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 denote a field. The affine -space over is the set of all -tuples where for all . For a general subset , the ideal is called the ideal of functions vanishing at and is defined to be
It’s easily verified that is indeed an ideal and that it is, by definition, the unique largest ideal of functions which are identically zero on all of . With that in mind, consider the aforementioned example:
Page 660, Example 2. Over any field , the ideal of functions vanishing at is a maximal ideal since it is the kernel of a surjective (ring) homomorphism from to the field given by evaluation at . It follows that
Proof. As mentioned above, is certainly an ideal. To show that it’s a maximal ideal in , first define by the action that sends to the constant . Verifying that this is a ring homomorphism is trivial, and given an element , where is a polynomial in . Moreover, the kernel of consists precisely of those elements in for which , whereby it follows that where .
Hence, is a surjective ring homomorphism with kernel . In particular, by the first (ring) isomorphism theorem, where by surjectivity of . Clearly, then, is a field, something that happens if and only if the ideal is maximal. Therefore, the first claim is proved.
For the second claim, note that the ideal is an ideal that vanishes precisely on . Clearly, then, . On the other hand, an element is an element that vanishes at the point , whereby it follows that
for some positive powers . Then, clearly, and so . This concludes the proof.