How Do D&F Derive \(a(x)(1 + x) + b(x)(x^3 - 2) = 1\)?

[Math Amateur]

I am reading Dummit and Foote Ch 13 on Field Theory.

On page 515-516 D&F give a series of basic examples on field extensions - see attachment.

The start to Example (4) reads as follows: (see attachment)

(4) Let $$F = \mathbb{Q}$$ and $$p(x) = x^3 - 2$$, irreducible by Eisenstein. (by Eisenstein?)

Denoting a root of p(x) by $$\theta$$ we obtain the field

$$ \mathbb{Q} [x] / (x^3 -2) $$ $$ \cong $$ $$ \{a + b \theta + c {\theta}^2 | a,b,c \in \mathbb{Q} \} $$

with $${\theta}^3 = 2$$ , an extension of degree 3.

To find the inverse of, say, $$1 + \theta$$ in this field, we can proceed as follows:

By the Euclidean Algorithm in $$\mathbb{Q}[x]$$ there are polynomials a(x) and b(x) with

$$a(x)(1 + x) + b(x)(x^3 - 2) = 1$$

... ... etc etc

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My problem is this:

How do D&F get the equation $$a(x)(1 + x) + b(x)(x^3 - 2) = 1$$?

It looks a bit like they are implying that there is a GCD of 1 between (1 + x) and $$(x^3 - 2)$$ and then use Theorem 4 on page 275 (see attached) relating the Euclidean Algorithm and the GCD of two elements of a Euclidean Domain, but I am not sure and further, not sure why the GCD is 1 anyway.

Can someone please clarify the above for me?

Peter

[This has also been posted on MHF]

 

[Fernando Revilla]

How do D&F get the equation $$a(x)(1 + x) + b(x)(x^3 - 2) = 1$$?

Using the Euclid algorithm: $$\begin{array}{r|*{3}{r}}{}&x^2-x+1&(-1/3)x-1/3\\\hline
{}x^3-2&x+1&-3\\\hline
{}-3&0&
\end{array}$$ So, $-3=(x^3-2)-(x^2-x+1)(x+1).$ We can express:
$$-\frac{1}{3}(x^3-2)+\frac{1}{3}(x^2-x+1)(x+1)=1$$ Then,
$$a (x)=\frac{1}{3}(x^2-x+1),\quad b(x)=-\frac{1}{3}$$

 

[Deveno]

In any Euclidean domain, which is a fortiori a Bezout domain, we can express the GCD gcd(a,b) of two elements d as:

ra + sb

for some elements r,s in the domain.

This is equivalent to saying:

(a) + (b) = (d)

(The sum of the ideal generated by a, and the ideal generated by b, is the ideal generated by d).

This GCD is only unique up to units, of course, but if we have a polynomial ring over a field, we can settle the uniqueness question of gcd(a(x),b(x)) by requiring it be monic as well (recall that in $F[x]$ the units are just the non-zero elements of $F$).

The Euclidean division algorithm (which holds in ANY Euclidean domain) gives a way to find such an r(x),s(x) (in your question these are the a(x) and b(x) polynomials).

In fact, we can find (explicily) an inverse for ANY non-zero element $p(\theta) \in \Bbb Q(\theta)$. By the IRREDUCIBILITY of $x^2 - 2$ over $\Bbb Q$;

$\text{gcd}(p(x),x^3 - 2) = 1$

we have $r(x),s(x) \in \Bbb Q[x]$ with:

$1 = r(x)p(x) + s(x)(x^3 - 2)$

Applying the ring-homomorphism $\Bbb Q[x] \to \Bbb Q(\theta) \cong \Bbb Q[x]/(x^3 - 2)$ which maps $x \mapsto \theta$, we get:

$1 = r(\theta)p(\theta) + s(\theta)(\theta^3 - 2) = r(\theta)p(\theta) + s(\theta)(0) = r(\theta)p(\theta)$, that is:

$p(\theta)^{-1} = r(\theta)$.

In your example, this gives:

$r(\theta) = \frac{1}{3}(1 - \theta + \theta^2)$.

We can verify, by direct computation:

$(1 + \theta)\frac{1}{3}(1 - \theta + \theta^2) = \frac{1}{3}(1 - \theta + \theta^2 + \theta - \theta^2 + \theta^3) = \frac{1}{3}(1 + \theta^3)$

But in $\Bbb Q(\theta), \theta^3 = 2$, so:

$\frac{1}{3}(1 + \theta^3) = \frac{1}{3}(1 + 2) = \frac{1}{3}(3) = 1$.

*********

On your question about the Eisenstein criterion, which is this:

In the ring $\Bbb Q[x]$ if:

$q(x) = a_0 + a_1x + \cdots + a_nx^n \in \Bbb Z[x]$

with $p$ a prime such that:

$p|a_0,a_1,...,a_{n-1}, p \nmid a_n$
$p^2 \nmid a_0$

then $q(x)$ is irreducible over $\Bbb Q$.

The way this is typically proved is reducing $q(x)$ mod $p$ and using the fact that if $q(x)$ factors over $\Bbb Q$ it factors over $\Bbb Z$ (Gauss' lemma).

In your example, we may take $p = 2$, since:

$x^3 - 2 = -2 + 0x + 0x^2 + x^3$, and:

$2|-2, 2|0, 2|0$ and $2 \nmid 1$
$4 \nmid 2$.

This criterion is useful for establishing the irreducibility of many polynomials of the form:

$x^n - a \in \Bbb Z[x]$ over $\Bbb Q$, which yields an extension of the form $\Bbb Q(\sqrt[n]{a})$

(For $n \leq 3$ this also provides a somewhat advanced proof (if the polynomial satisfies the Eisenstein criterion) that $\sqrt[n]{a}$ is irrational. For example, if the integer $a$ is square-free (and greater than 1), Eisenstein is satisfied).