Ideal of $\mathbb Q[x]$ which contains two polynomials

Question

Suppose $I$ is an ideal of $\mathbb Q[x]$ which contains $x^2 + 2x +4$ and $x^3 - 3$. Prove $I =\mathbb Q[x]$.

This is an exercise in my abstract algebra text book. I know the definition of an ideal. I know that every ideal of $\mathbb Q[x]$ is a principal ideal which means every ideal can be generated by a single polynomial in $\mathbb Q[x]$.

A hint given in the exercise says begin by showing that $x^3 - 3$ is irreducible in $\mathbb Q[x]$. I know that it is irreducible in $\mathbb Q[x]$ because it has no roots in $\mathbb Q[x]$. Likewise $x^2 + 2x + 4$ is irreducible in $\mathbb Q[x]$.

I also know that if I could show that $1$ is in $I$ then we are done. Otherwise I do not know how to get this proof started.

Answer 1

Notice that: $$\underbrace{(x^3-3)}_{\in I}-\underbrace{x}_{\in\mathbb{Q}[X]}\underbrace{(x^2+2x+4)}_{\in I}=-2x^2-4x-3.$$ Therefore, since $I$ is an ideal of $\mathbb{Q}[X]$, one has: $$2x^2+4x+3\in I.$$ Then, notice that: $$\underbrace{2x^2+4x+3}_{\in I}-\underbrace{2}_{\in\mathbb{Q}[X]}\underbrace{(x^2+2x+4)}_{\in I}=-1.$$ Finally, $1\in I$ and $I=\mathbb{Q}[X]$.

Answer 2

Hint: The ring $\mathbb Q[x]$ isn't just a principal ideal domain. It's also a Euclidean domain. And the fact that both polynomials are irreducible means, in particular, that they are coprime.

Answer 3

Let $p(x)$ be the generator of the ideal. Then $p(x)$ divides $x^2+2x+4$ and $x^3-3$. Thus $p(x)$ has degree $\le 2$. But $p(x)$ cannot have degree $1$ or $2$ since $x^3-3$ is irreducible. Thus $p(x)$ is a non-zero constant, and we are finished.

Remark: One can use less machinery. For example we can use the Euclidean Algorithm to express $1$ as a linear combination $p(x)(x^2+2x+4)+q(x)(x^3-3)$.