ is smaller than the ideal generated by $x$.

It is "harder" (given an element $a \in R$, and an ideal $I$) for $a$ to be in $I^2$ than it is to be in $I$, for example, every element of $I^2$ is of course in $I$, but not vice-versa. Adopting the convention $I^0 = (1) = R$, says that basically, no matter what ideal $I$ we're building the series of ideals:

$I,I^2,I^3,\dots$

out of, our "starting spot" is the ideal EVERY element ALWAYS belongs to: the entire ring. This is analogous (in my polynomial example) to building the series of ideals:

$(x^0) = R, (x),(x^2),(x^3),\dots$

*********

It might help, for your other question, to consider which polynomials AREN'T in:

$(x,y)^2 \subseteq R[x,y]$

For example, do we have $a + bx + cy \in (x,y)^2$?

Thanks Deveno.

Idea of "closer to 1" and the basic idea of ""harder" (given an element $a \in R$, and an ideal $I$) for $a$ to be in $I^2$ than it is to be in $I$" are very helpful in getting an intuitive idea of what is happening ...

I am assuming that my analysis on the exercise is basically correct? [I understand that the exercise is simply working with the definitions and notation ... but I think Sharp means it to be build confidence and to become familiar with the notation and the implications of the definitions]

I am now reflecting on the question you posed at the end of your post.

Peter

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#### Peter

##### Well-known member
MHB Site Helper

Typically, the "closer to 1" an element of an ideal is, the larger an ideal it generates:

For example, the ideal generated by $x^2$ in $R[x]$ is smaller than the ideal generated by $x$.

It is "harder" (given an element $a \in R$, and an ideal $I$) for $a$ to be in $I^2$ than it is to be in $I$, for example, every element of $I^2$ is of course in $I$, but not vice-versa. Adopting the convention $I^0 = (1) = R$, says that basically, no matter what ideal $I$ we're building the series of ideals:

$I,I^2,I^3,\dots$

out of, our "starting spot" is the ideal EVERY element ALWAYS belongs to: the entire ring. This is analogous (in my polynomial example) to building the series of ideals:

$(x^0) = R, (x),(x^2),(x^3),\dots$

*********

It might help, for your other question, to consider which polynomials AREN'T in:

$(x,y)^2 \subseteq R[x,y]$

For example, do we have $a + bx + cy \in (x,y)^2$?

Hi Deveno,

Just working from first principles on your suggestion:

" ... ... consider which polynomials AREN'T in:

$(x,y)^2 \subseteq R[x,y]$

For example, do we have $a + bx + cy \in (x,y)^2$?"

Taking things from first principles (and working from Sharp's definitions) to ensure I am understanding the nature and structure of things properly we have ...

Consider first the ideal generated by $$\displaystyle (x,y) \subseteq R[x,y]$$ ... ...

Let this ideal be denoted as I

Then $$\displaystyle I = \{ f_1x + f_2y \ | \ f_1, f_2 \in R[x,y] \}$$

and

$$\displaystyle I^2 =$$ the ideal generated by the set $$\displaystyle \{ a_ia_j \ : \ a_i \in I , a_j \in I \}$$

= the ideal generated by $$\displaystyle \{ (f_1x + f_2y)(g_1x + g_2y) \ | \ f_1, f_2, g_1, g_2 \in R[x,y] \}$$

= the ideal generated by $$\displaystyle \{ f_1g_1x^2 + (f_1g_2 + f_2g_1)xy + f_2g_2y^2 \ | \ f_1, f_2, g_1, g_2 \in R[x,y] \}$$

= the ideal generated by $$\displaystyle \{ h_1x^2 + h_2xy + h_3y^2 \ | \ h_1, h_2, h_3 \in R[x,y] \}$$

Is this correct?
... ... seems untidy ...

Just to check, I will follow Sharp on page 29 where he writes:

" ... ... a general element of $$\displaystyle I^m$$ (for positive m) has the form:

$$\displaystyle a_{11}a_{12} ... a_{1m} + a_{21}a_{22} ... a_{2m} + ... \ ... + a_{n1}a_{n2} ... a_{nm}$$

where $$\displaystyle n \in \mathbb{N}$$ and $$\displaystyle a_i \in I$$ for all $$\displaystyle i = 1,2, ... / ... n$$ and $$\displaystyle j = 1,2, ... / ... m$$.

