# Dummit and Foote Example

I found one particular example – namely Example 2 on page 660 of Dummit and Foote – to be a good exercise in the sense that it tied together a lot of ideas from earlier parts of the book. In order to share, though, I need to give a little background.

Throughout, let $k$ denote a field. The affine $n$-space $\mathbb{A}^n$ over $k$ is the set of all $n$-tuples $(k_1,k_2,\ldots,k_n)$ where $k_i\in k$ for all $i$. For a general subset $A\subset \mathbb{A}^n$, the ideal $\mathcal{I}(A)$ is called the ideal of functions vanishing at $A$ and is defined to be

$\mathcal{I}(A)=\{f\in k[x_1,\ldots,x_n]\,:\,f(a_1,\ldots,a_n)=0\text{ for all }(a_1,\ldots,a_n)\in A\}$.

It’s easily verified that $\mathcal{I}(A)$ is indeed an ideal and that it is, by definition, the unique largest ideal of functions which are identically zero on all of $A$. With that in mind, consider the aforementioned example:

Page 660, Example 2. Over any field $k$, the ideal of functions vanishing at $(a_1,\ldots,a_n)\in\mathbb{A}^n$ is a maximal ideal since it is the kernel of a surjective (ring) homomorphism from $k[x_1,\ldots,x_n]$ to the field $k$ given by evaluation at $(a_1,\ldots,a_n)$. It follows that

$\mathcal{I}((a_1,\ldots,a_n))=(x_1-a_1,x_2-a_2,\ldots,x_n-a_n)$.

Proof. As mentioned above, $\mathcal{I}((a_1,\ldots,a_n))$ is certainly an ideal. To show that it’s a maximal ideal in $k[x_1,\ldots,x_n]$, first define $\varphi:k[x_1,\ldots,x_n]\to k$ by the action that sends $f=f(x_1,x_2,\ldots,x_n)$ to the constant $f(a_1,a_2,\ldots,a_n)\in k$. Verifying that this is a ring homomorphism is trivial, and given an element $x$, $x=f(x,0,\ldots,0)$ where $f(x_1,\ldots,x_n)=x_1$ is a polynomial in $k[x_1,\ldots,x_n]$. Moreover, the kernel of $\varphi$ consists precisely of those elements in $f\in k[x_1,\ldots,x_n]$ for which $f(a_1,\ldots,a_n)=0$, whereby it follows that $\ker(\varphi)=\mathcal{I}(A)$ where $A=\{(a_1,\ldots,a_n)\}\subset\mathbb{A}^n$.

Hence, $\varphi$ is a surjective ring homomorphism with kernel $\mathcal{I}(A)$. In particular, by the first (ring) isomorphism theorem, $k[x_1,\ldots,x_n]/\mathcal{I}(A)\cong\text{Im}(\varphi)$ where $\text{Im}(\varphi)=k$ by surjectivity of $\varphi$. Clearly, then, $k[x_1,\ldots,x_n]/\mathcal{I}(A)$ is a field, something that happens if and only if the ideal $\mathcal{I}(A)$ is maximal. Therefore, the first claim is proved.

For the second claim, note that the ideal $(x_1-a_1,\ldots,x_n-a_n)$ is an ideal that vanishes precisely on $A$. Clearly, then, $\mathcal{I}(A)\supseteq (x_1-a_1,\ldots,x_n-a_n)$. On the other hand, an element $f\in\mathcal{I}(A)$ is an element that vanishes at the point $(a_1,\ldots,a_n)$, whereby it follows that

$\displaystyle f(x_1,\ldots,x_n)=g(x_1,\ldots,x_n)\prod_{i=1}^n(x_i-a_i)^{\alpha_i}$

for some positive powers $\alpha_i$. Then, clearly, $f\in (x_1-a_1,\ldots,x_n-a_n)$ and so $\mathcal{I}(A)\subseteq (x_1-a_1,\ldots,x_n-a_n)$. This concludes the proof.   $\square$