I wish to solve the following:
Determine the structure of $\mathbb{Z}[x]/(x^2+3,p)$ for $p=3$ and $p=5$
I've read the question "Determining the structure of the quotient ring $\mathbb{Z}[x]/(x^2+3,p)$" but I was thinking of a different way. Also I have only seen the first isomorphism theorem for rings.
Case A
Let $p=3$ then I want to find the structure of $\mathbb{Z}[x]/(x^2+3,3)$. I think this question implies to find a ring isomorphic to this quotient ring.
This implies that I need to find an morphism such that the kernel is $(x^2+3,3) = (x^2+3)\mathbb{Z}[x]+3\mathbb{Z}[x]$. Then the first isomorphism theorem would answer the question.
I was thinking of the following evaluation morpism:
$$\begin{align} \theta: &\mathbb{Z}[x] \to \mathbb{Z}/3\mathbb{Z}; \\ &f(x) \mapsto f(\sqrt{3}) \pmod{3} \end{align} $$
This clearly is a morphism with $\mathrm{im}(\theta) = \mathbb{Z}/3\mathbb{Z}$.
I'm unsure about the kernel though. For which polynomials $f(x)$ will $f(\sqrt{3}) \equiv 0 \pmod{3}$. Any polynomial which is multiple of 3 would do, $3\mathbb{Z}[x]$? And the polynomials with degree higher then 2, where 3 can be an extra constant, $(x^2+3)\mathbb{Z}[x]$?
So $\mathbb{Z}[x]/(x^2+3,3) \cong \mathbb{Z}/3\mathbb{Z}$?
Questions
- I wonder if for example evaluating the polynomial $x^3$ at $x=\sqrt{3}$ is even defined in $\mathbb{Z}/3\mathbb{Z}$ since $\sqrt{3}^3 = 3\sqrt{3} \not \in \mathbb{Z}$
- Would the above be right?
Case B
When $p=5$ the questions seems to be a lot harder, how should I find a morphism such that the kernel is $(x^2+3, 5)$... Recycling the idea from case A seems nog possible?
Your morphism $\theta$ is not well defined, since $f(\sqrt{3})$ does not need to be an integer.
There is no hope the quotient ring will be a field anyway. If $p=3$, $(x^2+3,3)=(x^2,3)\subsetneq (x,3)\subsetneq \mathbb{Z}[x]$, so your ideal is not maximal.
I would suggest you try to prove an isomorphism with $\mathbb{F}_3[T]/(T^2)$.
For $p=5$, you should try to see if $X^2+3$ is irreducible modulo $5$ or not. If it is, then your quotient ring is a field with ?? elements, so...
If not, you can split it into two factors, and try a Chinese remainder mumbo jumbo.
Good luck !