2.1. Finite-type classification

The central object of the present paper are cluster algebras of finite type. We end the section by recalling this notion as well as a classification theorem connecting cluster algebras of finite type with Dynkin diagrams. Precise references for more details are [Reference Fomin and Zelevinsky19],[Reference Fomin, Williams and Zelevinsky16], or [Reference Marsh34, 5.1], for example.

Recall that a Cartan matrix $ A = (a_{i,j}) $ is called of finite type if all its principal minors are positive. For the $ 2 \times 2 $ principal minors, this implies the condition $ a_{i,j} a_{j,i} \leq 3 $ for $ i \neq j $ .

Definition 2.17 ([Reference Fomin, Williams and Zelevinsky16, Definition 5.2.4]). Let $ A = (a_{i,j}) $ be an $ n \times n $ Cartan matrix of finite type. The Dynkin diagram of A is a graph with vertices $ \{ 1, \ldots, n \} $ , for which the edges are determined as follows: Let $ i, j \in \{ 1, \ldots, n \} $ with $ i \neq j $ . If $ a_{i,j} a_{j,i} \leq 1 $ , then the vertices i and j are joined by an edge if $ a_{i,j} \neq 0 $ . Whenever $ a_{i,j} a_{j,i} > 1 $ , the following rule is applied for the edge between i and j :

Here are two examples. The graph on the right hand side is the Dynkin diagram of the corresponding matrix on the left-hand side:

There is the following classification of cluster algebras of finite type, cf. [Reference Fomin and Zelevinsky19, Theorem 1.4].

Recall, that the Cartan matrices A(B) of finite type are classified by the Dynkin diagrams $ A_n $ , $ B_n $ , $ C_n $ , $ D_n $ , $ E_6, E_7, E_8, F_4, G_2 $ (for $ n \geq 1, 2, 3, 4 $ respectively), see [Reference Fomin, Williams and Zelevinsky16, Theorem 5.2.6] or [Reference Carter8, Section 6.4].

4.1 ${\textbf{A}}_{\textbf{n}}$ cluster algebras

Assume that Q is a simply laced Dynkin diagram of type $A_n$ with any orientation. Since all trees with the same underlying undirected graph are mutation-equivalent (Lemma 2.14), we may choose the following orientation:

Recall that we denote by $ \mathcal{A}(A_n) $ the corresponding cluster algebra.

Lemma 4.1. The cluster algebra $\mathcal{A}(A_n) $ is isomorphic to $ K[z_1, \ldots, z_{n+1}]/\langle f_n \rangle $ with

\begin{equation*}\begin{array}{rl}f_n(z_1,\ldots,z_{n+1})\;:\!=\; &P_{n+1}(z_1, \ldots, z_{n+1})-1 \ .\end{array}\end{equation*}

Here, $P_{n+1}$ is the continuant polynomial defined in Section 3. In particular, the variety $ \mathrm{Spec}(\mathcal{A}(A_n)) $ is isomorphic to a hypersurface in $ \mathbb{A}_K^{n+1} $ .

This result can also be found in [Reference Dupont11, Corollary 4.2]. We provide a simpler and shorter proof.

Proof. By Theorem 2.12, $\mathcal{A}(A_n) $ is isomorphic to a quotient $K[x_1, \ldots, x_n,y_1, \ldots, y_n]/I$ , where the ideal I is generated by:

\begin{equation*} x_1y_1-x_2-1, \ \ x_2y_2-x_1-x_3, \ \ x_3y_3-x_2-x_4, \ \ \ldots, \ \ x_{k}y_k-x_{k-1}-x_{k+1}, \ \ \ldots\end{equation*}

\begin{equation*} \ldots, \ \ x_{n-1}y_{n-1}-x_{n-2}-x_n, \ \ x_ny_n-x_{n-1}-1 \ .\end{equation*}

For $i \geq 2$ one can stepwise express each $x_i$ in terms of $x_1, y_1, \ldots, y_n$ : The first equation shows that $x_2=x_1y_1-1=P_2(x_1, y_1)$ . Substituting into the second equation yields

\begin{equation*}x_3=x_2y_2-x_1=P_2(x_1, y_1)y_2-P_1(x_1) \ , \end{equation*}

which is by Lemmas 3.4 and 3.3 equal to $P_3(x_1, y_1, y_2)$ . Recursively, we obtain for $k=2, \ldots, n$ :

\begin{equation*}x_k=P_k(x_1, y_1, \ldots, y_{k-1}) \ . \end{equation*}

Thus, the last generator $f_n\;:\!=\;x_ny_n-x_{n-1}-1$ becomes

\begin{equation*}P_n(x_1, y_1, \ldots, y_{n-1})y_n - P_{n-1}(x_1, \ldots, y_{n-2})-1=P_{n+1}(x_1, y_1, \ldots, y_n)-1 \ .\end{equation*}

In conclusion, we have $ K[x_1, \ldots, x_n, y_1, \ldots, y_n]/I \cong K[z_1, \ldots, z_{n+1}]/\langle f_n \rangle $ , where the generator on the right-hand side is $f_n(z_1, \ldots, z_{n+1})=P_{n+1}(z_1, \ldots,z_{n+1})-1$ .

