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

• Oct 19th 2013, 02:23 AM
Bernhard
Field Theory - Nicholson - Splitting Fields - Section 6.3 - Example 1
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 $\displaystyle E \supseteq \mathbb{Z}_2$ in which $\displaystyle f(x) = x^3 + x + 1$ factors completely into linear factors.

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The solution reads as follows:

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Solution. The polynomial f(x) is irreducible over $\displaystyle \mathbb{Z}_2$ (it has no root in $\displaystyle \mathbb{Z}_2$ ) so

$\displaystyle E = \{ a_0 + a_1 t + a_2 t^2 \ | \ a_i \in \mathbb{Z}_2 , f(t) = 0 \}$

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 $\displaystyle f(x) = (x+t) g(x)$ where $\displaystyle g(x) = x^2 + tx + (1 + t^2)$

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

Trial and error give $\displaystyle g(t^2) = 0$ so $\displaystyle g(x) = (x + t^2)(x + v)$ for some $\displaystyle v \in F$.

... ... etc (see attachment)

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

I would appreciate some help.

Peter