Ideals generated by polynomials

Definition

We repeat the definition of a ring ideal:

As a mini example consider the set $I\subseteq\R[x]$ given by

$$I=\{a_p x^p + a_{p+1}x^{p+1}+\ldots + a_{p+m}x^{p+m}, a_{p+i}\in \R, m\in \N, i=1,\ldots, m\}.$$

Clearly, $I$ satisfies the three conditions of the definition.

Another ideal is the set of even polynomials, which can have the form $p=a_0+a_2x^2+a_4x^4+\ldots+a_{2k}x^{2k}$.

We have the following immediate result

Example (trivial ideal). In $k[x]$, the set $\{0\}$ is an ideal. It is generated by the zero polynomial, i.e., $\langle 0\rangle = \{0\}$.

Some observations

When dealing with a system of polynomial equations

$$\begin{aligned} f_1(x_1, \ldots, x_n) {}={}& 0,\\ f_2(x_1, \ldots, x_n) {}={}& 0,\\ \vdots\;\,{}&\\ f_s(x_1, \ldots, x_n) {}={}& 0, \end{aligned}$$

we can obtain a new equations. For example, if we multiply the first by $h_1$, the second by $h_2$, etc and sum up, we get

$$h_1f_1+\ldots+h_sf_s = 0,\tag{2}$$

which means that if $(x_1, \ldots, x_n)$ solves the above equations and $q\in\langle f_1, \ldots, f_s\rangle$, then $q(x_1, \ldots, x_n)=0$.

Basis

Next we show that

Proof. Since $\langle f_1, \ldots, f_s\rangle = \langle g_1, \ldots, g_t\rangle$ every "polynomial combination" of the f's can be written as such a combination of the g's. This means that every one of the f's can be written as a polynomial combination of the g's and vice versa. Now take $x\in \mathbf{V}(f_1, \ldots, f_s)$, i.e., $f_1(x)=0$, $\ldots$, $f_s(x)=0$. We'll show that $\mathbf{V}(g_1, \ldots, g_t)$, i.e., we'll show that $g_j(x)=0$, which is easy. $\Box$

We can show that every polynomial ideal is finitely generated (this is closely related to Noetherian rings).

Exercises

These exercises are from[^1], p35-57. Note that Exercises 3a and 3b can be solved like Exercise 3c (better approach).

Exercise 3a. Show that $\langle x+y, x-y\rangle = \langle x, y\rangle$

Solution. It is

$$\begin{aligned} \langle x+y, x-y\rangle {}={}& \left\{h_1\cdot (x+y) + h_2 \cdot (x-y); h_1, h_2\in k[x, y]\right\} \\ {}={}& \left\{(h_1 + h_2)x + (h_1-h_2)y; h_1, h_2\in k[x, y]\right\} \\ {}={}& \left\{p_1 \cdot x + p_2\cdot y; p_1, p_2\in k[x, y]\right\}. \end{aligned}$$

Here we used the fact that for any polynomials $p_1, p_2$ there are polynomials $h_1, h_2$ st $p_1 = h_1+h_2$ and $p_2 = h_1 - h_2$. Simply take $h_1 = \tfrac{1}{2}p_1 + \tfrac{1}{2}p_2$, and $h_2 = \tfrac{1}{2}p_1 - \tfrac{1}{2}p_2$. $\bullet$

Exercise 3b. Show that $\langle x+xy,y+xy,x^2,y^2\rangle = \langle x, y\rangle$

Solution. We have

$$\begin{aligned} \langle x+xy,y+xy,x^2,y^2\rangle {}={}& \left\{ \left.\begin{array}{l} h_1 {\cdot} (x+xy)+h_2{\cdot} (y+xy) \\ \quad + h_3 {\cdot} x^2 + h_4{\cdot}y^2 \end{array}\;\right|\; {} h_i\in k[x,y]\right\} \\ {}={}& \{xp_x + y p_y; p_x, p_y\in k[x, y]\}, \end{aligned}$$

and we need to show that for every $p_x, p_y$ there are $h_1,\ldots, h_4$ such that $p_x = h_1\cdot(1+y) + yh_2 + xh_3$ and $p_y = xh_1 + (1+x)h_2 +yh_4$. Or,

$$\begin{aligned} p_x ={}& h_1 + xh_3 + y(h_1+h_2),\\ p_y ={}& h_2 + x(h_1+h_2) + yh_4, \end{aligned}$$

and we can choose $h_2=-h_1$, $h_3=h_4=0$, so the $p_x=h_1$, and $p_y=h_2$. $\bullet$