Lemmas 4.1, 3.7, and Proposition 3.8 immediately imply Theorem A in the $ A_n $ -case, where it states

4.2. ${\textbf{D}}_{\textbf{n}}$ cluster algebras

Next, we consider the quiver Q, whose underlying graph is a simply laced Dynkin diagram of type $ D_n $ , for some $ n \geq 4 $ . Since all orientations on a tree are mutation-equivalent (Lemma 2.14), we choose the following orientation and numbering of the vertices for Q:

The corresponding cluster algebra, which we denote by $ \mathcal{A}(D_n) $ , coincides with the lower cluster algebra of Q (Theorem 2.12) and the latter is completely described by its exchange relations by Lemma 2.10.

Lemma 4.4. The cluster algebra $ \mathcal{A}(D_n) $ is isomorphic to

\begin{equation*} K[u_1, u_2, u_3, u_4, z_1, \ldots, z_{n-2}]/ \langle h_1, h_2 \rangle , \end{equation*}

where

\begin{equation*} \begin{array}{ll} h_1 \;:\!=\; u_1 u_2 - u_3 u_4 - u_1 u_2 u_3 u_4 - u_2 u_4 P_{n-3} (z_1, \ldots, z_{n-3}), \\[5pt] h_2 \;:\!=\; u_3 u_4 - P_{n-2} (z_1, \ldots, z_{n-2}) - 1 . \end{array} \end{equation*}

In particular, the variety $ \mathrm{Spec}(\mathcal{A}(D_n)) $ is isomorphic to a subvariety of $ {\mathbb{A}}^{n+2}_K $ of codimension 2. (As before, $ P_{n-2} $ an $ P_{n-3} $ are the continuant polynomials discussed in Section 3.)

Proof. As mentioned before, we have $ \mathcal{A}(D_n) \cong K [x_1, \ldots, x_n, y_1, \ldots, y_n ]/ I, $ where I is the ideal generated by:

(4.1) \begin{eqnarray} x_1y_1-x_2-1, & x_k y_k - x_{k+1} - x_{k-1}, \mbox{ for } k \in \{ 2, \ldots, n-3\} \ , \end{eqnarray}

(4.2) \begin{equation} g_3 \;:\!=\; x_{n-2} y_{n-2} - x_{n-1} x_n - x_{n-3}, \ \ g_2 \;:\!=\; x_{n-1} y_{n-1} - x_{n-2} - 1, \ \ g_1 \;:\!=\; x_n y_n - x_{n-2} - 1 \ . \end{equation}

As in the proof of Lemma 4.1, we obtain from (4.1) $ x_k = P_k (x_1, y_1, \ldots, y_{k-1}) $ , for all $ k \in \{ 2, \ldots, n-2 \} $ . The last generator in (4.2) can be replaced by:

\begin{equation*} g'_{\!\!1} \;:\!=\; g_1 - g_2 = x_n y_n - x_{n-1} y_{n-1} \end{equation*}

If we substitute $ x_2, \ldots, x_{n-2} $ in the remaining two generators, we obtain

\begin{equation*} g_2 = x_{n-1} y_{n-1} - P_{n-2}(x_1, y_1, \ldots, y_{n-3}) -1 \end{equation*}

and

\begin{equation*} \begin{array}{rcl} g_3 & = & P_{n-2}(x_1, y_1, \ldots, y_{n-3}) y_{n-2} - x_{n-1} x_n - P_{n-3}(x_1, y_1 \ldots, y_{n-4}) \\ [5pt] & = & y_{n-2} (x_{n-1} y_{n-1} - 1 - g_2 ) - P_{n-3}(x_1, y_1, \ldots, y_{n-4}) - x_{n-1} x_n \\[5pt] & = & - x_{n-1} ( x_n - y_{n-2} y_{n-1}) - y_{n-2} - P_{n-3}(x_1, y_1, \ldots, y_{n-4}) - y_{n-2} g_2 \ . \end{array} \end{equation*}

