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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

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It's a degree two extension of $\mathbb F_2$. There's only one of these, right? –  Dylan Moreland Mar 8 '12 at 17:20
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Also, note that $\mathbf F_4$ is often defined as the splitting field of $X^4 - X$ over $\mathbf F_2$, and all elements of $\mathbf F_4$ are roots of this polynomial. And $X^4 - X = X(X^3 - 1) = X(X - 1)(X^2 + X + 1)$. The first two roots that you see there are already in $\mathbf F_2$, and the last factor is irreducible, as you've said. –  Dylan Moreland Mar 8 '12 at 17:26
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Kind of depends on how you define $\mathbb F_4$. –  Thomas Andrews Mar 8 '12 at 17:29

2 Answers 2

up vote 5 down vote accepted

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 $\mathbb F_4$ :)

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I feel rather silly now. Thanks a lot! –  Matt Mar 8 '12 at 17:22

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 $\ldots$

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