Exercise 3c. Show that $\langle 2x^2 +3y^2 −11,x^2 −y^2 −3\rangle = \langle x^2-4, y^2-1\rangle$

Solution. By Euclidean division of polynomials $$2x^2 +3y^2 −11 = 2(x^2-4) + 3(y^2-1).$$ and $$x^2-y^2 -3 = (x^2-4) - (y^2-1).$$ So, look: we define $f_1=2x^2 +3y^2 −11$, $f_2=x^2-y^2 -3$, $g_1=x^2-4$, $g_2=y^2-1$. We have

$$\begin{aligned} f_1 ={}& 2g_1 + 3g_2, \\ f_2 ={}& g_1 - g_2. \end{aligned}$$

We can solve the system and get

$$\begin{aligned} g_1 ={}& \tfrac{1}{5}f_1 + \tfrac{3}{5}f_2, \\ g_2 ={}& \tfrac{1}{5}f_1 - \tfrac{2}{5}f_2. \end{aligned}$$

For a more detailed exposition, see division of multivariate polynomials.

Exercise 6. Show that $$\mathbf{V}(x+xy,y+xy,x^2,y^2)=\mathbf{V}(x, y)$$

Solution. This is immediate from Exercise 3b and Proposition 1 above. The thing is that if we can find a simpler basis for an ideal, we can also solve simultaneous polynomial equations easier. $\bullet$

Ideal of variety

Let $V$ be an affine variety⁽¹⁾, that is, $V$ is the set of solutions of a set of polynomial equations. We ask what are the polynomial equations that have $V$ as their set of solutions (this is an ideal). We give the following definition.

It's easy to prove that $\mathbf{I}(V)$ is an ideal.

Proof. $0\in I(V)$ because $0(a_1, \ldots, a_n)=0$ for all $(a_1, \ldots, a_n)\in V$. If $f_1, f_2\in \mathbf{I}(V)$ then $(f_1+f_2)(a_1, \ldots, a_n) = f_1(a_1, \ldots, a_n) + f_2(a_1, \ldots, a_n) = 0$ for all $(a_1, \ldots, a_n)\in V$, so $f_1+f_2\in \mathbf{I}(V)$. Lastly, if $f\in \mathbf{I}(V)$ and $h\in k[x_1, \ldots, x_n]$, then $h(a_1, \ldots, a_n)f(a_1, \ldots, a_n) = 0$ because $f(a_1, \ldots, a_n)=0$ for all $(a_1, \ldots, a_n)\in V$. $\Box$

Example (Ideal of a simple variety). Consider the variety $V = \{(0, 0)\}$ in $\R^2$. This is the variety of the polynomials $p_1 = x$ and $p_2=y$. In this case, it turns out that the ideal of $V$ is the ideal generated by $p_1$ and $p_2$ (not always true). We will show that

$$\mathbf{I}(V) = \langle x, y\rangle.$$

Take $f\in \langle x, y\rangle$. This means that $f=a(x, y)x + b(x, y)y$. Then $f(0, 0)=0$, i.e., $f$ is zero on $V$, i.e., $f\in \mathbf{I}(V)$. Therefore, $\langle x, y\rangle \subseteq \mathbf{I}(V)$. For the converse, suppose $f\in \mathbf{I}(V)$ and say $f$ has the general form

$$f = a_{0,0} + \sum_{i,j\neq0, 0}a_{i, j}x^i y^j.$$

Since $f(0, 0)=0$, it follows $a_{0, 0}=0$, so

$$f = x\underbrace{\sum_{i>0, j}a_{i, j}x^{i-1} y^j}_{a(x, y)} + y \underbrace{\sum_{i,j>0}a_{i, j}x^i y^{j-1}}_{b(x, y)},$$

which means that $f\in \langle x, y\rangle$. $\bullet$

Warning: It is generally not true that $\mathbf{I}(\mathbf{V}(f_1, \ldots, f_s)) = \langle f_1, \ldots, f_s\rangle.$

However, we can prove the following inclusion

The proof goes like the first part of the previous example.