We introduce

\begin{equation*} (u_1, u_2, u_3, u_4) \;:\!=\; (x_n - y_{n-2} y_{n-1}, \, y_n, \, x_{n-1}, \, y_{n-1} ) \end{equation*}

\begin{equation*} (z_1, z_2, \ldots, z_{n-2} ) \;:\!=\; (x_1, y_1, \ldots, y_{n-3} ) \end{equation*}

and obtain

\begin{equation*} \begin{array}{l} h_1 \;:\!=\; g'_{\!\!1} = u_1 u_2 - u_3 u_4 + y_{n-2} u_2 u_4 , \\ h_2 \;:\!=\; g_2 = u_3 u_4 - P_{n-2} (z_1, \ldots, z_{n-2}) - 1 . \end{array} \end{equation*}

On the other hand, $ g_3 + y_{n-2} g_2 \in I$ yields that we may eliminate

\begin{equation*} y_{n-2} = - u_1 u_3 - P_{n-3}(z_1, \ldots, z_{n-3}), \end{equation*}

which provides $ h_1 = u_1 u_2 - u_3 u_4 - u_1 u_2 u_3 u_4 - u_2 u_4 P_{n-3} (z_1, \ldots, z_{n-3}) $ , as desired.

By the previous result, $ \mathrm{Spec} ( \mathcal{A}(D_n)) \cong V (h_1, h_2) \subset {\mathbb{A}}^{n+2}_K $ . Using this presentation, we determine the singular locus of $ \mathrm{Spec} ( \mathcal{A}(D_n)) $ .

Lemma 4.5. Let $\mathcal{A}(D_n)$ be the cluster algebra of type $D_n$ over a field K.

If $\mathrm{char}(K) \neq 2$ , then we have

\begin{equation*} \mathrm{Sing} (\mathrm{Spec} ( \mathcal{A}(D_n))) \cong \begin{cases} Y_0 \cup Y_1 \cup Y_2 \cup Y_3 \cup Y_4 & if\; n \equiv 0 \mod 4 , \\ Y_0 & otherwise , \end{cases} \end{equation*}

where, for $ i \in \{ 1, 2, 3, 4 \} $ , the component $ Y_i$ is the $ u_i $ -axis and so $ \dim(Y_i) = 1 $ , while

\begin{equation*} Y_0 \;:\!=\; V (u_1, \ldots, u_4, \ P_{n-2} (z_1, \ldots, z_{n-2}) + 1) \end{equation*}

and $ \dim (Y_0 ) = n - 3 $ . The only possible singular point of $Y_0$ is the origin,

\begin{equation*} \mathrm{Sing}(Y_0) = \begin{cases} \bigcap\limits_{i=1}^4 Y_i & if\; n \equiv 0 \mod 4 \\ \varnothing & otherwise . \end{cases} \end{equation*}

Observe that $ Y_0 $ has two irreducible components if $ n = 4 $ since $ P_2 (z_1,z_2) + 1 = z_1 z_2 $ .

On the other hand, if $ \mathrm{char}(K) = 2 $ , the same statement holds true if we replace every condition $ n \equiv 0 \mod 4 $ by $ n \equiv 0 \mod 2 $ .

Note that Lemmas 4.4 and 4.5 for $ n = 4 $ imply Theorem A(4)(a).

Proof of Lemma 4.5.By Lemma 4.4, $ \mathrm{Spec} ( \mathcal{A}(D_n)) $ is isomorphic to the subvariety of $ {\mathbb{A}}^{n+2}_K $ determined by:

\begin{equation*} \begin{array}{l} h_1 = u_1 u_2 - u_3 u_4 - u_1 u_2 u_3 u_4 - u_2 u_4 P_{n-3} (z_1, \ldots, z_{n-3}) = 0 , \\ h_2 = u_3 u_4 - P_{n-2} (z_1, \ldots, z_{n-2}) - 1 = 0 . \end{array} \end{equation*}

Observe that there is an ordering on the variables such that $ u_1 u_2 $ is the leading monomial of $ h_1 $ and $ u_3 u_4 $ is the one of $ h_2 $ . Hence, the dimension of $ \mathrm{Spec} ( \mathcal{A}(D_n)) $ is equal to n and by applying the Jacobian criterion for smoothness, the singular locus of $ \mathrm{Spec} ( \mathcal{A}(D_n)) $ is determined by the vanishing of the $ 2 \times 2 $ minors of the Jacobian matrix of $ ( h_1, h_2 ) $ . We abbreviate

\begin{equation*} \operatorname{Jac}(D_n) \;:\!=\; \operatorname{Jac}(h_1, h_2; u_1, u_2, u_3, u_4, z_1, \ldots, z_{n-2}). \end{equation*}

