There seems to be a deep relationship between nilpotence and prime ideals. The intersection of all prime ideals is the nilradical.

What is the geometric content of this theorem, in terms of thinking about $\operatorname{Spec}(R)$? Is there something we can say concretely for some reasonably concrete ring? [candidates for 'reasonably concrete' can be (i) polynomial rings, (ii) $\mathbb Z/ n \mathbb Z$].

My current intuition is that we think of $\operatorname{Spec}(R) \equiv \{ p \in Ideals(R) : p ~\text{is prime} \}$ as points, and the other ideals $F \subseteq R%$ as "functions over $\operatorname{Spec}(R)$, given by the evaluation map $F: p \mapsto F / p$ which takes a point $p \in \operatorname{Spec}(R)$ and evaluates $F$ on it by passing to the quotient. Then the closed sets of $\operatorname{Spec}(R)$ are generated by the zero of the evaluation map. That is, for every $F \subseteq R$, we get a closed set $\{ p \in \operatorname{Spec}(R): F(p) = 0 \}$.

The reason this viewpoint 'works' is that it's a direct generalization of the case of $\mathbb R[X]$. We have the reals $r \in \mathbb R$ which are in bijection to the prime ideals $(X - r)$. So $\operatorname{Spec}(R)$ will contain our prime ideals $(X - r)$. If we then want to evaluate a polynomial $f(x) \in \mathbb R[X]$ at a point $r$, we can think of this as $(i)$: substituting $x = r$ into $f(x)$, or $(ii)$: passing into the quotient $S = \mathbb R[X]/(X - r)$ and looking at the image of $f(x)$ inside $S$. This works because quotienting by $(X - r)$ is the same as imposing $X - r = 0$, which is the same as setting $X = r$. So this gives us back the classical algebraic geometry story of "varieties".

What I don't understand is how nilpotence fits into this. Going back to the general case, we call the set of points over which an ideal vanishesas $V(I)$: $V(I) \equiv \{ p \in \operatorname{Spec}(R): I/p = 0 \}$ . We can show relatively easily that $V(I) = V(\sqrt I)$. this "seems" to be a generalization to the observation that a real polynomial $f(x) \in \mathbb R[X]$ vanishes wherever $\sqrt{f(x)}$ does. However, I don't understand why this works out in general, for arbitrary rings (I can prove it using algebra, but that's hardly the same thing).

Is this intuition for nilpotence correct? Is there a better way to think about it?

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    $\begingroup$ Your interpretation of ideals as functions is strange. The usual story is that the "functions" on $\mathrm{Spec} (R)$ are the elements of $R$, and evaluating a "function" at a prime ideal $\mathfrak{p}$ is sending that element of $R$ to its residue class in $R / \mathfrak{p}$. Viewed this way, the nilradical of $R$ consists of all the elements of $R$ that vanish at every point – so, if you like, the nilradical is precisely the part of $R$ that has no geometrical signficance. $\endgroup$
    – Zhen Lin
    Jul 10, 2020 at 3:45
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    $\begingroup$ Section II.3 of Eisenbud and Harris's The Geometry of Schemes discusses nonreduced schemes and infinitesimals. $\endgroup$ Jul 10, 2020 at 16:42

1 Answer 1


Elaborating on the Zhen Lin's comment, we can interpret all rings as functions on a "space" $\operatorname{Spec}(R)$, for which we can intepret everything geometrically, and these spaces have the seemingly strange property that functions can be zero at every "point" but still be nonzero. This tells you that your notion of space will need to be broader than topological space, two frameworks which allow these more general spaces (affine schemes) are locally ringed spaces, and functors $Ring\rightarrow Set$, but both of these take some getting used to.

While a full understanding of this formalism would probably be the best for understanding, I'll try convince you that nilpotent elements have a natural geometric interpretation without this, as infinitesimals. The ring $\mathbb{R}[x]$ can be naturally thought of as functions on the real line $\mathbb{R}$, so we have a nice tangible geometric understanding here. What about $\mathbb{R}[x][\epsilon]$ where $\epsilon^2=0$ then? If we try view $\epsilon$ as a function on $\mathbb{R}$ we won't have much luck, but if we interpret it as a "infinitesmally" small function, then we can speak of the "infinitesmal" behaviour of a polynomial $p(x)$ by looking at $p(x+\epsilon)$, and comparing this to $p(x)$.

So lets now observe that we have the following equality of elements in $\mathbb{R}[x][\epsilon]$\begin{equation}p(x+\epsilon)-p(x)=\epsilon p'(x)\end{equation}

Where $p'(x)$ is the formal derivative of polynomials (which is also the usual derivative). So working in this ring with nilpotents allows up to naturally and algebraically see the "infinitesmal" behaviour of functions on $\mathbb{R}$, in the form of derivatives.

Another instance is to compare the following pairs of ideals in $\mathbb{R}[x,y]$. Our first pair is $(y),(y^2-x)$. These are the vanishing ideals of the $x$ axis, and a parabola intersecting the $x$ axis transversely at the point $(0,0)$ in $\mathbb{R}^2$.

The second pair is $(y),(y-x^2)$, geometrically, the $x$ axis and a parabola, intersecting tangentially at $(0,0)$. These are clearly different situations geometrically, but how does the algebra show this?

We look at the "intersection" of the two spaces, so the sum of the ideals. In the first case, we have the ideal $(x,y)$, the ideal that vanishes at $(0,0)$, as expected. In the second case we have $(y,x^2)$, which is not radical, as in, there are nilpotents in the quotient. The nilpotent element in the quotient is $\bar{x}$, indicating that this "intersection" is "thicker" in the $x$ direction, we can interpret $\bar{x}$ as measuring something infinitesimal in this direction.

  • $\begingroup$ Thanks a lot. Can you tell me where I can read more about this "nilpotents as infinitesimal" perspective? $\endgroup$ Jul 10, 2020 at 5:17
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    $\begingroup$ Unfortunately I don't have a reference really, since I learnt this by word of mouth, and I'm not that well versed in this stuff. The other main main geometric thing is interpreting the family $Spec(R/I^n)$ as "infinitesimal" thickenings of $Spec(R/I)$ within $Spec(R)$. Gluing all this infinitesimal data together yields formal schemes, and Hartshorne has a (lacklustre, I have been told) treatment of these, so hopefully another textbook that treats formal schemes would have the intuition spelled out well. I hope someone with more knowledge can give an actual source for you though. $\endgroup$
    – Chris H
    Jul 10, 2020 at 12:50

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