# Characterise ideals in a ring

I am trying to determine whether the following ideals of $\mathbb{Z}[X]$ are prime or maximal ideals:

1. $(X^2 - 3)$,

2. $(5,X^2 + 3)$.

I am trying to do this by establishing whether $\mathbb{Z}[X]/I$ is a field or domain. For (2), I have the following attempt so far:

$$\mathbb{Z}[X]/(5,X^2 + 3) \cong (\mathbb{Z}[X]/(5)) / ((5,X^2 + 3)/(5)) \cong (\mathbb{Z}/5 \mathbb{Z})[X] / (\overline{X}^2 + \overline{3})$$

I don't know how to proceed from here or if I'm on the right track.

For (1), I wanted to show that $\phi: \mathbb{Z}[X] \to \mathbb{R} : f \mapsto f(\sqrt{3})$ is a homomorphism with kernel $(X^2 - 3)$, but I am not sure if it is surjective onto $\mathbb{R}$.

Anyone can help me further on this?

• $\mathbb R$ has maaaaany more elements that $\mathbb Z[X]$. – Mariano Suárez-Álvarez Oct 15 '17 at 19:59
• As for (2), yes, you are on the right track. (It would be good to know how exactly you haver shown that the kernel of the map is what you say it is) – Mariano Suárez-Álvarez Oct 15 '17 at 20:00

Your ideas are good. For $(X^2-3)$, you will want to consider the map $f:Z[x]\to \mathbb{Z}[\sqrt{3}]$ to get an isomorphism with $\mathbb{Z}[x]/(x^2-3)$, but even if you consider the map to $\mathbb{R}$ that is ok, since the image is a subring (the map is not surjective) of $\mathbb{R}$ which is necessarily an integral domain.
For the next part, note that $\overline{X}^2+\overline{3}$ is irreducible as a polynomial in $\mathbb{F}_5$, since $\overline{2}$ is not a quadratic residue in that field. Can you go from there?
• For the second part, I don't quite see how it follows that $\overline{X}^2 + \overline{3}$ is an irreducible polynomial in the field $\mathbb{F}_5$, could you elaborate a bit on the proof of this? – Randolph Oct 15 '17 at 21:03
For $2$, you're on the right track. Just observe that $\mathbf Z/5\mathbf Z$ is a field, hence the polynomial ring $\mathbf Z/5\mathbf Z[X]$ is a P.I.D. so all you have to prove is that $X^2+3$ is irreducible in $\mathbf Z/5\mathbf Z[X]$. As it is a quadratic polynomial, it amounts to proving it has no roots mod. $5$, i.e. proving $-3$ is not a quadratic residue mod.5. Now the squares mod.5 are $0,1$ and $-1$.
For $1$, sme method, but simpler: if $X^2-3$ were not irreducible it would have integer roots, which is not the case. Now in a P.I.D. (more generally in a U.F.D.), irreducible elements generate a prime ideal. This ideal is not maximal because, for instance it is (strictly) contained in the ideal $(X^2-3,5)$ – which, incidentally, is maximal.