## Monthly Archives: January 2011

### 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.

### Homework 1

Since there may be some delay in obtaining the textbook for the course, I will postpone the due date of the first homework for M211 to 24 January.

### 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.