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Let $K_{1},K_{2},...K_{n}$ be a collection of finite fields. Now consider the direct product $J=K_{1} \times K_{2} \times ... K_{n}$. What are the subfields of $J$? must $J$ be isomorphic to $K_{i}$?

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In my book, for rings $R$ and $S$, for one to be a subring of the other, you need the unit of $R$ to be the same as the unit of $S$. So if $n>1$, none of the $K$'s is a subfield of $J$. But if $N=2$, for instance, and both fields are of the same characteristic, then $(1,1)$ generates a field isomorphic to the prime field, which is a subfield of $J$. At first glance, the situation looks fairly complicated to me. – Lubin Nov 25 '11 at 7:14
up vote 4 down vote accepted

The projections $p_i: J\to K_i$ are ring homomorphisms. If $K$ is a subfield of $J$, then the restriction $p_i|_{K} : K\to K_i$ is injective because $K$ is a field. Let $\sigma_i = p_i|_{K}$ and let $F_i$ be the subfield $F_i=\sigma_i(K)\subseteq K_i$. Then the $F_i$'s are all isomorphic to each other.

So the subfields of $J$ are exactly of the form $$K=\{ (x_1,\dots, x_n) \mid x_1\in F_1, x_i=\tau_i(x_1)\}$$ where $F_1$ is a subfield of $K_1$, and the $\tau_i : F_1\to K_i$ are ring homomorphisms. For such $K$ to exist, of course the $K_i$ must have the same characteristic as said Lubin in the comments.

Example: $J=K_1\times K_2$ with $[K_1: \mathbb F_p]=6$, $[K_2: \mathbb F_p]=4$. Then the possible $K$ are of degree over $\mathbb F_p$ dividing $\gcd(4,6)=2$. Then you get

  1. $\{(x,x) | x\in \mathbb F_p\}$;
  2. $\{(x, \tau(x)) | x\in F_1 \}$, where $F_1$ is the unique subextension of $K_1$ of degree $2$ over the prime field. There are two such subfields because there are two embeddings $\tau: F_1\to K_2$.

Bonus exercise: find the number of subfields of $J$ in the general case.

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Just the answer I hit on as I was trying to fall asleep last night. – Lubin Nov 25 '11 at 16:27

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