The first four columns of $ \operatorname{Jac}(D_n) $ are

\begin{equation*} \begin{pmatrix} u_2 (1-u_3 u_4)\;\;\;\; & u_1 (1-u_3 u_4) - u_4 P_{n-3}\;\;\;\; & -u_4 (1 + u_1 u_2)\;\;\;\; & -u_3 (1 + u_1 u_2 ) - u_2 P_{n-3} \\ 0\;\;\;\; & 0\;\;\;\; & u_4 \;\;\;\; & u_3\;\;\;\; \end{pmatrix} \end{equation*}

while the remaining columns are

(4.3) \begin{equation} \begin{pmatrix} - u_2 u_4 \dfrac{\partial P_{n-3}}{\partial z_1}\;\;\;\; & \cdots \;\;\;\; & - u_2 u_4 \dfrac{\partial P_{n-3}}{\partial z_{n-3}}\;\;\;\; & 0\;\;\;\; \\[8pt] -\dfrac{\partial P_{n-2}}{\partial z_1}\;\;\;\; & \cdots\;\;\;\; & -\dfrac{\partial P_{n-2}}{\partial z_{n-3}}\;\;\;\; & -\dfrac{\partial P_{n-2}}{\partial z_{n-2}} \end{pmatrix}. \end{equation}

(Here, we use the obvious abbreviations $ P_{n-2} $ and $ P_{n-3} $ .) Clearly, the maximal minors of the first matrix are of the form $ u_3 ( \ldots ) $ and $ u_4 (\ldots ) $ . Suppose $ u_3 = u_4 = 0 $ . The vanishing of $ h_1 $ and $ h_2 $ provides that for a singular point, we have to have $ u_1 u_2 = 0 $ and $ P_{n-2} + 1 = 0 $ . Further, the first row of the first matrix becomes $ \begin{pmatrix} u_2 & u_1 & 0 & - u_2 P_{n-3} \end{pmatrix} $ and every entry of the first row of the second matrix is zero. Thus, if $ u_1 = u_2 = 0 $ , we obtain the irreducible component:

\begin{equation*} Y_0 = V (u_1, \ldots, u_4, P_{n-2} + 1 ) \end{equation*}

in the singular locus. On the other hand, suppose that $ u_1 \neq 0 $ . Since $ u_1 u_2 = 0 $ has to vanish, we get $ u_2 = 0 $ . The minors corresponding to the derivatives with respect to $ ( u_2, z_i ) $ , for $ i \in \{ 1, \ldots, n - 3 \} $ provide that

\begin{equation*} \dfrac{\partial P_{n-2}}{\partial z_{1}} = \ldots = \dfrac{\partial P_{n-2}}{\partial z_{n-2}} = 0 . \end{equation*}

This yields the irreducible component $ Y_1 = V (u_2, u_3, u_4 ) \cap \mathrm{Sing} (V(P_{n-2} + 1)) $ of the singular locus of $ \mathrm{Spec}( \mathcal{A}(D_n) ) $ . Analogously, we get $ Y_2 = V (u_1, u_3, u_4 ) \cap \mathrm{Sing} (V(P_{n-2} + 1)) $ if $ u_2 \neq 0 $ .

Next, suppose that $ u_3 \neq 0 $ or $ u_4 \neq 0 $ . Then, the minors corresponding to the derivatives $ (u_1, u_4) $ and $ ( u_2, u_4 ) $ (resp. $ (u_1, u_3) $ and $ ( u_2, u_3 ) $ ) provide that we have to have

(4.4) \begin{equation} u_2 (1-u_3 u_4) = 0 \ \ \ \mbox{ and } \ \ \ u_1 (1-u_3 u_4) - u_4 P_{n-3} = 0 \end{equation}

for a singular point. If $ 1 - u_3 u_4 = 0 $ , we get that $ P_{n-3}(z_1, \ldots, z_{n-3}) = 0 $ . On the other hand, the vanishing of $ h_2 $ yields $ P_{n-2}(z_1, \ldots, z_{n-2}) = 0 $ , which is a contradiction as we have seen at the beginning of the proof of Lemma 3.7.

Therefore, we get $ u_2 = 0 $ if $ u_3 \neq 0 $ or $ u_4 \neq 0 $ . This implies that all entries in the first row of the matrix (4.3) are zero. Moreover, $ h_1 = h_2 = 0 $ is equivalent to $ u_3 u_4 = P_{n-2} + 1 = 0 $ . We have two cases:

