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.

  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Jun 18 '14 at 16:39
  • $\begingroup$ @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.. $\endgroup$ – cactus314 Jun 18 '14 at 18:36
  • $\begingroup$ @johnmangual: $(1)$ is never a prime, and $(x \pm 1)$ is not a zerodivisor (nor nilpotent) $\endgroup$ – zcn Jun 18 '14 at 18:37
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    $\begingroup$ @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") $\endgroup$ – zcn Jun 18 '14 at 18:53
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    $\begingroup$ 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 $$(x+1)^6\equiv 2^3(x-3)^3=8(x-3)^3\equiv0.$$ $\endgroup$ – 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 $\mathbb{Z}/2\mathbb{Z}$.

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    $\begingroup$ Can you explain from this picture of $\mathrm{Spec}$ why $x^2 + 7$ should have extra roots? $\endgroup$ – cactus314 Jun 18 '14 at 18:55
  • $\begingroup$ @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 $\endgroup$ – zcn Jun 18 '14 at 19:04

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