Ideals of Zero Locus of function is generated by function.

[ay46]

Let $f(x) \in k[x]$. Show that $\mathcal{I}(\mathcal{Z}(f)) = \langle f \rangle$ if and only if $f$ is the product of distinct linear factors in $k[x]$.

Here, $\mathcal{Z}$ is the zero locus and $\mathcal{I}$ is the ideal generated. This is a question from Dummit and Foote Chapter 15.1.19.

 

[jvicens]

Thank you for your question. I am happy to provide a proof of this statement for you.

First, let us define the notation $\mathcal{I}(\mathcal{Z}(f)) = \langle f \rangle$ to mean that the ideal generated by the zero locus of $f$ is equal to the ideal generated by $f$ itself. In other words, every element in the zero locus of $f$ can be written as a multiple of $f$, and every element in the ideal generated by $f$ can be written as a linear combination of elements in the zero locus of $f$.

Now, let us assume that $f$ is the product of distinct linear factors in $k[x]$. This means that $f$ can be written as $f = (x-a_1)(x-a_2)\cdots(x-a_n)$, where $a_i \in k$ are distinct elements. We want to show that $\mathcal{I}(\mathcal{Z}(f)) = \langle f \rangle$.

First, let us consider an element $g \in \mathcal{I}(\mathcal{Z}(f))$. This means that $g$ is a polynomial that vanishes on the zero locus of $f$. In other words, $g(a_i) = 0$ for all $a_i \in \mathcal{Z}(f)$. Since $f$ is the product of distinct linear factors, we can write $g$ as $g = (x-a_1)^{m_1}(x-a_2)^{m_2}\cdots(x-a_n)^{m_n}$, where $m_i \geq 1$ are the multiplicities of $a_i$ in the zero locus of $f$.

Now, we can use the fact that $a_i$ are distinct to rewrite $g$ as $g = (x-a_1)^{m_1}(x-a_2)^{m_2}\cdots(x-a_n)^{m_n} = (x-a_1)^{m_1}(x-a_2)^{m_2-1}\cdots(x-a_n)^{m_n-1}(x-a_i)$ for some $i \in \{1,2,\dots,n\}$. Therefore, we can see that $g$