• • $ u_3 = 0 $ and $ u_4 \neq 0 $ . Then $ \dfrac{\partial h_1}{\partial u_3} = - u_4 \neq 0 $ . The minors corresponding to the derivatives with respect to $ (u_3, z_i) $ provide that all derivatives of $ P_{n-2} $ have to vanish. Since $ \dfrac{\partial P_{n-2}(z_1,\ldots, z_{n-2})}{ \partial z_{n-2}} = P_{n-3}(z_1,\ldots, z_{n-3}) $ , the second equality of (4.4) and $ u_3 = 0 $ imply $ u_1 = 0 $ . Hence, we get the irreducible component $ Y_4 = V (u_1, u_2, u_3 ) \cap \mathrm{Sing} (V(P_{n-2} + 1)) $

• • $ u_3 \neq 0 $ and $ u_4 = 0 $ . Via the analogous arguments as in the previous case, we obtain the irreducible component $ Y_3 = V (u_1, u_2, u_4 ) \cap \mathrm{Sing} (V(P_{n-2} + 1)) $ in the singular locus.

Note that this covers all cases, where the minors of $ \operatorname{Jac}(D_n) $ vanish. Hence, we determined all components of the singular locus. Furthermore, observe that

(4.5) \begin{equation} \bigcap_{i=1}^4 Y_i = \mathrm{Sing} (Y_0). \end{equation}

First, assume $ \mathrm{char}(K) > 2 $ . By Lemma 3.7 and Proposition 3.8, we have

\begin{equation*} \mathrm{Sing}( P_{n-2} (z_1, \ldots, z_{n-2} ) + 1 ) = \begin{cases} V(z_1, \ldots, z_{n-2}) & \text{ if } n \equiv 0 \mod 4 \\ \varnothing & \text{ otherwise} . \end{cases} \end{equation*}

This implies

\begin{equation*} \mathrm{Sing} (\mathrm{Spec} ( \mathcal{A}(D_n))) \cong \begin{cases} Y_0 \cup Y_1 \cup Y_2 \cup Y_3 \cup Y_4 & \text{ if } n \equiv 0 \mod 4 \\ Y_0 & \text{ otherwise} , \end{cases} \end{equation*}

where $ \mathrm{Sing}(Y_0) = V (u_1, \ldots, u_4, z_1, \ldots, z_{n-2} ) = \bigcap_{i=1}^4 Y_i $ is the origin if $ n \equiv 0 \mod 4 $ , while $ Y_0 $ is regular in the second case. Observe that $ Y_i $ is the $ u_i $ -axis and so $ \dim(Y_i) = 1 $ , for $ i \in \{ 1, \ldots, 4 \} $ , and $ \dim (Y_0 ) = n - 3 $ .

Let us turn to the case $ \mathrm{char}(K) = 2 $ . The same arguments apply and the only difference appears, when we apply Proposition 3.8, which leads to the condition $ n \equiv 0 \mod 2 $ instead of $ n \equiv 0 \mod 4 $ .

As before in the $ A_n $ -case, we can classify the singularities and construct a desingularization from this.

Proof. By Lemma 4.4, $ \mathrm{Spec} ( \mathcal{A}(D_n)) $ is isomorphic to the subvariety of $ {\mathbb{A}}^{n+2}_K $ given by:

\begin{equation*} \begin{array}{l} h_1 = u_1 u_2 - u_3 u_4 - u_1 u_2 u_3 u_4 - u_2 u_4 P_{n-3} (z_1, \ldots, z_{n-3}) = 0 \ , \\ h_2 = u_3 u_4 - P_{n-2} (z_1, \ldots, z_{n-2}) - 1 = 0 \ . \end{array} \end{equation*}

First, suppose $ \mathrm{Sing} (\mathrm{Spec} ( \mathcal{A}(D_n))) \cong Y_0 $ , where

\begin{equation*} Y_0 = V ( u_1, \ldots, u_4, \ P_{n-2} (z_1, \ldots, z_{n-2}) + 1) \ , \end{equation*}

is regular and of dimension $ n - 3 $ . Moreover, Lemma 3.7 and Proposition 3.8 provide that $ H \;:\!=\; \mathrm{Spec}(K[u_1, \ldots,u_4, z_1, \ldots, z_{n-2}]/\langle h_2 \rangle ) $ is regular. Locally at $ Y_0 $ , the element $ 1 - u_3 u_4 $ is invertible and thus we may introduce the local variable $ w_{1} \;:\!=\; u_1 (1 - u_3 u_4) - u_4 P_{n-3} $ . Using the latter, we get $ h_1 = w_1 u_2 - u_3 u_4 $ locally. Therefore, locally at $ Y_0 $ , the variety $ \mathrm{Spec}(\mathcal{A}(D_n)) $ is isomorphic to an intersection of a cylinder over an $ A_1 $ -hypersurface singularity and a regular variety H, which is transversal to the cylinder. In particular, the blowup of $ Y_0 $ is a desingularization of $ \mathrm{Spec}(\mathcal{A}(D_n)) $ . This ends the proof of part (1).

