## 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.