# Field Theory - Nicholson - Splitting Fields - Section 6.3 - Example 1

#### Peter

##### Well-known member
MHB Site Helper
I am reading Nicholson: Introduction to Abstract Algebra, Section 6.3 Splitting Fields.

Example 1 reads as follows: (see attachment)

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Example 1. Find an extension [TEX] E \supseteq \mathbb{Z}_2 [/TEX] in which [TEX] f(x) = x^3 + x + 1 [/TEX] factors completely into linear factors.

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Solution. The polynomial f(x) is irreducible over [TEX] \mathbb{Z}_2 [/TEX] (it has no root in [TEX] \mathbb{Z}_2 [/TEX] ) so

[TEX] E = \{ a_0 + a_1 t + a_2 t^2 \ | \ a_i \in \mathbb{Z}_2 , f(t) = 0 \} [/TEX]

is a field containing a root t of f(x).

Hence x + t = x - t is a factor of f(x)

The division algorithm gives [TEX] f(x) = (x+t) g(x) [/TEX] where [TEX] g(x) = x^2 + tx + (1 + t^2) [/TEX]

, so it suffices to show that g(x) also factors completely in E.

Trial and error give [TEX] g(t^2) = 0 [/TEX] so [TEX] g(x) = (x + t^2)(x + v) [/TEX] for some [TEX] v \in F[/TEX].

... ... etc (see attachment)

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My problem is that I cannot show how [TEX] g(t^2) = 0 [/TEX] implies that [TEX] g(x) = (x + t^2)(x + v) [/TEX] for some [TEX] v \in F[/TEX].

I would appreciate some help.

Peter

[Note; This has also been posted on MHF]

#### Deveno

##### Well-known member
MHB Math Scholar

One way is simply to divide $g(x)$ by $x + t^2$:

$g(x) = x^2 + tx + (1+t^2) = (x + t^2)(x + (t+t^2)) + (1 + t^2 + t^3 + t^4)$

$= (x + t^2)(x + (t+t^2)) + (1 + t^3) + t^2 + t(t^3)$

$= (x + t^2)(x + (t+t^2)) + t + t^2 + t(t + 1)$

(since $t^3 + t + 1 = 0$ means $t^3 = -t - 1 = t + 1$, and similarly $t^3 + 1 = -t = t$)

$= (x + t^2)(x + (t+t^2)) + t + t^2 + t + t^2 = (x + t^2)(x + (t+t^2))$

which immediately gives $v = t + t^2$.

There seems to be a typo in the book, it should read: "for some $v \in E$".

A more abstract approach is this:

We have that $g(x)$ has a root in $\Bbb Z_2(t)$, namely $t^2$, so it follows by the division algorithm that $g(x) = (x + t^2)(x + c_0)$ for some $c_0 \in \Bbb Z_2(t)$ (this is because:

$a \in F$ is a root of $f(x) \in F[x] \iff (x - a)|f(x)$ for any field $F$. The other factor must be linear because the degree of $g$ is 2, and the factor $x + t^2$ is linear).

By the uniqueness of this factorization, we can take $v = c_0$.

Multiplying this out, we obtain:

$x^2 + tx + (1 + t^2) = g(x) = x^2 + (t^2 + v)x + t^2v$ yielding:

$t = t^2 + v$
$v = t - t^2 = t + t^2$.

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It might be helpful to keep in mind this basic fact:

for any polynomial $p(x) \in F[x]$ for any field $F$, and any $a \in F$ we have:

$p(x) + q(x)(x - a) + r(x)$ with deg(r) < deg($x-a$) = 1, or $r(x) = 0$, that is:

$r$ is a CONSTANT polynomial. Which one? Clearly, $p(a)$, that is:

$p(x) = (x - a)q(x) + p(a)$.