If $f (x) = x^3 + x^2 + x +1$ and $g(x) = x^3 – x^2 + x -1$, then which of the followings are true? Let$ \langle p(x)\rangle $ denote the ideal generated by the polynomial $p(x)$ in $\mathbb Q[x]$. If  $f (x) = x^3 + x^2 + x +1$  and $g(x) = x^3 – x^2 + x -1$, then which of the followings are true?
1. $ \langle f (x)\rangle + \langle g (x)\rangle = \langle x^3 + x\rangle$
2. $ \langle f (x)\rangle + \langle g (x)\rangle =  \langle f (x)\cdot g (x)\rangle$
3. $ \langle f (x)\rangle + \langle g (x)\rangle = x^2 +1$
4. $ \langle f (x)\rangle + \langle g (x)\rangle = \langle x^2 -1\rangle$   
Here gcd of them is $x^2+1$ so 3 is true and 4 is false. but I am not sure about the others. Can anybody help me.
 A: Obviously $x^2+1\in\langle x^2+1\rangle$.  Can you show that it's not in $\langle x^3+x\rangle$?  (Hint: think about degrees of polynomials under division — what are the possible degrees of polynomials in $\langle x^3+x\rangle$?)  That should answer part 1 of the question, and with a bit of additional thought you should be able to use a similar technique to solve part 2 of the question.
A: $(1)$ is an incorrect choice. Reason: $x^3+x^2+x+1\notin \langle x^3 + x\rangle$ since $\langle x^3 + x\rangle$$=\{p(x)(x^3 + x):p(x):\mathbb Q[x]\}$ and no polynomial of degree $0$ multiplied by $(x^3 + x)$ leaves $x^3+x^2+x+1.$
$(2)$ is an incorrect choice. Reason: $x^3+x\in(f(x))+(g(x))$ but not in $(f(x)g(x))$ since $(f(x)g(x))=\{p(x)(x^6-x^4-x^2+1):p(x)\in\mathbb Q[x]\}$ and multiplying $(x^6-x^4-x^2+1)$ by a nonzero polynomial at least leave a polynomial of degree $6$ [$\mathbb Q[x]$ is an integral domain]
I think a better argument for showing $4$ is incorrect goes as follows: If $4$ be true then $\langle x^2 + 1\rangle=\langle x^2 - 1\rangle$ which gives a contradiction since $c(x^2+1)\ne(x^2-1)~\forall~c\in\mathbb Q.$
