# Why is the Kernel of the Evaluation Map Principal?

• MHB
• Euge
In summary, the kernel of the evaluation map is principal because it represents the ideal element in the ring of functions that vanishes on a given subset of the space. This ideal element is unique and generates the kernel, making it principal. This relationship between the kernel and principal ideals is essential in understanding the structure of algebraic varieties. The principal nature of the kernel of the evaluation map allows for a straightforward representation of the algebraic variety and provides a useful tool for studying the properties and behavior of functions on the variety. However, in some cases, the kernel of the evaluation map can be non-principal, occurring when the algebraic variety has singular points and the evaluation map is not surjective. The principal nature of the kernel of the evaluation map
Euge
Gold Member
MHB
POTW Director
Here is this week's POTW:

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Let $F$ be the field of fractions of a unique factorization domain $A$, and let $L$ be an algebraic extension field of $F$. Fix $c\in L$. Prove that the kernel of the evaluation map $\operatorname{ev}_c : A[x] \to L$ is principal.-----

No one answered this week’s problem. You can read my solution below.

Since $L$ is an algebraic extension of $F$, $c$ has a minimal polynomial $p(x) \in F[x]$. Fix $f \in \operatorname{ker}(\operatorname{ev}_c)$. By the Euclidean algorithm, $f(x) = p(x) q(x) + r(x)$ for some polynomials $q(x)$ and $r(x)$ in $F[x]$ with $\operatorname{deg} r < \operatorname{deg} q$. Now $r(c) = f(c) - p(c)q(c) = 0 - 0q(c) = 0$, so by minimality of $p$, $r \equiv 0$. Thus $f(x) = p(x)q(x)$ in $F[x]$. Write $p(x) = u p’(x)$ where $u$ is nonzero in $F$ and $p’(x)\in A[x]$ is primitive. By Gauss’s lemma, $f$ is in the principal ideal (in $A[x]$) generated by $p’$. As $p’(c) = 0$, $\operatorname{ker}(\operatorname{ev}_c) = (p’)$.

## 1. Why is the Kernel of the Evaluation Map Principal?

The kernel of the evaluation map is principal because it represents the ideal element in the ring of functions that vanishes on a given subset of the space. This ideal element is unique and generates the kernel, making it principal.

## 2. How does the Kernel of the Evaluation Map relate to principal ideals?

The kernel of the evaluation map is a principal ideal generated by a single element, which is the identity function. This relationship between the kernel and principal ideals is essential in understanding the structure of algebraic varieties.

## 3. What is the significance of the Kernel of the Evaluation Map being principal?

The principal nature of the kernel of the evaluation map allows for a straightforward representation of the algebraic variety. It also provides a useful tool for studying the properties and behavior of functions on the variety.

## 4. Can the Kernel of the Evaluation Map ever be non-principal?

Yes, in some cases, the kernel of the evaluation map can be non-principal. This occurs when the algebraic variety has singular points, and the evaluation map is not surjective. In these cases, the kernel is not generated by a single element and is therefore non-principal.

## 5. How does the principal Kernel of the Evaluation Map affect the theory of algebraic geometry?

The principal nature of the kernel of the evaluation map has a significant impact on the theory of algebraic geometry. It allows for a more straightforward understanding and study of algebraic varieties and their properties, leading to a deeper understanding of the underlying mathematical concepts.

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