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I'm not sure how to show this:

Show the field $\mathbb{F}_4[X] /(X^2 +X + \alpha)$ is isomorphic to the field $\mathbb{F}_{16}=\mathbb{F}_2[x]/(x^4 + x^3 + 1)$ by explicitly constructing an isomorphism between the two fields. Let $\alpha$ be the element (0 1) in $\mathbb{F}_4$, i.e., $\alpha$ = $x$ in polynomial form.

Any help or hint would be helpful.

mathmajor9015
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  • You need to tell us what $\alpha$ is. We can guess, but... Here I factor $x^4+x+1$ over $\Bbb{F}_4$ (see the last part of the answer). My $\beta$ is probably your $\alpha$. Observe that $x^4+x+1$ and $x^4+x^3+1$ are reciprocals of each other. Do you know how to take advantage of that? – Jyrki Lahtonen Nov 08 '16 at 05:01

1 Answers1

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Writing

Let $a$ be the element $(0,1)$

is more confusing than anything. Just write $\alpha$ and explicitly state that $\mathbb{F}_4$ has teh elements $0,1,\alpha, \alpha+1$ or the likes, it is clearer.

Now onto the question, the question is about showing an explicit homomorphism, let $\beta$ be the root to $X^2+X+\alpha=0$, as such we have all elements in $\Bbb F_4/(X^2+X+\alpha)$ in the form of $n\beta+m$ with $m,n\in\Bbb F_4$. Which we can make into $$n_1+n_2\alpha+n_3\beta+n_4\alpha\beta$$ with $n_i\in\Bbb F_2$. From this do we have a basis and now we just to mape it properly. For $\Bbb F_{16}$ we have $\gamma^i$ being the basis, with $\gamma^4=\gamma+1$. Then we just need to find which element has that property and map it to our $\gamma$. A quick check gives us that $\beta^4=\beta+1$ and as such we have that $\beta\mapsto\gamma$. Next we remember that $\beta^2=\beta+\alpha$ which gives $\beta^2+\beta=\alpha$ and as such $\alpha\mapsto\gamma^2+\gamma$ and finally $\alpha\beta\mapsto\gamma^3+\gamma^2$.

You need to show this is a homomorphism. We can do this in iterations of all permutations which I do for simplicity, Addition is trivial and I'll do one of the multiplication pairs. $$\varphi(\alpha\cdot\alpha\beta)=\varphi(\alpha\beta+\beta)=\gamma^3+\gamma^2+\gamma=$$ $$\gamma^3+\gamma(\gamma+1)=\gamma^3\gamma^4+\gamma^5+\gamma^4=(\gamma^2+\gamma)(\gamma^3+\gamma^2)=\varphi(\alpha)\cdot\varphi(\alpha\beta)$$

Alternatively can you show that the powers of $\beta$ gives the others.

After that it is trivial showing it is surjective and injective.

Zelos Malum
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  • Otherwise you get to the point (+1), but the OP specifically asked for an isomorphism with $\Bbb{F}_{16}=\Bbb{F}_2[\delta]$, where $\delta$ has minimal polynomial $x^4+x^3+1$. The remedy is to (see my comment under the OP) observe that we have an isomorphism $\gamma\to\delta^{-1}$, and compose your isomorphism with that. – Jyrki Lahtonen Nov 16 '16 at 11:23
  • It's a neat way yes, I wanted to show a brute force reasoning method :) – Zelos Malum Nov 18 '16 at 07:04