The Ideal of a Set of Points is a Radical Ideal

# The Ideal of a Set of Points is a Radical Ideal

Recall from the The Ideal of a Set of Points page that if $K$ is a field then the ideal of $X$ is defined as:

(1)\begin{align} \quad I(X) = \{ F \in K[x_1, x_2, ..., x_n] : F(\mathbf{p}) = 0, \: \forall \mathbf{p} \in X \} \end{align}

We will shortly observe that the ideals $I(X)$ are always radical ideals. We first define what it means for an ideal to be a radical ideal.

Definition: Let $R$ be a ring and let $I$ be an ideal. The Radical of $I$ is $\mathrm{Rad} (I) = \{ a \in R : a^n \in I, \: \mathrm{for \: some \:} n \in \mathbb{N} \}$. An ideal $I$ is said to be a Radical Ideal if $I = \mathrm{Rad} (I)$. |

We first verify that a radical ideal is indeed an ideal.

Theorem 1: Let $R$ be a ring and let $I$ be an ideal. Then $\mathrm{Rad} (I)$ is an ideal. |

**Proof:**Let $x, y \in \mathrm{Rad} (I)$. Then there exists $m, n \in \mathbb{N}$ such that $x^m \in I$ and $y^n \in I$. Let $t = m + n + 1$. Then we have that:

\begin{align} \quad (x + y)^t = (x + y)^{m + n + 1} \quad (*) \end{align}

- Observe that each term in the expansion $(*)$ has a term of the form $x^iy^j$ where either $i > m$ or $j > n$ (for if $i < m$ and $j < n$ then $t = i + j < m + n < m + n + 1 = t$ which is a contradiction). So each term $x^iy^j \in I$ since $I$ is an ideal. So $(x + y)^t \in I$. Hence $(x + y) \in \mathrm{Rad} (I)$.

- Now let $x \in \mathrm{Rad} (I)$ and $y \in R$. Then there exists an $n \in \mathbb{N}$ such that $x^n \in I$. But $(xy)^n = x^ny^n$ where $x^n, y^n \in I$, so $(xy)^n \in I$. Thus $xy \in \mathrm{Rad} (I)$.

- Therefore $\mathrm{Rad} (I)$ is an ideal. $\blacksquare$

Theorem 2: Let $K$ be a field and let $X \subseteq \mathbb{A}^n(K)$. Then $I(X)$ is a radical ideal. |

**Proof:**We aim to show that $I(X) = \mathrm{Rad} (I(X))$. Clearly by definition we have that $I(X) \subseteq \mathrm{Rad}(I(X))$ since if $F \in K[x_1, x_2, ..., x_n]$ we have for $n = 1$ that $F^1 \in I$, so $F \in \mathrm{Rad}(I(X))$.

- Now let $F \in \mathrm{Rad}(I(X))$. Then for some $n \in \mathbb{N}$ we have that $F^n \in I(X)$.So $F^n(\mathbf{p}) = 0$ for all $\mathbf{p} \in X$. But this can only happen if $F(\mathbf{p}) = 0$ for all $\mathbf{p} \in X$. So $F \in I(X)$. Hence $I(X) \supseteq \mathrm{Rad}(I(X))$. Thus:

\begin{align} \quad I(X) = \mathrm{Rad} (I(X)) \end{align}

- Therefore $I(X)$ is a radical ideal $\blacksquare$.