Let $I$ be an ideal generated by a polynomial in $\mathbb Q[x]$. When is $\mathbb Q[x] / I$ a field? I was looking at my old exam papers and I was stuck on the following problem:  

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 $\mathbb Q[x]$. If $F_1=\mathbb Q[x]/I_1$ and $F_2=\mathbb Q[x]/I_2$, then which of the following options is/are  correct?     
1) $F_1$ and $F_2$ are fields     
2) $F_1$ is a field but $F_2$ is not a field     
3) $F_2$ is a field but $F_1$ is not a field       
4) neither $F_1$ nor $F_2$ is a field    

Can someone explains in details how to tackle this.
 A: The key points here are:


*

*Since $\mathbb Q$ is a field, $\mathbb Q[x]$ is a Euclidean domain.

*In a principal ideal domain, a nonzero prime ideal is maximal.

*$R / I$ is a field iff $I$ is maximal (where $R$ is any commutative ring and $I$ an ideal in $R$).

*In a unique factorization domain, an element is prime iff it's irreducible.


Using these points, we find that $Q[x]/I_j$ is a field iff $I_j$ is irreducible.
$x^3 + 1$ is clearly reducible since it has a root ($x = -1$).
By the rational roots theorem, $x^4 + 3x^2 + 2$ doesn't have roots in $\mathbb Q$, but it may be factored into two polynomials of degree $2$. Write $x^4 + 3x^2 + 2 = (x^2 + ax +b)(x^2 + cx + d)$ and solve for $a, b, c, d$ to see if this is the case. Solving this can be simplified by noting that the coefficients of $x^4 + 3x^2 + 2$ are integers. By Gauss's lemma, if it's irreducible over $\mathbb Z$, it's also irreducible over $\mathbb Q$.
A: This is essentially about asking if the polynomials are irreducible or not. For instance, for $I_2$ you can see that it has $-1$ as a root, so certainly isn't irreducible. 
A: Hint: $\ \color{#c00}{x\!-\!(-1)}\mid x^3\!-(-1)^3\ $ and $\ x^4\!+3x^2\!+2 = (\color{#0a0}{x^2\!+\!2})(x^2\!+\!1),\,$ or $\,(x^2\!+\!2)^2\!-(x^2\!+\!2).\ $ Hence both quotient rings have zero-divisors, $\ \color{#c00}{x\!+\!1},\ $ and $\ \color{#0a0}{x^2\!+\!2},\,$ so neither is a field.
