Monthly Archives: February 2011

Algebraic geometry sheet 3

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.

Algebraic geometry coursework

Selected solutions to sheet one and sheet two are now available on the coursework webpage for algebraic geometry.

Cubic polynomial

There was a question about a classic exercise: Let f(x)\in Q[x] be a cubic irreducible polynomial, which we take to be monic, and let a, b, c be the roots. Let F=Q(a) and K=Q(a,b,c). Clearly, [F:Q]=3, but the exercise is about the degree of K.

Fact: [K:F] is 1 or 2.

Proof: We have f(x)=(x-a)(x-b)(x-c), so that we can divide f by x-a in F[x]. Therefore, b and c are roots of the quadratic polynomial g(x)=f(x)/(x-a) \in F[x]. Since a+b+c\in Q, we have K=F(b,c)=F[b]. Thus, if b\in F, then [K:F]=1. If b\notin F, then [K:F]=2.

By the tower theorem corresponding to these two possibilities, we have that [K:Q] is 3 or 6.

Fact: [K:Q]=3 if and only if \delta=(b-a)(c-b)(c-a)\in Q.

Proof: Note that \delta^2=(b-a)^2(c-b)^2(c-a)^2\in Q, since it is a symmetric polynomial in a,b,c. Furthermore, Q(\delta)\subset K. If \delta\notin Q, then [Q[\delta):Q]=2 and this degree must divide [K:Q]. So we have [K:Q]=6. On the other hand, suppose \delta\in Q. The polynomial h(x)=f(x+a)\in F[x] has the roots 0, b'=b-a, and c'=c-a. So h(x)/x has roots b' and c'. Therefore,
b'c'\in F. But we are assuming that b'c'(c'-b')=b'c'(c-b)\in Q. So c-b\in F. Since b+c\in F as well, we have b,c\in F and K=F, so that [K:Q]=3.

Minimal polynomial

I am asking a question about Sheet 2 Problem 9 (solutions are provided for up to 8).

Why \cos(\frac{4pi}{7}) and \cos(\frac{6pi}{7}) are conjugates of \alpha=\cos(\frac{2pi}{7}) ?

I’ve managed to find minimal polynomial using DeMoivre’s Theorem and I’ve shown that \cos(\frac{4pi}{7}) , cos(\frac{6pi}{7}) satisfy it (m_{\alpha}=8x^3+4x^2-4x-1). But the question asks to do the opposite: show that they are conjugates and hence find the minimal polynomial of \alpha over \mathbb{Q}.

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In the problem, you have K=Q(\alpha)\subset L=Q(\beta), where \beta=\exp(2\pi i/7). Thus, \alpha= [\beta+\bar{\beta}]/2, where we use the bar to denote the complex conjugate. Now, the key point is this: every embedding of K into C is the restriction of an embedding of L into C, except some of the embeddings will become equal when restricted. Now, L has six embeddings \sigma_j for j=1,\ldots, 6 such that \sigma_j(\beta)=\beta^j. This can be seen by considering the roots of the minimal polynomial for \beta. Since \bar{\beta}=\beta^6, we see that for each embedding, \sigma_j(\bar{\beta})=\sigma_j(\beta^6)=\beta^{6j}=\bar{\beta^j}. Therefore, the conjugates of \alpha are simply [\beta^j+\bar{\beta^j}]/2=\cos(2\pi j/7). Hence, you get exactly the three numbers indicated.

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