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

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One thought on “Dummit and Foote Example

  1. Pingback: Verifying Easy Properties, or Nowhere, Going Nowhere | riemannian hunger

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