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Consider the $5$ variables polynomial ring over the complex field, $\mathbb{C}[x,y,u,v,w]$ and $J$ the ideal generated by the set $\{vxy, vwy, uwy, uwx, uvx\}$. Then, how could we define the powers of the ideal $J^n$?

Typically, is the ideal $J^2$ equal to the ideal generated by the pairwise products of any two generators of $J$? Does a similar reasoning apply for any power of the ideal? What if we want $J^d$ for $d\ge 6$? Note that the symbolic power given in Wikipedia article is not the definition I am looking for, as it assumes the power of an ideal beforehand.

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Let $R$ be a commutative ring. The notation $J^n$ (where $J$ is an ideal of $R$ and $n$ is a nonnegative integer) stands for the $n$-th power of $J$ in the monoid of ideals of the ring.

What is this monoid? Well, here is the definition: If $U$ and $V$ are two ideals of $R$, then we define their product $UV$ to be the ideal of $R$ generated by elements of the form $uv$ with $u \in U$ and $v \in V$. Thus, explicitly, $UV$ is the set of all $R$-linear combinations $r_1 u_1 v_1 + r_2 u_2 v_2 + \cdots + r_k u_k v_k$ with $k \in \mathbb{N}$ and $r_1, r_2, \ldots, r_k \in R$ and $u_1, u_2, \ldots, u_k \in U$ and $v_1, v_2, \ldots, v_k \in V$. It is easy to see that $UV$ is also the set of all sums $u_1 v_1 + u_2 v_2 + \cdots + u_k v_k$ with $k \in \mathbb{N}$ and $u_1, u_2, \ldots, u_k \in U$ and $v_1, v_2, \ldots, v_k \in V$ (because if $r_i \in R$ and $u_i \in U$, then $r_i u_i \in U$).

So now we have defined a product operation on the set of all ideals of $R$ (sending a pair $\left(U,V\right)$ of ideals to the ideal $UV$). This product operation has a neutral element, namely the ideal $R$ (check this). Furthermore, this operation is associative: i.e., if $U$, $V$ and $W$ are three ideals of $R$, then $\left(UV\right) W = U \left(VW\right)$ (and moreover, this ideal $\left(UV\right) W = U \left(VW\right)$ is the ideal generated by all elements of the form $uvw$ with $u \in U$, $v \in V$ and $w \in W$).

Thus, equipping the set of ideals of $R$ with this product operation, we obtain a monoid, which is called the monoid of ideals of $R$.

It is not hard to show that if $U_1, U_2, \ldots, U_n$ are $n$ ideals of $R$, then their product $U_1 U_2 \cdots U_n$ (in this monoid) is the ideal of $R$ generated by all elements of the form $u_1 u_2 \cdots u_n$ with $u_1 \in U_1$, $u_2 \in U_2$, $\ldots$, $u_n \in U_n$. Moreover, if $n > 0$, then these latter elements not only generate $U_1 U_2 \cdots U_n$ as an ideal, but even generate it as an additive group (so each element of $U_1 U_2 \cdots U_n$ is not only an $R$-linear combination of products of the form $u_1 u_2 \cdots u_n$, but actually a sum of such problems). Somewhat distractingly, this is false for $n = 0$.

When you have an ideal $J$ of $R$ and a nonnegative integer $n$, you can take the $n$-th power of $J$ in the monoid of ideals of $R$ (since $n$-th powers are defined in any monoid); this is the ideal called $J^n$. This should answer your question.

For a simple example, you can check how principal ideals behave under products and powers. For example, if $a$ and $b$ be two elements of $R$, then $\left(aR\right) \left(bR\right) = \left(ab\right)R$. If $a$ is an element of $R$ and $n$ is a nonnegative integer, then $\left(aR\right)^n = a^n R$. Another illustrative example is the case when $R$ is a polynomial ring $k\left[x_1, x_2, \ldots, x_t\right]$ over a commutative ring $k$, and when $\mathfrak{m}$ is the ideal of $R$ generated by all $t$ indeterminates $x_1, x_2, \ldots, x_t$. In this case, the $n$-th power of $\mathfrak{m}$ (for any given $n \geq 0$) is the ideal of $R$ generated by all monomials of degree $n$, so it consists of all polynomials that contain no monomials of degree $< n$. (Such polynomials are said to have a "singular point of multiplicity $\geq t$ at $0$".)

