Prove that ideal $I = \langle X^2,X+1\rangle$ is principal or not (in $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$ )

Question

I tried the following $$I = \langle X^2,X+1\rangle =\langle X^2,X+1,X^2+2(X+1)\rangle =\langle X^2,X+1,(X+1)^2+1 \rangle$$

Yet no matter how I arrange it, I cannot obtain $1$. Can someone help me out?

Answer 1

$-(X+1)^2+X^2+2(X+1)=1$. So, your ideal contains $1$.

Answer 2

Note that $I$ contains $$X^2+(X+1)(-X+1) =1$$

Answer 3

From your prior question $(x+1,f(x)) = (1)\iff f(-1)\mid 1$ which is true for $\,f(x) = x^2.$

Remark $ $ The other answers are essentially a special case of the proof in the prior question.

If you wish to gain further algorithmic insight on how to perform such calculations then follow the link I gave here on Hermite Normal Form.

Answer 4

I am using $\langle f,g \rangle = \langle r,g \rangle$ when $f=h\,g+r$ for $h,r \in K[X]$.

So lets do the polynomial division:

X^2        : X + 1     = X - 1
X^2 + X
-------
    - X
    - X - 1
    -------
          1

So we get $\langle 1,g \rangle$ for both $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$, or simply $\langle 1 \rangle$.