Exercise 4.3.18 (Elements of field extension)

Question

How many elements does the field $\mathbb{F}_3(\sqrt 2)$ have? What about $\mathbb{F}_2(\alpha)$, where $\alpha$ is a root of $1+t+t^2$?

Hassaan Naeem · Accepted Answer

extbf{Solution:}
We know that $\mathbb{F}_3(\sqrt 2)$ can be constructed as $\mathbb{F}_3[t]/\langle t^2 - 2 angle$. Hence, any element of the field has the form
$a_0 + a_1t + \langle t^2 - 2angle$ with $a_i \in \mathbb{F}_3$. Hence, there are $3^2 = 9$ elements.
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In a similar manner, we know that $\mathbb{F}_2(\alpha)$ can be constructed as $\mathbb{F}_2[t]/\langle t^2+t+1 angle$. Hence any element of the field has the form
$a_0 + a_1t + \langle t^2+t+1angle$ with $a_i \in \mathbb{F}_2$. Hence there are $2^2 = 4$ elements.

Exercise 4.3.18 (Elements of field extension)

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Exercise 4.3.18 (Elements of field extension)

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