Let $I_1$ be the ideal generated by $x^4+3x^2+2$ and $I_2$ be the ideal generated by $x^3+1$ in $\Bbb Q[x]$. If $ F_1 = \frac{\Bbb Q[x]}{I_1}$ and $F_2 = \frac{\Bbb Q[x]}{I_2}$, then

1) $F_1$ and $F_2$ are fields,

2) $F_1$ is a field and $F_2$ is not a field,

3) $F_1$ is not a field while $F_2$ is a field,

4)neither $F_1$ nor $F_2$ is a field.

Consider the polynomial $x^4 + 3x^2+2$, substituting $t = x^2$ we get a new polynomial in $t$ given by $t^2 + 3t + 2 = (t+1)(t+2)$. This shows that $x^4+3x^2+2 = (x^2 + 1)(x^2 + 2)$ and hence this polynomial is reducible over $\Bbb Q$.

Similarly, since $-1$ is a root of the polynomial $x^3+1$, we have $x^3 + 1$ is also reducible over $\Bbb Q$.

1) $F_1$ and $F_2$ are fields,

2) $F_1$ is a field and $F_2$ is not a field,

3) $F_1$ is not a field while $F_2$ is a field,

4)neither $F_1$ nor $F_2$ is a field.

**Solution**:Consider the polynomial $x^4 + 3x^2+2$, substituting $t = x^2$ we get a new polynomial in $t$ given by $t^2 + 3t + 2 = (t+1)(t+2)$. This shows that $x^4+3x^2+2 = (x^2 + 1)(x^2 + 2)$ and hence this polynomial is reducible over $\Bbb Q$.

Similarly, since $-1$ is a root of the polynomial $x^3+1$, we have $x^3 + 1$ is also reducible over $\Bbb Q$.

**Result**: The quotient ring $\frac{F}{<f(x)>}$ is a field if and only if $f(x)$ is irreducible over $F$.**This result shows that neither $F_1$ nor $F_2$ is a field**.**Share to your groups:**

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