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How can I prove that $\mathbb F_4 \otimes_{\mathbb F_2}\mathbb F_8$ is isomorphic to $\mathbb F_{64}.$ I don't even know how to approach. I need some help. Thanks.

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    I want to write $\mathbb F_4\otimes_{\mathbb F_2}\mathbb F_2[x]/(p(x))\cong \mathbb F_4[x]/(p(x))\cong\mathbb F_{64}$ for a degree $3$ polynomial in $\mathbb F_2[x]$ which is irreducible... but I don't have any reason why it should remain irreducible in $\mathbb F_4[x]$. Maybe that approach does not work... – 2017-01-24
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    @ rschwieb Sir can you please explain why the 1st isomorphism holds ? – 2017-01-24
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    @rschwieb Hmm, I think one can instead mimic the complex numbers here by regarding this as $2$-dimensional over $\mathbb{F}_8$ by choosing a suitable basis for $\mathbb{F}_4$ over $\mathbb{F}_2$. – 2017-01-24
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    @TobiasKildetoft I thought I *was* mimicking the complex version :( – 2017-01-24
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    @rschwieb See my answer for a more detailed explanation of what I meant. Unfortunately I was not able to finish it quite. – 2017-01-24
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    @user371231 There's a standard result that says if $F\subset E$ is a field extension, then $E\otimes_F F[x]\cong E[x]$, and it extends to quotients on the right too. – 2017-01-24
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    @TobiasKildetoft Ah, we were thinking of two different things. – 2017-01-24
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    @ rschwieb sir can you kindly give me some reference for that theorem . – 2017-01-24

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This is a corollary of how we construct compositum fields in general (without assuming a common field contains the two of them). For two fields algebraic over a common field, $k\subseteq K, L$ and $[L:k]$ coprime to $[K:k]$, we have that

$$K\otimes_k L\cong KL\subseteq \overline{k}$$

But then in your case it's a simple matter of noting that since $\Bbb F_4\cap\Bbb F_8=\Bbb F_2$--because the degrees are coprime--that the compositum field has degree $2\cdot 3=[\Bbb F_4:\Bbb F_2][\Bbb F_8:\Bbb F_2]$ over the base field, and this of course corresponds to the field $\Bbb F_{64}$.

There are not a lot of details to the general theorem I quote, but far too long to include in a MSE post when they're basically textbook. The interested parties can read them here.

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    This is incorrect, and only works if $ K $ and $ L $ are linearly disjoint (over $ k $). Let $ k $ be a perfect field, and $ K, L $ finite extensions of $ k $. Let $ f $ be the minimal polynomial of a primitive element of $ K/k $. Then, $ K \otimes_k L \cong k[x]/(f) \otimes_k L \cong L[x]/(f) $, which is a field if and only if $ f $ remains irreducible in $ L[x] $; which happens if and only if $ K $ and $ L $ are linearly disjoint extensions. The fields in the question are linearly disjoint, so it does work in this specific case. – 2017-02-04
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    @starfall yes that's included in the link as the last result. I agree it should be clearer in the answer, however: I'll edit that in. – 2017-02-04
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    $ K \cap L = k $ does not guarantee that $ K $ and $ L $ are linearly disjoint, unless one of $ K $ or $ L $ is Galois over $ k $. However, $ \gcd([K : k], [L : k]) = 1 $ *does* guarantee linear disjointness. (Alternatively, one may use the fact that any algebraic extension of a finite field is always Galois, thus it suffices to look at the intersection in OP's case.) – 2017-02-04
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    @Starfall Oh so it is, I got too caught up in the context--thanks for the check! – 2017-02-04
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As a vector space over $\mathbb{F}_2$ we have a basis for $\mathbb{F}_4$ consisting of the two elements $1$ and $\alpha$ where $\alpha^2 = -\alpha - 1$ and we have an element $\overline{\alpha}$ such that $\alpha + \overline{\alpha} = -1$ and $\alpha\overline{\alpha} = 1$.

The given tensor product can then be identified with the set of $\mathbb{F}_8$-linear combinations of $1$ and $\alpha$ with multiplication induced from this.

Now, note that $(a + b\alpha)(a + b\overline{\alpha}) = a^2 + b^2 - ab$ so the claim that this is isomorphic to $\mathbb{F}_{64}$ is equivalent to the fact that the only solution to $a^2 + b^2 - ab = 0$ in $\mathbb{F}_8$ is $a = b = 0$.

Unfortunately, as I write this I realize that I don't have a good argument for why this should be, but I will make this community wiki and maybe someone else can finish it.