How can I show that the ideal $(2x, x^2+1)$ does not generate all of $\mathbb Z[x]$?

Question

How can I show that the ideal $(2x, x^2+1)$ does not generate all of $\mathbb Z[x]$?

Not sure of what to try first.

Answer 1

Hint:

$(2x, x^2+1)\subset (2, x^2+1)$ and $$\mathbf Z[x]/(2, x^2+1)\simeq (\mathbf Z/2\mathbf Z)[x]\big/(x^2+1)\ne \{0\}.$$

Answer 2

First guess an element that is not inside, for example $x$. Then prove that is not inside by contradiction.

If $x\in(2x,x^2+1)$ then there are polynomials $p(x)=a_0+a_1x+...+a_nx^n$, $q(x)=b_0+b_1x+...+b_kx^k$ such that $x=p(x)\cdot 2x+q(x)\cdot (x^2+1)$ so $$x=(a_0+a_1x+...+a_nx^n)2x+(b_0+b_1x+...+b_kx^k)(x^2+1)$$

Equating the coefficients at both sides

$1)$ $b_0=0$

$2)$ $2a_0+b_1=1$

$3)$ $2a_{r+1}+b_{r+2}+b_r=0$ for $r\geq 0$ or equivalently $b_r=-2a_{r+1}-b_{r+2}$

Using $3)$ lots of times on $2)$, as $b_r=0$ for $r$ big enough we get

$\begin{align} 1&=2a_0+b_1 \\ &= 2a_0-2a_2-b_3 \\ & =2a_0-2a_2+2a_4+b_5 \\ &\dots \\ &=2a_0-2a_2+2a_4+\dots \pm 2a_{\lfloor n/2\rfloor} \end{align}$

And as the left hand side is $1$ and the right hand side is even we get a contradiction. This shows $(2x,x^2+1)\neq \mathbb{Z}[x]$.

Answer 3

Both generators evaluate to $2$ at $x=1$. It follows that $f(1)$ is an even integer for all $f(x)$ in the ideal. Therefore the ideal cannot be all of $\Bbb{Z}[x]$.