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 $K$.

  • $\begingroup$ I haven't studied Hilbert Nullstellensatz yet. Do you have any equivalent argument of the above result that doesn't involve Nullstellensatz? $\endgroup$ – Dbchatto67 Feb 11 at 20:23
  • $\begingroup$ 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 $\endgroup$ – Alan Muniz Feb 11 at 20:27
  • $\begingroup$ Why does $F_1 \cdot F_2 \cdots F_s \mid G\ (\text {over}\ L)?$ $\endgroup$ – Dbchatto67 Feb 11 at 20:29
  • $\begingroup$ Because $G$ belongs to the ideal $\sqrt{(F)} = (F_1 \cdots F_s)$ $\endgroup$ – Alan Muniz Feb 11 at 20:37

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