# Prove that $\mathcal {I} (H) = (F_1 \cdot F_2 \cdots F_s).$

Let $$\Bbb A^n (L)$$ be an $$n$$-dimensional affine space over the field $$L.$$ Let $$K (\subseteq L)$$ be a subfield of $$L.$$ For any $$V \subseteq \Bbb A^n(L)$$ define $$\mathcal I (V)$$ as the set of all $$F \in K[X_1,X_2,\cdots,X_n]$$ with $$F(X_1,X_2,\cdots,X_n) = 0$$ for all $$(X_1,X_2,\cdots,X_n) \in V.$$ Then $$\mathcal I(V)$$ is an ideal of $$K[X_1,X_2,\cdots,X_n],$$ known as the vanishing ideal of $$V$$ in $$K[X_1,X_2,\cdots,X_n].$$

Proposition $$:$$

Let $$L$$ be an algebraically closed field and $$n \geq 1.$$ Let $$H \subseteq \Bbb A^n (L)$$ be a $$K$$-hypersurface defined by an equation $$F=0$$ and let $$F=c \cdot {F_1}^{\alpha_1} \cdot {F_2}^{\alpha_2} \cdots F_s^{\alpha_s}$$ be a decomposition of $$F$$ into powers of different unassociated irreducible polynomials $$F_i\ (c \in K^{\times}).$$ Then $$\mathcal I(H) = (F_1\cdot F_2 \cdots F_s).$$

Now it is easy to prove that $$(F_1\cdot F_2 \cdots F_s) \subseteq \mathcal I (H).$$ But I find difficulty to prove the other way round. How can I prove the reverse inclusion? Please help me in this regard.

Thank you very much.

Using the Nullstellensatz for $$L$$ (since it is algebraically closed) we have that the ideal of $$H$$ is generated by $$F_1\dots F_s$$ over $$L$$. However these are polynomials over $$K$$ and the result follows.
Indeed, suppose that $$G \in K[X_, \dots, X_n]$$ belongs to $$\mathcal{I}(H)$$ then we look to $$G$$ as a polynomial over $$L$$ and we have that $$F_1\dots F_s \mid G$$ (over $$L$$) and the same holds over $$K$$.
• I only have used that, over an algebraic closed field, if you have variety $V$ defined by an ideal $J$ then $\mathcal{I}(V) = \sqrt{J}$. You can prove this for principal ideals only and you are done – Alan Muniz Feb 11 at 20:27
• Why does $F_1 \cdot F_2 \cdots F_s \mid G\ (\text {over}\ L)?$ – Dbchatto67 Feb 11 at 20:29
• Because $G$ belongs to the ideal $\sqrt{(F)} = (F_1 \cdots F_s)$ – Alan Muniz Feb 11 at 20:37