## Category Archives: 3704

### Lecture notes on class groups

I was wondering whether the notes online are the complete set? I have noticed that towards the end there are certain chapters which do not have any notes e.g. Lecture 24, 25, 26. Is the content of those topics
essential and if yes, then will there be material provided to cover those areas?

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Yes. The material for those chapters are in the optional coursework sheet 6 and the supplementary material `Some sample class groups.’

### Revision sessions, 3704 and M211

There will be revisions sessions for Math 3704 and Math M211 as follows:

3704
Wednesday, 20 April, 10:00-12:00 in Math 706

M211
Wednesday, 20 April, 14:00-16:00 in Math 706.

Because of the many holidays towards the end of the month, it was hard to find any other days possible. I hope many of you can come.

### Question on coursework sheet 5, number 6

A question regarding the optional coursework, and my reply.

### 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.

### Inverses in algebraic number fields

There is one thing that you haven’t discussed in detail in the lectures is the computation of the inverse of any given nonzero element alpha in an algebraic number field. In question 2 of Exercise Sheet 2, where the number field is $k = \mathbb{Q}[X]/(x^3-2),$ isomorphic to $F = \mathbb{Q}(2^{1/3}),$ I managed to work out that the inverse of the element $X + (f) \in k$, is $(1/2)x^2 + (f),$ by calculating in $F,$ where the basis is $\{1, 2^{1/3}, 2^{2/3}\}.$ But for the other one, $x + x^2 + (f),$ I could not find it. In this field $k$, which we can think of also as $F,$ we know any element beta is some linear combination $a + b(2^{1/3}) + c(2^{2/3})$ of the basis elements, for unique rationals $a,b,c.$ The element $x + x^2 + (f) \in k$ uniquely corresponds to the element $2^{1/3} + 2^{2/3} \in F,$ and to find the inverse of $x + x^2 + (f)$ we need to find rationals $a,b,c$ such that the inverse $(2^{1/3}+2^{2/3})^{-1},$ which is also an element of $F,$ is the linear combination $a +b2^{1/3}+ c2^{2/3}.$ I stopped there, but could not continue further.

The same kind of question is asked in question 3. Is there a general method for computing the inverse of a given element in algebraic number fields $Q[x]/(f)$?

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The general method uses Bezout’s lemma. Given two relatively prime polynomials $f$ and $g$, there are polynomials $a$ and $b$ such that

$af+bg=1$

Furthermore, $a$ and $b$ can be computed explicitly by using Euclid’s algorithm for polynomials. Once you know how to do this, you can take a non-zero class $[g]\in Q[x]/(f)$. Since $f$ is irreducible, $f$ and $g$ must be relatively prime. Then computing $a$ and $b$ as above will give the equation

$[b][g]=1$

in $Q[x]/(f)$.

Try the problems out again from this point of view, and ask again if you get stuck. The ease of doing this kind of polynomial algebra is one of the main reasons that the ‘quotient of polynomial ring’ approach is often more useful than the ‘subfield of $\mathbb{C}$‘ approach to algebraic number fields.

### Beginning questions in number theory

I have just a couple of quick questions about the lecture yesterday. In the notes you have lectures 1 to 3, and are these the notes for the 3 hour lecture yesterday?
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The material is more or less the same, but I am presenting the material in different order. Eventually, I will put up a new set where the order is more consistent with the lectures.

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Also, I am a bit unclear about the link between roots, factors and reducibility of polynomials in $Q[x].$ Does reducibility basically mean existence of roots and vice versa?

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A non-constant polynomial $f \in Q[x]$ is reducible if it can be written as $f=gh$ where both $g$ and $h$ have degree strictly less than the degree of $f$. This will be true if $f$ has a root $a\in Q$, since then $f(x)=(x-a)h(x)$ for some $h$. However, this isn’t the only way to be reducible. For example,

$(x^2+1)(x^2-2)$

is clearly reducible in $Q[x]$, but it has no root.

That is, if $f$ is irreduble, then it has no root. But the converse is false in general.

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For degree 2 or 3 polynomials $p(x)$ in $Q[x]$ you said that $p(x)$ is irreducible if it has no roots in $Q$. Is this also true for higher degree? We know by Fundamental Theorem of Algebra that any nonconstant polynomial in $C[x]$ has roots in $C$, and therefore the only irreducible polynomials in $C[x]$ are those which are nonconstant of degree 1. If a polynomial $p(x)$ of some degree $d >1$ in $Q(x)$ has all roots in $Q$, then it can be written as the product of some $k \leq d$ factors with roots in $Q$, and is therefore reducible in $Q[x]$.

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As mentioned above,

$f\in Q[x]$ is irreducible if and only if it has no root in $Q$

is *false* in general, as the degree 4 example I’ve given above illustrates. But

$f\in Q[x]$ of degree 2 or degree three is irreducible if and only if it has no root in $Q$

is true. This is because in this case, the only way to be reducible if for it to have a linear factor, and hence, a root.

As you note, over an algebraically closed field like $C$, the only irreducible polynomials are the linear ones.

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Also in $R[x]$, some polynomials like $p(x) = x^3+x^2+x+1 = (x + 1)(x^2 + 1)$ have one real root, but two complex roots. So in this situation, if the polynomial has some of roots as real, then we cannot still say that it is reducible in $R[x]$?

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The polynomial you give is indeed reducible in $R[x]$. In fact, you can check easily that any polynomial of degree $\geq 3$ is reducible in $R[x]$. For odd degree polynomials, this is easily seen because of the existence of a real root. For even degree ones without a real root, factorize first over $C$, and observe that the roots occur in complex conjugate pairs, which can be combined to form quadratic factors.

### Addition to the revision session

A number of students have expressed concern about an exam during the 3704 revision session tomorrow. For them, I will also have an extra office hour tomorrow from 5 to 6 pm in the 5th floor common room.

### 3704 Office hours

Because people haven’t been coming to my office hours, I thought I’d have the one on the coming Tuesday, 26 April, in the 5th floor common room. I’ll be there from 5 to 6.

### Algebraic number theory revision

There will be a revision session for algebraic number theory on Tuesday, 4 May, 2-4 pm in room 505. By this time, you should have read the notes several times and seriously attempted all practice questions.