Thus for m = 2, elements of $$\displaystyle I^2$$ will have the form

$$\displaystyle a_{11}a_{12} + a_{21}a_{22} + ... \ ... + a_{n1}a_{n2}$$

So elements of $$\displaystyle I^2$$ will indeed have the form of finite sums of $$\displaystyle h_1x^2 + h_2xy + h_3y^2$$ ... ... (1)

So ... ... to return to your suggestion/question it seems that elements like $$\displaystyle a + bx + cy \notin (x, y)^2$$ since all
terms in $$\displaystyle (x, y)^2$$ are finite sums of terms of the form shown in (1) - and these terms have no constant, x or y terms ... ...

Can you comment on the above ... just confirm it is OK ...

"Note however that I have some trouble with reconciling the last statement: "a typical element of $$\displaystyle I_1I_2 ... \ ... I_n$$ is a sum of finitely many elements of L." with the equation (1) above.

BUT ... from one it appears to me that an element of RL would be of the form $$\displaystyle r (a_1, a_2, ... \ ... , a_n)$$ where $$\displaystyle r \in R$$ - however this is not a finite sum ...

Can someone please clarify this issue for me .."

I can now see that $$\displaystyle {\prod}_{i=1}^{n} I_i = I^n = I_1I_2 ... \ ... I_n$$ has finitely many elements .. I think my problem was not being fully familiar with the notation $$\displaystyle RL$$

Regarding this notation Sharp writes on page 24:

"2.17 GENERATION OF IDEALS

Let $$\displaystyle R$$ be a commutative ring ... ...

... ...

Let $$\displaystyle H \subseteq R$$. We define the ideal of R generated by H, denoted by (H) or RH or HR to be the intersection of the family of all ideals of R which contain H.

Sharp then in Proposition 2.18 on page 24 shows that $$\displaystyle (H)$$ (or $$\displaystyle RH$$ or $$\displaystyle HR$$) is a set of finite sums of the form $$\displaystyle {\sum}_{i=1}^{n} r_ih_i$$ where $$\displaystyle r_1, r_2, ... r_n \in R$$ and $$\displaystyle h_1h_2 ... h_n \in H$$.

Can you please comment on the above? Essentially please just confirm the reasoning is basically OK.

Peter

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#### Deveno

##### Well-known member
MHB Math Scholar
Yes it looks to me that you're on the right track.

As Turgul and I have pointed out before, whenever you are "generating" something from a set, there is an implicit "minimality" condition (sometimes phrased in terms of intersection).

In the example I posed for you, you can see that such a polynomial in $R[x,y]$ belonging to $(x,y)^2$ must have 0 constant term, and 0 linear terms.

So, loosely speaking, what happens with $(x_1,\dots,x_n)^m$ is that the only things that are in it are polynomials with certain terms that are below a certain degree having 0 coefficients. As $m$ gets larger, fewer and fewer polynomials qualify.

I could be wrong, but I think where your text is heading with all of this is an exploration of ascending chain conditions on ideals in rings, culminating in Hilbert's basis theorem:

If $R$ is Noetherian, so is $R[x_1,x_2,\dots,x_n]$.

This, in turn leads to the study of Grobner bases, which contain the "answer" to an earlier question you had about generalizing Euclidean division to multivariate polynomials.

#### Peter

##### Well-known member
MHB Site Helper
Yes it looks to me that you're on the right track.

As Turgul and I have pointed out before, whenever you are "generating" something from a set, there is an implicit "minimality" condition (sometimes phrased in terms of intersection).

In the example I posed for you, you can see that such a polynomial in $R[x,y]$ belonging to $(x,y)^2$ must have 0 constant term, and 0 linear terms.

So, loosely speaking, what happens with $(x_1,\dots,x_n)^m$ is that the only things that are in it are polynomials with certain terms that are below a certain degree having 0 coefficients. As $m$ gets larger, fewer and fewer polynomials qualify.

I could be wrong, but I think where your text is heading with all of this is an exploration of ascending chain conditions on ideals in rings, culminating in Hilbert's basis theorem:

If $R$ is Noetherian, so is $R[x_1,x_2,\dots,x_n]$.

This, in turn leads to the study of Grobner bases, which contain the "answer" to an earlier question you had about generalizing Euclidean division to multivariate polynomials.
Thanks Deveno ... most helpful ... ...

You are right regarding Sharp's text, "Steps in Commutative Algebra". After chapters on Primary Decomposition, Rings of Fractions and Modules we have:

Chapter 7: Chain Conditions on Modules
Chapter 8: Commutative Noetherian Rings.

R.Y. Sharp does not cover Grobner Bases, but another text I am using, namely Topics in Commutative Ring Theory by John J Watkins, which also has a chapter on Noetherian Rings, does have a chapter devoted to Grobner Bases.

Thanks again for all your help with this exercise - as usual, I learned a good deal from your posts,

Peter