what are the “points” of the scheme $\mathbb{Z}_8[x] /(x^2 + 7)$

I noticed modulo 8 the quadratic $$x^2 + 7$$ is zero for four separate values $$x = 1,3,5,7 \in \mathbb{Z}_8$$. The number of zeros exceeds the degree.

I would like to define the "variety" $$\mathbb{Z}_8[x]/(x^2 + 7)$$, but $$\mathbb{Z}_8$$ is not a field, so I have to use schemes. What is the geometric meaning of this quadratic vanishing at 4 points?

I know the points are the prime ideals in the ring I have constructed. In other worse, if $$ab \in P$$ then $$a \in P$$ or $$b \in P$$. Not sure how to compute those in this ring.

The topology on this scheme is given by the Zariski topology (I think) but I don't know what the open sets look like here.

Basic understanding of Spec$(\mathbb Z)$

Diophantine applications of Spec?

EDIT Mumford seem to have drawn groovy images of prime ideals in the scheme. • I would begin by making the observation that as a nilpotent element $2$ belongs to all the prime ideals. Thus the prime ideals of your ring are in 1-1 correspondence with the prime ideals of $\Bbb{Z}_2[x]/(x^2+1)$. Here, again, $x+1$ is nilpotent. Looks like there is a unique prime ideal in there. – Jyrki Lahtonen Jun 18 '14 at 16:39
• @JyrkiLahtonen The "primes" would be $(1), (x+c)$ where $c \in \{ 1,3,5,7\}$. The nilpotent $(2)$ acts like an infinitesimal, but $(x\pm1)$ is a zero divisor, not nilpotent.. – cactus314 Jun 18 '14 at 18:36
• @johnmangual: $(1)$ is never a prime, and $(x \pm 1)$ is not a zerodivisor (nor nilpotent) – zcn Jun 18 '14 at 18:37
• @johnmangual: No problem. $(0)$ is not prime in this ring though - $(0)$ is prime iff the ring is a domain, which this is not. You may be thinking of generic points, which this scheme also does not have (being zero-dimensional, although since the ring is not reduced, the single point does have "fuzz") – zcn Jun 18 '14 at 18:53
• And $x+1$ is nilpotent. After all $$(x+1)^2=x^2+2x+1\equiv 2x-6=2(x-3)\pmod{x^2+7}.$$ This in turn implies that $$(x+1)^6\equiv 2^3(x-3)^3=8(x-3)^3\equiv0.$$ – Jyrki Lahtonen Jun 18 '14 at 18:58

Edit: As a topological space, $\operatorname{Spec}(\mathbb{Z}/8\mathbb{Z}[x]/(x^2+7)) = \{(2, x+1)\}$. Notice that modulo $8$, $x^2 + 7 = x^2 - 1 = (x + 1)(x - 1)$. As pointed out by Jyrki Lahtonen, $2$ is nilpotent in $\mathbb{Z}/8\mathbb{Z}[x]$, hence is in every prime ideal, so every prime of $\mathbb{Z}/8\mathbb{Z}[x]/(x^2+7)$ must contain $(2, x+1) = (2,x-1)$, and this is maximal, with residue field $\mathbb{Z}/2\mathbb{Z}$.
• Can you explain from this picture of $\mathrm{Spec}$ why $x^2 + 7$ should have extra roots? – cactus314 Jun 18 '14 at 18:55
• @johnmangual: Being a root (in $\mathbb{Z}/8\mathbb{Z}$) of this particular polynomial is equivalent to being a unit in the base coefficient ring (strictly speaking, an inversion, but every unit has order $2$ in $\mathbb{Z}/8\mathbb{Z}$). The fact that this polynomial has "extra" roots is just saying that $\mathbb{Z}/8\mathbb{Z}$ has "many" units – zcn Jun 18 '14 at 19:04