Identify $\mathbb{Z}[x]/(2x^2+1,2x-3)$.

Question

Identify $\mathbb{Z}[x]/(2x^2+1,2x-3)$.

I tried this: $$ \mathbb{Z}[x]/(2x^2+1,2x-3)=:R \\ (2x^2+1,2x-3)=:I \\ \because (2x^2+1)-x(2x-3)=3x+1, \\ 2(3x+1)-3(2x-3)=11 \\ \therefore I=(2x^2+1,2x-3,11) \\ \therefore R \cong \mathbb{Z}_{11}[x]/(2x^2+1,2x-3) \\ \cong \mathbb{Z}_{11}[x]/(x^2+6,x-7) \\ \cong \mathbb{Z}_{11} $$

but I'm totally unsure. Thank you for your help :)

Answer 1

That's good. A way to then work backwards is to show that:

$$2x-3,2x^2+1\in\langle 11,x-7\rangle$$

You get: $$\begin{align}2x-3&=2(x-7)+11\cdot1\\ 2x^2+1&=2(x-7)(x+7)+11\cdot 9 \end{align}$$

Answer 2

$2x-3=0 \implies x=\frac 3 2$. Substituting $\frac {11} 2=0$, thus of course $11=0$. Our ring becomes $\mathbb Z_{11}[\frac 3 2]$. Now we have to check that we already have an element in $\mathbb Z_{11}$ such that multiplying it by $2$ it becomes $3$: $$2y \equiv 3 (mod \hspace{0,2cm}11)$$ we multply by $6$ and we get $$y \equiv 18 \equiv 7(mod \hspace{0,2cm}11)$$ and we already have $7 \in \mathbb Z_{11}$, thus $\mathbb Z_{11}[\frac 3 2] \simeq \mathbb Z_{11}$.

In the same way you could add the two equations and get $x^2 + x - 1=0$ and get $\mathbb Z_{11}[\frac{-1+\sqrt5}{2} ]$. The inverse of $2$ is already in $\mathbb Z_{11}$, it is $6$, and the square roots of $5$ are $(\pm 4)^2 \equiv 5$, so again $\mathbb Z_{11}[\frac{-1+\sqrt5}{2} ] \simeq \mathbb Z_{11}$.