'Finitely presented' implies 'always finite presented' for algebraic theories

Question

In this MO question it is proven the answer is yes for modules. The proof given relies on the snake lemma, which does not generally make sense in the category of rings, groups, monoids, etc.

It seems strange to me that this implication would hold for modules but not for rings, groups, monoids, lattices etc. Maybe naively, I think it's true for any algebraic theory - if you can present something finitely, then you cannot contrive an infinite presentation. I really have no idea how to prove it though.

So, in which algebraic theories does 'finitely presented' impliy 'always finitely presented'?

In more detail, I am asking for which Lawvere theories do the categories of models satisfy the following property: If there exists a finite presentation of a model, then every presentation on finitely many generators of this model is finite.

Answer 1

I will work in a category $\mathsf{Alg}(\tau)$ of algebraic structures of a given type $\tau$ (in the sense of universal algebra).

Let $A$ be a finitely presented algebra. Choose some surjective homomorphism $\phi : \langle x_1,\dotsc,x_n \rangle \to A$ such that the kernel is a finitely generated congruence. Let $y_1,\dotsc,y_m$ be another generating set of $A$ and consider the surjective homomorphism $\psi : \langle y_1,\dotsc,y_m \rangle \to A$; we want to prove that the kernel of $\psi$ is finitely generated, too.

We define $\alpha : \langle x_1,\dotsc,x_n,y_1,\dotsc,y_m \rangle \to \langle x_1,\dotsc,x_n \rangle$ as follows: We map $x_i \mapsto x_i$, and we map $y_j$ to some preimage of $y_j \in A$ under $\phi : \langle x_1,\dotsc,x_n \rangle \twoheadrightarrow A$. We also obtain $\lambda : \langle x_1,\dotsc,x_n,y_1,\dotsc,y_m \rangle \twoheadrightarrow A$, which is also $\phi \circ \alpha$.

The kernel of $\alpha$ is generated (as a congruence) by the elements $(y_j,\alpha(y_j))$, hence it is finitely generated. The kernel of $\lambda$ is generated by these elements and lifts of the generators of $\ker(\phi)$; hence it is finitely generated. Since the kernel of $\lambda$ surjects onto the kernel of $\psi$, we are done.

(The proof is a bit sketchy. Perhaps I will add some details later.)

Answer 2

Let me elaborate the proof from Martin's answer.

Suppose $A = B[X_1, \cdots , X_n]/(f_1, \cdots , f_r)$, where $f_i \in B[X_1, \cdots , X_n]$. Suppose also that $y_1, \cdots, y_m$ generate $A$ as a $B$-algebra, so we have a surjection $\psi: B[Y_1, \cdots, Y_m] \twoheadrightarrow A$ where $Y_j \mapsto y_j$. Show that the ideal $\ker(\psi) \subset B[Y_1, \cdots, Y_m]$ is finitely generated.

To simplify notations we write $B[X]$ for $B[X_1, \cdots, X_n]$, and denote $\phi: B[X]\twoheadrightarrow A$ the quotient map. Choose $g_j\in B[X]$ any lifts of $y_j$, hence $\phi(g_j) = y_j = \psi(Y_j)$. Define a homomorphism of algebras $\alpha: B[X, Y] \to B[X]$ fixing $B[X]$ and sending $Y_j \mapsto g_j$. This is an evaluation of $B[X][Y]$ at $Y_j=g_j$ and the kernel is

$$\ker\alpha=(Y_1-g_1, \cdots, Y_m-g_m).$$

Composing $\alpha$ with $\phi$ we obtain $\lambda: B[X, Y]\xrightarrow{\alpha} B[X]\xrightarrow{\phi} A$, whose kernel is finitely generated:

$$\ker\lambda=\alpha^{-1}\ker\phi=(f_1, \cdots, f_r, Y_1-g_1, \cdots, Y_m-g_m).$$

Similarly we can choose $h_i\in B[Y]$, the lifts of $\phi(X_i)$, hence $\psi(h_i)=x_i=\phi(X_i)$, and construct an evaluation map $\beta: B[X, Y]\to B[Y]$ fixing $Y_j$ and sending $X_i\mapsto h_i$, with a kernel

$$\ker\beta=(X_1-h_1, \cdots, X_n-h_n).$$

Composing with $\psi$ to obtain $\mu: B[X, Y]\xrightarrow{\beta} B[Y]\xrightarrow{\psi} A.$ Note that $\lambda$ and $\mu$ are actually the same algebra homomorphism since they both send $X_i\mapsto x_i, Y_j\mapsto y_j$, hence they have the same kernel $\ker\mu=\ker\lambda$. Since $\ker\mu=\beta^{-1}\ker\psi$, and $\beta$ is surjective, we get that $$\beta(\ker\mu)=\ker\psi,$$ which shows that $\ker\psi$ is also finitely generated as the image of a finitely generated ideal.