# Generator of Power of Ideal

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.

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.

• 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? Jul 24, 2019 at 12:02
• 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). Jul 24, 2019 at 12:25