Proving that $\mathbb{Q}[X] = (X-1) + (X+1)$

Question

Prove that $I+J = \mathbb{Q}[X]$ with $I$ and $J$ ideals in $\mathbb{Q}[X]$ where $I=(X-1)$, $J=(X+1)$.

I was thinking something along the lines of $I+J=(1)$ because they are coprime. That means $$ I+J = \{1 \cdot a \mid a \in \mathbb{Q}[X]\} = \{a \mid a \in \mathbb{Q}[X]\} = \mathbb{Q}[X]. $$

I'm not sure if this is a proper proof for this problem.

This is the proof suggested by my professo, but I can't really figure where the $1/2$ comes from. I should've formated it on the website but it would take me a long time to do so.

Answer 1

The number $\frac{1}{2}$ is an element of $\mathbb{Q}[x]$. Hence, you can multiply any element of an ideal contained in $\mathbb{Q}[x]$ by $\frac{1}{2}$ and the result will remain in the ideal by definition of ideal.

What your professor has proved is that $1=-\frac12(x-1)+\frac12(x+1)$, which is an element of $I+J$ by definition of the ideal. This is sufficient to show $I+J=\mathbb{Q}[x]$ since the inclusion $I+J\subseteq \mathbb{Q}[x]$ is rather trivial.

Answer 2

Observe simply that

$$-\frac12(x-1)+\frac12(x+1)=1\implies I+J=\Bbb Q[x]\;$$

I think your professor was doing something like this, but I don't quite follow in romanian...:)