How many fields are there (upto isomorphism) of order 6.
I dont know how to proceed. I don't know how to proceed. Please help me. Thank you in advance.
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1The order of a finite field is always a power of a prime. – Pipicito Jun 10 '16 at 02:04
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How many fields do you know of order $6$? Is there any of them? – Kushal Bhuyan Jun 10 '16 at 02:06
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@Pipicito according to you there cannot exist any field of order 6.So,there is no question of getting field isomorphism.Isn't it? – Jun 10 '16 at 02:08
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Your question makes sense. And the answer to it is "up to isomorphism, there are zero fields of order six" – Pipicito Jun 10 '16 at 02:10
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Let $F$ be a finite field. Consider the unique ring homomorphism $\phi : \mathbb{Z}\to F$ given by $1 \mapsto 1$. You have $\operatorname{Im}(\phi) \cong \mathbb{Z} / \operatorname{Ker}(\phi)$ as rings because of the isomorphism theorem. The ring $\operatorname{Im}(\phi)$ is integral because of being subring of the integral domain $F$. Thus, $\operatorname{Ker}(\phi)$ is a prime ideal. Because we know the prime ideals of $\mathbb{Z}$, there is a positive prime $p \in \mathbb{Z}$ such that $\operatorname{Ker}(\phi) = \langle p \rangle $. So $$\operatorname{Card}(\operatorname{Im}(\phi)) = \operatorname{Card}(\mathbb{Z} / \operatorname{Ker}(\phi)) = \operatorname{Card}(\mathbb{Z} /\langle p \rangle ) = p$$ Observe that if $F_0 := \operatorname{Im}(\phi)$ is a subfield of $F$ because it is a subring of $F$ that is isomorphic to the field $\mathbb{Z} /\langle p \rangle$ when we consider in $F_0$ the operations of $F$ restricted to $F_0$. So, we can consider $F$ as a vector space over $F_0$. Because $F$ is finite, there exists an $n \in \mathbb{N}$ with $F \cong F_0\,^n$ as vector spaces over $F_0$. Finally, $\operatorname{Card}(F) = p^n$ for some positive prime $p$ and some $n \in \mathbb{N}$.
Because $6$ is not a power of a prime, then there is no field of order $6$.
Pipicito
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