F_2(α) isomorphic to F_2[x]/<f(x)>

[rukawakaede]

Homework Statement

Show that $$\mathbb{F}_2(\alpha)$$ is isomorphic to $$\mathbb{F}_2[x]/ <f(x)>$$.

where $$f(x)\in\mathcal{F}_2[x]$$ and $$E\supseteq\mathcal{F}_2$$ is an extension of $$\mathbb{F}_2$$ such that $$f(x)$$ has a root $$\alpha\in E$$. Also $$\mathbb{F}_2(\alpha)$$ is the subfield $$E$$ generated by $$\mathbb{F}_2$$ and $$\alpha$$.

Homework Equations

The Attempt at a Solution

Could anyone give me some direction on how to start this proof?

 

[micromass]

What about the first isomorphism theorem?

Show that $$\mathbb{F}_2[X]\rightarrow \mathbb{F}_2[\alpha]$$ is a surjection and find it's kernel...

 

[rukawakaede]

What about the first isomorphism theorem?

Show that $$\mathbb{F}_2[X]\rightarrow \mathbb{F}_2[\alpha]$$ is a surjection and find it's kernel...

is it ok to say $$\mathbb{F}_2[\alpha]=\mathbb{F}_2(\alpha)$$?

 

[micromass]

is it ok to say $$\mathbb{F}_2[\alpha]=\mathbb{F}_2(\alpha)$$?

I'm sorry, I missed that. But now that you mention it, it seems that it is not correct what you're trying to prove: Take f(X)=(X-1)². Then F[X]/(X-1)² is not a field and can thus not be isomorphic to a field.