Are those rings fields? Let $f(x) = x^4+x^2+1 \in Z_{2}$ and $A = Z_{2}/f(x)$


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*Is $A$ a field?


Let $f(x) = x^4-x^2+1 \in Z_{7}$ and $B = Z_{7}/f(x)$


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*Is $B$ a field?


I know that a ring $A = Z_{n}/f(x)$ is a field $\Leftrightarrow$ $f(x)$ has no roots in $Z_{n}$ . But this happens only for 2nd-degree and 3rd-degree polynoms
What if you have a 4th-degree polynom?

Good luck allowed this argument in answer to the second question.
In characteristic $7$, we get $(a+b)^6=a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6$. In $\mathbb F_7[x]$, $(x^2+1)^6=x^{12}-x^{10}+x^8-x^6+x^4-x^2+1$, which is clearly congruent to $1$ modulo your polynomial $x^4-x^2+1$. Thus, in $B$, $x^2+1$ is a sixth root of unity. But all the six nonzero elements of $\mathbb F_7$ are sixth roots of unity, and we have just found a seventh. So $B$ is not a field.
