Confirm/Critique Solution to Dummit and Foote, Section 13.2: Exercise 2

[Math Amateur]

I am unsure of my approach to Exercise 2 Dummit and Foote, Section 13.2 : Algebraic Extensions ..

I am therefore posting my solution to the part of the exercise dealing with the polynomial \(\displaystyle g(x) = x^2 + x + 1\) and the field \(\displaystyle F = \mathbb{F}_2\) ... ...

Can someone please confirm my solution is correct or critique my solution pointing out shortcomings ... Exercise 2 of Dummit and Foote, Section 13.2 reads as follows:
View attachment 6610Now, we need to obtain a field by adjoining a root of \(\displaystyle g(x)\) [see above] to \(\displaystyle \mathbb{F}_2\) ...We have that the polynomial \(\displaystyle g(x) = x^2 + x + 1\) is irreducible in \(\displaystyle \mathbb{F}_2 [x]\) since it does not have a root in \(\displaystyle \mathbb{F}_2\) ( since \(\displaystyle g(0) = 1\) and \(\displaystyle g(1) = 1\))Adjoin a root \(\displaystyle \theta\) to \(\displaystyle \mathbb{F}_2\) by considering an extension \(\displaystyle K\) to \(\displaystyle \mathbb{F}_2\) ... ...Take \(\displaystyle K = \mathbb{F}_2 [x] / (g(x))\) ... then \(\displaystyle \theta = x \text{ mod } ( g(x) )\) is a root of \(\displaystyle g(x)\) in \(\displaystyle K\) ... and we have [Corollary 7, D&F, Section 13.1] ...

\(\displaystyle K = \{ a + b \theta \ | \ a,b \in \mathbb{F}_2 \}\) ... ...

... where \(\displaystyle \theta^2 + \theta + 1 = 0 \)

or

\(\displaystyle \theta^2 = - \theta - 1 = \theta + 1\)
Now by giving \(\displaystyle a,b\) their possible values in \(\displaystyle \mathbb{F}_2\) we find that the elements of \(\displaystyle K\) are \(\displaystyle \{ 0, 1, \theta, \theta + 1 \}\)

Proceeding to calculate the multiplication table we get ...

\begin{tikzpicture}
\usetikzlibrary{matrix}
\matrix (m) [nodes={minimum width=3em,minimum height=2ex},matrix of nodes]
{
$\times$ & 0 & 1 & $\theta$ & $\theta + 1$ \\
0&0&0&0&0\\
1&0&1& $\theta$ & $\theta + 1$ \\
$\theta$ & 0 & $\theta$ & $\theta + 1$ & 1 \\
$\theta$ +1 & 0 & $\theta + 1$ & 1 & $\theta$ \\
};
\draw[very thick] (m-1-1.north east) -- (m-5-1.south east);
\draw[very thick] (m-1-1.south west) -- (m-1-5.south east);
\foreach \x in {2,...,5}{
\draw (m-1-\x.north east) -- (m-5-\x.south east);
\draw (m-\x-1.south west) -- (m-\x-5.south east);
}
\end{tikzpicture}

Can someone please confirm that the above analysis/calculations are correct and/or point out errors or shortcomings i.e. critique the analysis ...Peter
NOTE: If someone can help me with pointing out the error in the Latex for the display of my multiplication table I would be most gratefulSpecial thanks ! to I like Serena for editing this post so that the multiplication table displayed properly!

 

[jvicens]

Hello Peter,

Thank you for sharing your solution to Exercise 2 of Dummit and Foote, Section 13.2. Your approach seems to be correct and your calculations are also accurate. However, I would like to point out a few things for your consideration:

1. In your solution, you have written that g(x) = x^2 + x + 1 is irreducible in \mathbb{F}_2 [x] because it does not have a root in \mathbb{F}_2. While this is true, it would be helpful to also mention that g(x) is a quadratic polynomial and hence, it cannot be irreducible in \mathbb{F}_2 [x] if it has a linear factor. This can be easily seen by using the fact that any linear polynomial over a field is irreducible if and only if it does not have a root in that field.

2. In your solution, you have taken K = \mathbb{F}_2 [x] / (g(x)) to be the field obtained by adjoining a root of g(x) to \mathbb{F}_2. However, it would be more accurate to say that K is an extension of \mathbb{F}_2 obtained by adjoining a root of g(x) to \mathbb{F}_2. This is because K may not necessarily be a field, unless g(x) is a irreducible polynomial in \mathbb{F}_2 [x].

3. In the multiplication table that you have given, the element \theta + 1 should be in the second row and third column, and not in the fourth column. This is because \theta + 1 is the product of \theta and \theta + 1, and not \theta + 1 and 1.

Overall, your solution seems to be correct and well-written. I hope my suggestions are helpful to you. Keep up the good work!