I'm working through An Introduction to Synthetic Differential Geometry and I have found myself a bit stuck.

## Context

Recall that we are working without the law of excluded middle, and that there is a distinguished $\mathbb{Q}$-algebra $R$ known as the "geometric line".

Presumably, all of these axioms are meant to apply in the internal logic of a topos (or at the very least a locally cartesian closed category).

From the book:

Definition 2.7 Let $R[X_1, \ldots, X_n]$ be a commutative ring with $n$ generators. Let $p_1(X_1, \ldots, X_n), \ldots, p_m(X_1, \ldots, X_m)$ be polynomials with coefficients from $R$, and let $I$ be the ideal generated from these polynomials. A

finitely presented $R$-algebrais an $R$-algebra $$R[X_1, \ldots, X_n] / I = R[X_1, \ldots, X_n] / (p_1(X_1, \ldots, X_n), \ldots, p_m(X_1, \ldots, X_n))$$

So far so good. This is a familiar definition. I wouldn't have phrased it in exactly this way, but it seems clear enough.

Definition 2.9 A

Weil algebraover $R$ is an $R$-algebra $W$ (denoted sometimes as $R \otimes W$) such that:

- There is an $R$-bilinear multiplication map $\mu : R^n \times R^n \to R^n$, making $R^n$ an $R$-algebra with $(1, 0, \ldots, 0)$ as a multiplication unit.
- The object ('set') $I$ of elements of $R^n$ with first coordinate equal zero is a nilpotent ideal.
- There is an $R$-algebra map $\pi : W \to R$ given by $(x_1, \ldots, x_n) \mapsto x_1$, called the
augmentation(its kernel is $I$, and it's called theideal of augmentation).

The notation here is somewhat vague. I am making a few assumptions here.

The first is that $R^n$ is to be given the obvious structure as an $R$-module. We then equip it with $\mu$ which should be compatible with the aforesaid $R$-module structure and make $R^n$ into an $R$-algebra.

The second is that $(1, 0,\ldots, 0)$ is supposed to be the multiplicative identity and not merely a unit.

The third is that $W$ is supposed to consist of the triple $(R^n, \mu, \pi)$. This was never explicitly stated, but it's the only thing that makes sense in context.

## Problem

The following is a quote from immediately after Definition 2.9:

Moreover, it is easy to see, [sic] that each Weil algebra is a finitely presented $R$-algebra.

My question is: how does one prove this?

It's definitely clear that each Weil-algebra is finitely generated. Indeed, a Weil algebra on $R^n$ is clearly generated by $n – 1$ elements – that is, the $R$-algebra homomorphism $R[X_1, \ldots, X_{n – 1}] \to R^n$ sending $X_i$ to the unit vector $e_{i + 1}$ is surjective. But I don't see how the kernel of this map is a finitely generated ideal.

Perhaps I am missing something obvious, but it's really bugging me.

## Best Answer

Ah, I figured it out. The following theorem holds:

Proof: suppose that $x_1, \ldots, x_n$ are generators of $W$ as an $R$-module which come with a presentation.

Then in particular, consider the unique $R$-algebra homomorphism $\phi : R[X_1, \ldots, X_n] \to W$ such that for all $1 \leq i \leq n$, we have $\phi(X_i) = x_i$. The image of such a map is a sub-$R$-algebra, hence a sub-$R$-module, which contains all the $x_i$, and thus is all of $W$. So $\phi$ is surjective; by the isomorphism theorem, $\phi$ thus gives rise to an isomorphism $\tilde{\phi} : R[X_1, \ldots, X_n] / \ker \phi \to W$. It therefore suffices to show that $\ker \phi$ is finitely generated.

Choose values $w_{i, j, k}$ such that for all $1 \leq i, j \leq n$, we have $x_i x_j = \sum\limits_{k = 1}^n w_{i, j, k} x_k$. Define $P_{i, j} = X_i X_j - \sum\limits_{k = 1}^n w_{i, j, k} X_k$. Note that $\phi(P_{i, j}) = 0$.

Finally, we write $1 \in W$ as $\sum\limits_{i = 1}^n \omega_i x_i$ for $\omega_1, \ldots, \omega_n \in R$. Let $Q = 1 - \sum\limits_{i = 1}^n \omega_i X_i$; then $\phi(Q) = 0$.

Let $I_1$ be the ideal generated by the $P_{i, j}$ and by $Q$. Note that $I_1 \subseteq \ker \phi$.

