February 18, 2011 – 1:12 pm
Since sheet 3 of the algebraic geometry coursework is a bit difficult, I am setting a new due date of 28 February. We will discuss some of the problems on Monday, 21 February.

February 16, 2011 – 1:56 pm
Selected solutions to sheet one and sheet two are now available on the coursework webpage for algebraic geometry.

February 10, 2011 – 3:45 pm
There was a question about a classic exercise: Let be a cubic irreducible polynomial, which we take to be monic, and let be the roots. Let and . Clearly, , but the exercise is about the degree of .

Fact: is 1 or 2.

Proof: We have , so that we can divide by in . Therefore, and are roots of the quadratic polynomial . Since , we have . Thus, if , then . If , then .

By the tower theorem corresponding to these two possibilities, we have that is 3 or 6.

Fact: if and only if .

Proof: Note that , since it is a symmetric polynomial in . Furthermore, . If , then and this degree must divide . So we have . On the other hand, suppose . The polynomial has the roots , and . So has roots and . Therefore,

. But we are assuming that . So . Since as well, we have and , so that .

February 1, 2011 – 10:50 am
I am asking a question about Sheet 2 Problem 9 (solutions are provided for up to 8).

Why and are conjugates of ?

I’ve managed to find minimal polynomial using DeMoivre’s Theorem and I’ve shown that , satisfy it (). But the question asks to do the opposite: show that they are conjugates and hence find the minimal polynomial of over .

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In the problem, you have , where Thus, , where we use the bar to denote the complex conjugate. Now, the key point is this: every embedding of into is the restriction of an embedding of into , except some of the embeddings will become equal when restricted. Now, has six embeddings for such that . This can be seen by considering the roots of the minimal polynomial for . Since , we see that for each embedding, . Therefore, the conjugates of are simply . Hence, you get exactly the three numbers indicated.

By the way, the minimal polynomial needs to be monic.