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Element $a$ generates a multiplicative group of a field F of $343$ elements. Is the polynomial $x^2+ax-a+2a^2$ irreducible in the polynomial ring $F[x]$?

As I understand, $F[x]$ is the ring of all polynomials, which can be generated by the coefficient from $F$. How then can I check a reducibility of a polynomial $p(x)$ in $F[x]$? Should I try to decompose it into the $p(x)=g(x)*h(x)$, where "=" means equality modulo 343?

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    $\begingroup$ Note that the field with $7^3=343$ elements is not the ring of integers mod $343$. $\endgroup$ – hardmath May 10 '18 at 15:27
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Since it has degree $2$, $x^2+ax-a+2a^2$ is reducible in $F[x]$ iff $x^2+ax-a+2a^2$ has a root in $F$ iff its discriminant is a square in $F$.

The discriminant is $4 a - 7 a^2 = 4a$, since $F$ has characteristic $7$ because $343=7^3$.

Therefore, $x^2+ax-a+2a^2$ reducible in $F[x]$ iff $a$ is a square in $F$.

But $a$ is not a square in $F$ because $a$ is a generator of $F^\times$, which is a cyclic group of even order.

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We can verify that $$ {\bf f }(x)=(x^2+ax+2a^2)-a=2\, \left( a+2\,x \right) ^{2}-a $$ Assume that there is an $x \in\operatorname{GF}(7^3)$ such that ${\bf f}(x)=0$, which result in $$ 2\, \left( a+2\,x \right) ^{2}=a \Rightarrow \left( a+2\,x \right) ^{2}=4\, a $$ The rest of proof is the same as the nice proof that is given by @lhf.

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