Counterexample. We will show that

$$\langle x^2, y^2\rangle \subsetneq \mathbf{I}(\mathbf{V}(x^2, y^2)),$$

where

$$\mathbf{V}(x^2, y^2) = \{(x, y): x^2=0, y^2=0\} = \{(0, 0)\},$$

so $\mathbf{I}(\mathbf{V}(x^2, y^2)) = I(\{0, 0\}) = \langle x, y\rangle$ as we saw in the previous example. It is clear that $x\neq \langle x^2, y^2\rangle$. $\bullet$

The following result is very important:

Proof. (1) We have

$$\begin{aligned} \mathbf{I}(V)={}&\{f\in k[x_1, \ldots, x_n]: f(a_1, \ldots, a_n)=0, \forall (a_1, \ldots, a_n)\in V\} \\ ={}& \bigcap_{(a_1, \ldots, a_n)\in V} \{f\in k[x_1, \ldots, x_n]: f(a_1, \ldots, a_n)=0\}, \end{aligned}$$

from which it is clear that $V\subseteq W$ implies $\mathbf{I}(V) \supseteq \mathbf{I}(W)$. For the converse, assume $\mathbf{I}(V) \supseteq \mathbf{I}(W)$, and $W$ is the set of points $x\in k^n$ such that $g_1(x)=0,\ldots, g_s(x)=0$. Then, $g_1, \ldots, g_s\in \mathbf{I}(W) \subseteq \mathbf{I}(V)$, so the $g_i$ vanish on $V$. In other words, $V$ is an affine variety defined by these $g_i$ and possibly additional polynomials, so $V \subseteq W$.

(2) This is a consequence of (1): $V = W$ means $V \subseteq W$ and $W \subseteq V$; then we apply (1) to show that $\mathbf{I}(V) \supseteq \mathbf{I}(W)$ and $\mathbf{I}(W) \supseteq \mathbf{I}(V)$, so $\mathbf{I}(V) = \mathbf{I}(W)$. $\Box$

Proof. Immediate from Proposition 4.

Univariate polynomials and ideals

Things are particularly simpler in the case of univariate polynomials over a field. There, all ideals are generated by a single element^[2]. Such ideals are called principal.

Proof. If $I$ is trivial, i.e., $I=\{0\}$ we know that $I=\langle 0\rangle$, so let us assume that $I$ is nontrivial, i.e., it contains a nonzero polynomial, $f$, of least degree. We will show that $I = \langle f\rangle$.

Since $f\in I$, it follows that $\langle f\rangle \subseteq I$. We need to show that every $g\in I$ can be written as $g=hf$. By the division algorithm, we can write

$$g = qf + r,$$

where $r=0$, or $\deg r < \deg f$. Since $I$ is an ideal and $f\in I$, $qf\in I$. Therefore, $r = g - qf\in I$. If $r\neq 0$ this would mean that we have found a polynomial $r\in I$ with degree less than the degree of $f$. But we assumed that $f$ is polynomial of least degree. This would be a contradiction, so $r=0$. This means that $g = qf$, so $g\in \langle f\rangle$, and $I = \langle f\rangle$.

References

1. David A. Cox, John Little, Donal O’Shea, Ideals, Varieties and Algorithms: An Introduction to Computational Algebraic Geometry and Commutative Algebra, Springer, 2015
2. TheoremDep, Every ideal of a univariate polynomial ring is principal, accessed on 3 July 2024

Endnotes

1. Let $F$ be a field and $p_1,\ldots, p_s\in F[x_1,\ldots, x_n]$ be multivariate polynomials on $F$. The affine variety of $p_1,\ldots, p_s$ is the set $$\mathbf{V}(p_1,\ldots, p_s) = \{(a_1,\ldots, a_n)\in F^n: p_i(a_1,\ldots, a_n)=0, \forall i\}$$