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 $\mathbb{Z}[x]/(x^2+3,3) \cong \mathbb{Z}/3\mathbb{Z}$?


  • 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?

  • 1
    $\begingroup$ I'm not sure to understand your question. You have that $$\mathbb Z[X]/(X^2+3,3)\cong \mathbb F_3[X]/(X^2)$$ and $$\mathbb Z[X]/(X^2+3,5)\cong \mathbb F_5[X]/(X^2+3).$$ What else do you need ? $\endgroup$ – Surb Dec 30 '17 at 18:37
  • $\begingroup$ I wondered why the morphism I propoposed does not yield the same result. The answer from GreginGre answers that. Like I thought, $f(\sqrt{3})$ is not always well defined. $\endgroup$ – dietervdf Dec 30 '17 at 19:37
  • $\begingroup$ Of course it's not defined. If $f(x)=x$ how would you interpret $\sqrt 3\pmod 3$ ? $\endgroup$ – Surb Dec 30 '17 at 20:07

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 !

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