Describe $R=\mathbb{Z}[X]/(X^2-3,2X+4)$

Solution 1:

it looks to me like it is $\mathbb Z_2[X]/\left<X^2+1\right>$.

$2$ is in your ideal because:

$$2 = 2(X^2-3)-(2X+4)(X-2)$$

Also $X^2+1$ is in your ideal because: $$\begin{align}X^2+1 &= (X^2-3) + 4 = (X^2-3) + 4(X^2-3) - (2X+4)(2X-4) \\&= 5(X^2-3) - (2X+4)(2X-4) \end{align}$$

So we know that $\left<2,X^2+1\right>$ is contained in your ideal.

But it is obvious that the generators for your ideal are in $\left<2,X^2+1\right>$.

So we are done.

Note: Given that $(X+1)^2=X^2+1$ in $\mathbb Z_2$, this can be rewritten as isomorphic to $\mathbb Z_2[Y]/\left<Y^2\right>$ with $Y$ corresponding to $X+1$.

Solution 2:

I have another answer. $\mathbb Z[x]/(x^2-3,2x+4) = (\mathbb Z[x]/(x^2-3))/(x^2-3,2x+4)/(x^2-3)$(by third isomorphism theorem) now define a map $\mathbb Z[x]$ to $\mathbb Z[\sqrt{3}], $ $F(f(x))=f(\sqrt{3})$( evaluated at root $3)$ then $F$ is surjective homomorphism.By 1st isomorphism theorem $\mathbb Z[x]/(x^2-3)=\mathbb Z[\sqrt{3}]$ and $(x^2-3,2x+4)/(x^2-3)=(2x+4+(x^2-3))$ and when it undergoes $F$,resultant ideal will be $(2\sqrt{3}+4)$ so finally description will be $\mathbb Z[\sqrt{3}]/(2\sqrt{3}+4)$