TITLE: Cardinality of the quotient ring $\mathbb{Z}[x]/(x^2-3,2x+4)$
QUESTION [10 upvotes]: This problem is from a practice exam I was working on.

What is the cardinality of the quotient $\mathbb{Z}[x]/(x^2-3,2x+4)$ ?

Thoughts. If I find a ring that is easier to handle then this then I can go from there. So I think this is isomorphic to $\mathbb{Z}_2[x]/(x^2-3,x+2)$. And then I am stuck.
Am I correct so far? What should I be looking to do from here?
Thanks for your time and your answers.

REPLY [5 votes]: Hint $\ $ Note $\,(2\!-\!x)(2\!+\!x) = (2\!-\!\sqrt 3)(2\!+\sqrt 3)=1\,$ in $\ \Bbb Z[x]/(\color{#0a0}{x^2\!-\!3})\cong \,\Bbb Z[\sqrt 3],\, $ therefore, since $\ 2\!+\!x\,$ in a unit $\!\bmod \color{#0a0}{x^2\!-\!3},\,$ we can cancel  it from the generator $\,\color{#c00}2(2\!+\!x)\, $ of $I,\,$ yielding
$\qquad\ \ I = (\color{#0a0}{x^2\!-\!3},\,\color{#c00}2(2\!+\!x)) = (x^2\!-\!3,\color{#c00}2) = ((x\!+\!1)^2,2),\  $ so $\,\ \Bbb Z[x]/I = \Bbb Z_2[x]/((x\!+\!1)^2)$

Remark $\ $ Alternatively we can do the cancellation purely equationally as below
$\bmod I\!:\,\ \color{#0a0}{x^2\equiv 3},\,\ \color{}2(\color{}{2\!+\!x)\equiv 0} \,\Rightarrow\,0\equiv \color{}2(2\!+\!x)(2\!-\!x) \equiv 2(4\!-\!\color{#0a0}{x^2})\equiv 2(4\!-\!\color{#0a0}3) \equiv \color{#c00}2\ $
More generally when studying algebraic number rings we can often deduce much  by taking norms (e.g. above we used: $\,\alpha = 2+\sqrt{3}\,$ is a unit since its norm $\,N(\alpha ) = \alpha \bar\alpha  = 1)$.

The resulting quotient ring is $\,\cong \Bbb Z_2[\epsilon]/(\epsilon^2),\,$ the ring of dual numbers over $\,\Bbb Z_2.\,$ Such dual number rings prove useful for algebraically modelling derivatives, tangent spaces, etc.

The hint amounts to Euclid's Lemma in ideal form, i.e.
Lemma $\,\ \color{#0a0}{(a,b)=1}\,\Rightarrow\, (a,bc) = (a,c)\ $
Proof $\ \ \ (a,bc) = (a,ac,bc) = (a,\color{#0a0}{(a,b)}c) = (a,c)$