We will use $I_1$ to "kill the non-linear terms". I claim the following lemma:

Proof: Let $S$ be the set of all $P$ which can be written in such a way. Note that for all $1 \leq i \leq n$, we have $X_i = 0 + \sum\limits_{j = 1}^n \delta_{ij} X_j \in S$. Therefore, to show that $S = R[X_1, \ldots, X_n]$, it suffices to show that $S$ is a sub-algebra.

We first note that $S$ is closed under scalar multiplication and addition, and that $0 \in S$; therefore, $S$ is a sub-module. We further note that $1 = Q + \sum\limits_{i = 1}^n \omega_i X_i \in S$. It therefore suffices to show that $S$ is closed under multiplication.

Further note that $I_1 \subseteq S$, since given $Q \in I_1$, we have $Q = Q + \sum\limits_{i = 1}^n 0 X_i \in S$.

We see that

$$(T_1 + \sum\limits_{i = 1}^n p_{1, i} X_i)(T_2 + \sum\limits_{i = 1}^n p_{2, i} X_i) = T_1 (T_2 + \sum\limits_{i = 1}^n p_{2, i} X_i) + T_2 (\sum\limits_{i = 1}^n p_{1, i} X_i) + (\sum\limits_{i = 1}^n p_{1, i} X_i)(\sum\limits_{i = 1}^n p_{2, i} X_i)$$

The terms $T_1 (T_2 + \sum\limits_{i = 1}^n p_{2, i} X_i)$ and $T_2 (\sum\limits_{i = 1}^n p_{1, i} X_i)$ are in $I$, hence in $S$. Since $S$ is a sub-module, it suffices to show that $(\sum\limits_{i = 1}^n p_{1, i} X_i)(\sum\limits_{i = 1}^n p_{2, i} X_i) \in S$.

To do this, we apply double distribution:

$$(\sum\limits_{i = 1}^n p_{1, i} X_i)(\sum\limits_{i = 1}^n p_{2, i} X_i) = \sum\limits_{i = 1}^n \sum\limits_{j = 1}^n p_{1, i} p_{2, j} X_i X_j$$

Since $S$ is a sub-module, it suffices to show that $X_i X_j \in S$ for all $i, j$. This is straightforward, since $X_i X_j = P_{i, j} + \sum\limits_{k = 1}^n w_{i, j, k} X_k$. The lemma is thus proved. $\square$

Let us now back up. We have a surjective map of $R$-modules $\kappa : R^n \to W$ sending $e_i$ to $x_i$. Take $w_1, \ldots, w_m \in R^m$ which generate the submodule $\ker \kappa$.

Take the $R$-module map $\gamma : R^n \to R[X_1, \ldots, X_n]$ sending $e_i$ to $X_i$. Let $W_i = \gamma(w_i)$. Note that $\phi \circ \gamma = \kappa$ since for all $i$, we have $\phi(\gamma(e_i)) = \phi(X_i) = x_i = \kappa(e_i)$ and $\phi \circ \gamma$ is $R$-linear. Therefore, we have $\phi(W_i) = \phi(\gamma(w_i)) = \kappa(w_i) = 0$, so $W_i \in \ker \phi$.

Now let $I_2$ be the ideal generated by the $P_{i, j}$, $Q$, and also the $W_k$. Note that $\ker \kappa \subseteq \gamma^{-1}(I_2)$, since $\gamma(w_i) \in I_2$ for all $i$. I claim that

Clearly, $I_2 \subseteq \ker \phi$, since we've established that all the $P_{i, j}$, $Q$, and $W_k$ are in the kernel. So it suffices to show that $I_2 \subseteq \ker \phi$.

Indeed, suppose $P \in \ker \phi$. Write $P = T + \sum\limits_{i = 1}^n p_i X_i$ where $T \in I_1$ and $p_1, \ldots, p_n \in R$ (this can be done by the lemma). Let $x = (p_1, \ldots, p_n) \in R^n$; then $P = T + \gamma(x)$.

Then we see that $0 = \phi(T + \gamma(x)) = \phi(T) + \phi(\gamma(x)) = \kappa(x)$. So $\kappa(x) = 0$. Then $x \in \ker \phi \subseteq \gamma^{-1}(I_2)$. So $\gamma(x) \in I_2$. Since $T \in I_1 \subseteq I_2$, we have $T \in I_2$. So $P = T + \gamma(x) \in I_2$. We have thus proved that $I_2 = \ker(\phi)$. $\square$

We have proved a rather general and convenient result which is substantially more powerful than needed here. In fact, for this problem, we only need $W$ to be finite free, so we can actually dispense with $I_2$ altogether and show that $I_1 = \ker \phi$. But there's no reason not to go as general as possible.