Question

Determine whether the given congruence-class ring is a field. Justify your answer.

1. ${\mathbf{\mathbb{Z}}}_{\mathbf{3}}\mathbf{\left[}\mathit{x}\mathbf{\right]}\mathbf{/}\mathbf{(}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}$
   
2. ${\mathbf{\mathbb{Z}}}_{\mathbf{5}}\mathbf{\left[}\mathit{x}\mathbf{\right]}\mathbf{/}\mathbf{(}\mathbf{2}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}$
   
3. ${\mathbf{\mathbb{Z}}}_{\mathbf{2}}\mathbf{\left[}\mathit{x}\mathbf{\right]}\mathbf{/}\mathbf{(}{\mathbf{x}}^{\mathbf{4}}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}$
   

Short answer

c)${\mathrm{\mathbb{Z}}}_2\left[x\right]/({\mathrm{x}}^4+{\mathrm{x}}^2+1)$ is not a field.

Step-by-step solution

Statement of Theorem 5.10

Theorem 5.10 states consider F as a field and p (x)a nonconstant polynomialin F[x]. Then the statements are equivalent as follows:

1. p (x) is irreducible in F[x].
2. $F\left[x\right]/(p(x\left)\right)$will be a field.
3. $F\left[x\right]/(p(x\left)\right)$will be an integral domain.

Determine whether the given congruence-class ring is a field c)

According to theorem 5.10, a ring $F\left[x\right]/(p(x\left)\right)$will be a field if p (x) is irreducible.

${\mathrm{\mathbb{Z}}}_2\left[x\right]/({\mathrm{x}}^4+{\mathrm{x}}^2+1)$ is not a field, because $x^4+x^2+1={\left(x^2+x+1\right)}^2$in ${\mathrm{\mathbb{Z}}}_2\left[x\right]$.

Hence, ${\mathrm{\mathbb{Z}}}_2\left[x\right]/({\mathrm{x}}^4+{\mathrm{x}}^2+1)$is not a field.