The Ideal of a Set of Points is a Radical Ideal

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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

$I$

\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:

(2)

\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 $xy \in \mathrm{Rad} (I)$ .

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:

(3)

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

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