Let us come to the case $ \mathrm{Sing} (\mathrm{Spec} ( \mathcal{A}(D_n))) \cong Y_0 \cup Y_1 \cup Y_2 \cup Y_3 \cup Y_4 $ . By Lemma 4.5, this can only happen if $ n \equiv 0 \mod 2 $ . (Note that $ n \equiv 0 \mod 4 $ implies $ n \equiv 0 \mod 2 $ .) Here, the singular locus of $ \mathrm{Spec}(\mathcal{A}(D_n)) $ has five components, $ Y_0 $ above and the $ u_i $ -axes $ Y_i $ for $ i \in \{ 1, 2, 3, 4 \} $ . The only singular point of $ \mathrm{Sing} (\mathrm{Spec}(\mathcal{A}(D_n))) $ is the origin 0, which is also the singular locus of $ Y_0 $ , as well as the intersection of the four other components $ Y_1, \ldots, Y_4 $ .

The same argument as above shows that, locally at a singular point, which is contained in $ Y_0 \setminus \{0\} $ , the variety $ \mathrm{Spec}(\mathcal{A}(D_n)) $ is isomorphic to a cylinder over an $ A_1 $ -singularity intersected with a regular hypersurface, which is transversal to the cylinder.

Let us consider the other components. In the proof of Proposition 3.8, we have seen that there is a local coordinate transformation $(z_1, \ldots, z_{n-2}) \mapsto (t_1, \ldots, t_{n-2})$ , such that

(4.6) \begin{equation} P_{n-2}(t_1, \ldots, t_{n-2}) +1 =\sum_{i=1}^{m}t_{2i-1}t_{2i} \ , \end{equation}

where $ m \;:\!=\; \dfrac{n-2}{2} \in {\mathbb{Z}}_+ $ , which is an integer since $ n \equiv 0 \mod 2 $ .

Along the $ u_1 $ -axis without the origin, $ Y_1 \setminus \{ 0 \} $ , the term $ u_1( 1 - u_3 u_4) - u_4 P_{n-3} $ is invertible and hence we may introduce $ w_2 \;:\!=\; h_1 = u_2 ( u_1 ( 1 - u_3 u_4) - u_4 P_{n-3}) - u_3 u_4 $ to replace $ u_2 $ . Thus, locally at a point of $ Y_1 \setminus \{ 0 \} $ , we get that $ \mathrm{Spec}(\mathcal{A}(D_n)) $ is isomorphic to the hypersurface:

\begin{equation*} \mathrm{Spec}( K [u_1, u_3, u_4, t_1, \ldots, t_{n-2}]/ \langle u_3 u_4 - \sum_{i=1}^{m}t_{2i-1}t_{2i} \rangle ) \ , \end{equation*}

which is a cylinder over an $ A_1 $ -hypersurface singularity. The analogous situation appears for $ Y_2 \setminus \{ 0 \} $ .

Let us consider the local situation at $ Y_3 \setminus \{ 0 \} $ . There, $ u_3 $ is invertible so that it makes sense to define $ w_1 \;:\!=\; u_1 u_3, w_2 \;:\!=\; u_2 u_3^{-1}, w_4 \;:\!=\; u_3 u_4 $ . Using this, we obtain

\begin{equation*} h_1 = u_1 u_2 - u_3 u_4 - u_1 u_2 u_3 u_4 - u_2 u_4 P_{n-3} = w_2 (w_1 ( 1 - w_4) - w_4 P_{n-3}) - w_4 \ . \end{equation*}

Furthermore, we may introduce $ v_1 \;:\!=\; w_1 ( 1 - w_4) - w_4 P_{n-3} $ so that the vanishing of $ h_1 $ allows to eliminate the variable $ w_4 $ in $ h_2 = w_4 - P_{n-2} - 1 $ . Hence, locally at a point of $ Y_3 \setminus \{ 0 \} $ , the variety $ \mathrm{Spec}(\mathcal{A}(D_n)) $ is isomorphic to

\begin{equation*} \mathrm{Spec}( K [v_1, w_2, u_3, t_1, \ldots, t_{n-2}]/ \langle v_1 w_2 - \sum_{i=1}^{m}t_{2i-1}t_{2i} \rangle ) \ . \end{equation*}

In other words, we are in the same situation as for $ Y_1 \setminus \{ 0 \} $ . The analogous argument (and using Lemma 3.6) provide the same result for $ Y_4 \setminus \{ 0 \} $ . Hence, we have shown (2)(a).

