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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
in Mathematics by Diamond (89,043 points) | 119 views

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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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