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Every field with 6 elements is isomorphic to $\Bbb Z_2\times \Bbb Z_3$

I think it's false because:

$\Bbb Z_6[x] / (3x +2)$ is such that $3x+2$ is irreducible in $\Bbb Z_6$ so $\Bbb Z_6 [x]/ (3x +2)$ is a field with 6 elements.

This is isomorphic to $\Bbb Z_2\times \Bbb Z_3$ $\approx \Bbb Z_6$, as the elements are actually the same. We also know that if $f:A\rightarrow B$ is an isomorphism, if A is field then B is field and as $\Bbb Z_6 [x]/ (3x +2)$ $\approx \Bbb Z_6$, with $\Bbb Z_6 [x]/ (3x +2)$ field, then $\Bbb Z_6$ is a field (contradiction as 6 isn't prime).

Then not every field with 6 elements is isomorphic to $\Bbb Z_2\times \Bbb Z_3$, because $\Bbb Z_6[x] / (3x +2)$ is not isomorphic to $\Bbb Z_2\times \Bbb Z_3$.

puradrogasincortar
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    There is no field with $6$ elements. – lulu Sep 11 '17 at 12:36
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    Are you certain that $\Bbb Z_6[x] / (3x +2)$ is a field? – Arthur Sep 11 '17 at 12:36
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    ... because the number of elements of a finite field is a primary number ($p^n$ with $p$ prime). – Jean Marie Sep 11 '17 at 12:42
  • but 3x+2 is irreducible so the quotient is a field right?? what am i missing here? – puradrogasincortar Sep 11 '17 at 12:44
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    If $p(x)$ is an irreducible polynomial over a field $K$, then $K[x]/(p(x))$ is also a field. But $\Bbb{Z}_6$ is not a field, so that result does not apply. At least not in general. – Jyrki Lahtonen Sep 11 '17 at 12:46
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    Here actually $\Bbb{Z}_6[x]/(3x+2)$ is a field, but the field has only two elements. Let $I$ be the ideal generated by $(3x+2)$. As $2(3x+2)=4$ we see that $4=-2\in I$. So the quotient ring has characteristic two. Therefore also $$x=(3x+2)-2x-2\in I.$$ The conclusion is that $\Bbb{Z}_6[x]/(3x+2)\simeq \Bbb{Z}_2.$ – Jyrki Lahtonen Sep 11 '17 at 12:50
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    Curiously the answer the answer to the question is affirmative! Every field with six elements is isomorphic to $\Bbb{Z}_2\times\Bbb{Z}_3$. is a so called vacuously true statement :-) – Jyrki Lahtonen Sep 11 '17 at 12:53

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