It remains to study the situation at the origin, which is the singular locus of $ Y_0 $ and also equal to $ \bigcap_{i=1}^4 Y_i $ . By (4.6), $ Y_0 $ is isomorphic to a hypersurface singularity in $ \mathbb{A}_K^{n-2} $ of type $ A_1 $ . In particular, we get (2)(b) and blowing up the origin resolves the singularities of $ Y_0 $ .

Finally, for (2)(c), the same argument as above (for $ Y_0 $ ) provides that $ h_1 = w_1 u_2 - u_3 u_4 $ locally at the origin. In particular, $ h_1 $ and $ h_2 $ are both homogeneous of degree 2. This implies, if we blow up the origin, then the singular locus of the strict transform of $ \mathrm{Spec}( \mathcal{A}(D_n)) $ is equal to $ Y'_{\!\!0}\cup Y'_{\!\!1}\cup \ldots \cup Y'_{\!\!4} $ , where $ Y'_{\!\!i} $ denotes the strict transform of $ Y_i $ . Furthermore, for every $ i \neq j $ , we have $ Y'_{\!\!i} \cap Y'_{\!\!j} = \varnothing $ . Therefore, $ Z\;:\!=\; \bigcup_{i=0}^4 Y'_{\!\!i} $ is regular and after blowing up with center Z all singularities are resolved by (2)(a).

4.3. $\textbf{E}_6, \textbf{E}_7, \textbf{E}_8$ cluster algebras

Let us now turn our attention to the missing skew-symmetric cluster algebras of finite type, which are those arising from orientations on $ E_6, E_7, E_8 $ Dynkin diagrams. As before, we fix a field K.

Proof. We begin with $ E_8 $ . By Lemma 2.14, we may choose any orientation for the quiver, whose underlying graph is the $ E_8 $ Dynkin diagram. We choose

Using the analogous arguments as as in the $D_n$ -case, we obtain

\begin{equation*} \mathcal{A}(E_8) \cong K[x_1, x_6, x_8, y_1, \ldots, y_8] / \langle h_1, h_2 , h_3 \rangle , \end{equation*}

where we define

\begin{equation*} \begin{array}{ll} h_1 \;:\!=\; P_5 (x_1, y_1, y_2, y_3, y_4) - P_2 (x_6, y_6) , \\[5pt] h_2 \;:\!=\; P_2 (x_6, y_6) y_5 - x_6 P_2 (x_8, y_8) - P_4(x_1, y_1, y_2, y_3), \\[5pt] h_3 \;:\!=\; P_3 (x_8,y_8,y_7) - P_2(x_6, y_6) . \end{array} \end{equation*}

The singular locus of $ \mathrm{Spec}(\mathcal{A}(E_8)) $ is determined by the vanishing locus of the $ 3 \times 3 $ minors of the Jacobian matrix:

\begin{equation*} \operatorname{Jac}(E_8) \;:\!=\; \operatorname{Jac}(h_1, h_2, h_3; x_1, x_6, x_8, y_1, \ldots, y_8) . \end{equation*}

First, observe that $ h_3 = P_3 (x_8,y_8,y_7) - P_2(x_6,y_6) = x_8 y_8 y_7 - x_8 - y_7 - x_6 y_6 + 1 $ . We get

(4.7) \begin{equation} \dfrac{\partial h_3}{\partial x_8} = y_7 y_8 - 1 \neq 0, \mbox{ or } \dfrac{\partial h_3}{\partial y_7} = x_8 y_8 - 1 \neq 0, \mbox{ or } \dfrac{\partial h_3}{\partial y_8} = x_8 y_7 \neq 0, \end{equation}

where the nonvanishing of at least one derivative can be seen by setting two derivatives equal to zero which leads to the third derivative being nonzero. We fix $ z \in \{ x_8, y_7, y_8 \} $ such that $ \dfrac{\partial h_3}{\partial z} \neq 0 $ . The minor determined by the derivatives with respect to $ ( y_5, y_6, z ) $ provides that we must have $ x_6 (x_6 y_6 - 1 ) = 0 $ at a singular point of $ \mathrm{Spec}(\mathcal{A}(E_8)) $ . If $ x_6 y_6 - 1 = 0 $ , then vanishing of $ h_3 $ and of the minor of $ \operatorname{Jac}(E_8) $ determined by the derivatives with respect $ ( x_6, y_6, z) $ lead to a contradiction. Thus, we get $ x_6 = 0 $ . The columns of $ \operatorname{Jac}(E_8) $ corresponding to z and $ y_5 $ become the transpose of the vectors $ (0, 0, \dfrac{\partial h_3}{\partial z} ) $ and $ (0,1, \ast) $ , for some entry $ \ast $ . Since $ \dfrac{\partial h_3}{\partial z} $ is nonzero, we can only have a singular point if $ \operatorname{Jac}(h_1; x_1, x_6, y_1, \ldots, y_4, y_6 ) $ is the zero vector. Note that $ \dfrac{\partial h_1}{\partial x_6} = - y_6 $ and $ \dfrac{\partial h_1}{\partial y_6} = - x_6 = 0 $ . Using $ x_6 = 0 $ , the vanishing of $ h_1 $ can be reformulated as:

\begin{equation*} Q_5 \;:\!=\; P_5 (x_1, y_1, y_2, y_3, y_4) + 1 = 0. \end{equation*}

