0
$\begingroup$

What's the easiest way to see why $\mathbb F_2 [X] / \langle X^2 + X + 1 \rangle \cong \mathbb F_4$?

The polynomial is irreducible in $\mathbb F_2 [X]$, but that's about the only observation I've made...

Thanks

  • 1
    Kind of depends on how you define $\mathbb F_4$.2012-03-08

2 Answers 2

5

Since the polynomial is irreducible, the quotient is a field. Since it has degree two, the dimension over $\mathbb F_2$ of the quotient is two.

Now, there is exactly one extension of $\mathbb F_2$ of dimension two over $\mathbb F_2$, namely $\mathbb F_4$, so the quotient has to be $\mathbb F_4$ :)

  • 0
    I feel rather silly now. Thanks a lot!2012-03-08
1

Hint $\ $ Map $\rm\mathbb F_2[X]\:$ into $\mathbb F_4$ by evaluating $\rm\:X\:$ at $\mathbb \alpha \not\in \mathbb F_2$. The kernel is generated by the minimal polynomial of $\alpha,\:$ which must be an irreducible quadratic over $\mathbb F_2,\:$ so must be $\ldots$