Let me return to the general case. While you haven't asked, let me mention a few more properties of the set of ideals of $R$.

First of all, the monoid of ideals of $R$ is commutative, i.e., any two ideals $U$ and $V$ of $R$ satisfy $UV = VU$.

Second, there is not only a product operation on the set of ideals, but also a sum operation. It is defined as follows: If $U$ and $V$ are two ideals of $R$, then we define their sum $U + V$ to be the ideal of $R$ consisting of all elements of the form $u + v$ with $u \in U$ and $v \in V$. Yes, this is an ideal, as you can easily check. In order to make this definition more similar to the definition of the product $UV$, we could replace the words "consisting of all elements" by "generated by all elements", but this would just needlessly complicate it: We would get the same ideal, because the set of all elements $u + v$ with $u \in U$ and $v \in V$ is already an ideal of $R$.

We have thus defined a sum operation on the set of ideals of $R$. This operation, too, makes this set into a monoid (whose neutral element is the zero ideal $0R = 0$). Again, this monoid is commutative. Better yet: The sum operation and the product operation satisfy the distributivity laws $\left(U+V\right) W = UW + VW$ and $U\left(V+W\right) = UV + UW$ for any three ideals $U$, $V$ and $W$ of $R$; thus, the set of ideals of $R$ (equipped with these two operations) becomes a semiring. This is well-known and often used tacitly when computing with ideals. One consequence of this fact is that, e.g., the binomial formula holds for ideals of $R$ (since it holds in any semiring). That is, if $I$ and $J$ are two ideals of $R$, and if $n$ is a nonnegative integer, then \begin{align} \left(I+J\right)^n = \sum_{k=0}^n \dbinom{n}{k} I^k J^{n-k} \label{darij1.eq.binf1} \tag{1} \end{align} (where the expression "$\dbinom{n}{k} I^k$" means the sum $I^k + I^k + \cdots + I^k$ with $\dbinom{n}{k}$ addends, as in any semiring; this is not the same as $\left\{ \dbinom{n}{k} i \mid i \in I^k \right\}$). Note that the sum operation on the ideals of $R$ is idempotent: i.e., any ideal $U$ of $R$ satisfies $U + U = U$ and therefore $mU = U$ for every positive integer $m$. Thus, the $\dbinom{n}{k} I^k$ on the right hand side of \eqref{darij1.eq.binf1} simplifies to $I^k$. Hence, \eqref{darij1.eq.binf1} rewrites as follows: \begin{align} \left(I+J\right)^n = \sum_{k=0}^n I^k J^{n-k} . \label{darij1.eq.binf2} \tag{2} \end{align}

Let me finally remark that all of this can be generalized. If $A$ is an $R$-algebra, then we can replace ideals of $R$ by $R$-submodules of $A$. These form a monoid with respect to product (with neutral element $R \cdot 1_A$) and a commutative monoid with respect to sum, where products and sums are defined as above. The product operation will be commutative when $A$ is commutative (and sometimes even when it isn't); the distributivity laws also hold.

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  • $\begingroup$ so then, the definition I stated for the power, that is , product of any $d$ monomials in the generating set of the ideal is right, as the degree of the product of $d$ monomials is equal to $d$ times the degree of individual monomials, right? $\endgroup$ – vidyarthi Jul 24 '19 at 12:02
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    $\begingroup$ Yes, assuming that the original monomials all have the same degree. Note that you have to include products that contain equal factors (and for $d \geq 6$, all your products will contain at least some equal factors). $\endgroup$ – darij grinberg Jul 24 '19 at 12:25

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