I am facing following problem, and would be very thankful for any input:

I am supposed to prove that two finite fields with same number of elements are isomorphic. I know the usual proof via splitting fields, but I am supposed to try another way: Say $L_{1}$ and $L_{2}$ are finite extensions of the finite field $K$. First, I need to construct common extension $L \geq L_{1},L_{2}$, and then show, that if they have same size, they are identical as subfields of $L$.

The second part, I can manage (at least I think so), but how would I go about constructing said extension $L$? Thanks to anyone who finds his/her time to help.

  • $\begingroup$ If $L_1=K(\alpha_1,\dots,\alpha_r)$ and $L_2=K(\beta_1,\dots,\beta_s)$, then you can take $L$ to be $K(\alpha_1,\dots,\alpha_r,\beta_1,\dots,\beta_s)$. $\endgroup$ – Gerry Myerson Sep 16 '14 at 7:13
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    $\begingroup$ Well, the $\alpha_i$, $\beta_j$ may not lie in the same field. $\endgroup$ – Orest Bucicovschi Sep 16 '14 at 7:19

Well, construct an extension $L$ of $K$ such that there exist morphisms of $K$ algebras $L_i \to L$ ( necessarily injective). One sees that we can find such morphisms when $L$ is an algebraic closure of $K$. Now we can see $L_1$, $L_2$ as subfields of $L$. Let us show that they coincide as subsets of $L$. Indeed, let $q = |L_1| = |L_2|$.

Fact: every element $x$ of a field $F$ with $q$ elements satisfies the equation $x^q - x = 0$. True if $x$ is $0$.Otherwise $x$ is in in the group of invertible elements and this group has order $q-1$ and so $x^{q-1}=1$, $x^q = x$.

Fact: Let $L$ a field and $P$ a polynomial of degree $n$ with coefficients in $L$. Then $P$ has at most $n$ roots in $F$.

We are done now: both $L_1$ and $L_2$ equal the set of $q$ distinct roots in $L$ of the polynomial $x^q - x$.


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