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Find all $c$ in $Z_3$ such that $Z_3[x] / x^3+x^2+c$  is a field. Which of the following is the right choice? Explain your answer.

(A) 0 (B) 2 (C) 1 (D) 3
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For $Z_3[x] /\left(x^3+x^2+c\right)$ to be a field, the polynomial $x^3+x^2+c$ must be irreducible over $Z_3$, meaning it cannot be factored into non-constant polynomials of smaller degree. We'll consider each choice for $c$ and determine if the resulting polynomial is irreducible.

(A) $c=0$ : The polynomial is $x^3+x^2$, which factors as $x^2(x+1)$. It is reducible, so $c=$ 0 is not a valid choice.

(B) $c=2$ : The polynomial is $x^3+x^2+2$. There are no linear factors in $Z_3$, so it is irreducible. $c=2$ is a valid choice.

(C) $c=1$ : The polynomial is $x^3+x^2+1$, which factors as $(x+1)\left(x^2+1\right)$ in $Z_3$. It is reducible, so $c=1$ is not a valid choice.

(D) 3 is not an element of $Z_3$, so this choice is invalid.

Thus, the correct choice is (B) $c=2$. When $c=2$, the polynomial $x^3+x^2+2$ is irreducible over $Z_3$, and the quotient ring $Z_3[x] /\left(x^3+x^2+2\right)$ forms a field.
by Diamond (89,043 points)

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