Since $ \dfrac{\partial h_1}{\partial x_1} = \dfrac{\partial Q_5 }{\partial x_1} $ and $ \dfrac{\partial h_1}{\partial y_i} = \dfrac{\partial Q_5 }{\partial y_i}, $ for $ i \in \{ 1, \ldots, 4 \}, $ we get the inclusion:

\begin{equation*} \mathrm{Sing}(\mathrm{Spec} ( \mathcal{A}(E_8))) \subseteq \mathrm{Sing} (V(Q_5)) = \varnothing \ , \end{equation*}

where the last equality holds by Lemma 3.7. This concludes the proof for $ E_8 $ .

The statement for $ E_6 $ follows by applying the analogous arguments for the quiver:

We leave the details as an easy exercise for the reader.

Proof. Analogous to the proof of Proposition 4.7, we choose the quiver:

and we obtain $ \mathcal{A}(E_7) \cong K[x_1, x_5, x_7, y_1, \ldots, y_7] / \langle h_1, h_2 , h_3 \rangle , $ where we define

\begin{equation*} \begin{array}{ll} h_1 \;:\!=\; P_4(x_1, y_1, y_2, y_3) - P_2 (x_5, y_5), \\[5pt] h_2 \;:\!=\; P_2 (x_5, y_5) y_4 - x_5 P_2 (x_7, y_7) - P_3(x_1, y_1, y_2), \\[5pt] h_3 \;:\!=\; P_3 (x_7,y_7,y_6) - P_2(x_5, y_5) . \end{array} \end{equation*}

The same arguments as in the $ E_8 $ case provide that $ \mathrm{Sing} (\mathrm{Spec} ( \mathcal{A}(E_7))) $ is isomorphic to a subvariety of $ V (x_5) \cap \mathrm{Sing} (V(Q_4)) $ , where $ Q_4 \;:\!=\; Q_4 (z_1,z_2,z_3,z_4)\;:\!=\; P_4(z_1,z_2,z_3,z_4) + 1 $ and $ (z_1, z_2, z_3, z_4) \;:\!=\; (x_1, y_1, y_2, y_3 ) $ . Since $ \operatorname{char}(K) \neq 2 $ , we have $ \mathrm{Sing} (V(Q_4)) = \varnothing $ , by Proposition 3.8, which concludes the the proof of (1).

(2) Suppose $ \operatorname{char}(K) = 2 $ . We choose $ z \in \{ x_7, y_6, y_7 \} $ such that $ \dfrac{\partial h_3}{\partial z} \neq 0 $ . Then the minor of $ \operatorname{Jac}(E_7) \;:\!=\; \operatorname{Jac}(h_1, h_2, h_3; x_1, x_5, x_7, y_1, \ldots, y_7) $ corresponding to the derivatives with respect to $ (y_4, y_4,z) $ and $ (x_5, y_5, z) $ provide that at a singular point, we have $ x_5 = 0 $ . We get that $ P_2 (x_5, y_5) = 1 $ and hence

\begin{equation*} h_1 = P_4(x_1, y_1, y_2, y_3) - P_2 (x_5, y_5) = f_3 (x_1, y_1, y_2, y_3) . \end{equation*}

The minors corresponding to $ (*, y_4, z) $ , where $ * \in \{ x_1, x_5, y_1, y_2, y_3 \} $ lead to the equality:

\begin{equation*} \mathrm{Sing} (\mathrm{Spec} ( \mathcal{A}(E_7))) = \mathrm{Sing} (V(f_3)) \cap V(x_5, h_2, h_3) = V (x_1, x_5, y_1, \ldots, y_5, P_3 (x_7, y_7,y_6)+1) \ , \end{equation*}

which is regular by Lemma 3.7.

Let us consider the situation locally at the singular locus. Then, $ P_2 (x_5,y_5) $ and $ P_2 (y_2,y_3) = 1 + y_2 y_3 $ are units. Thus, we may introduce the local coordinates $ z_1 \;:\!=\; y_1 (1 + y_2 y_3) + y_3 $ and $ z_4 \;:\!=\; h_2 = P_2 (x_5, y_5) y_4 - x_5 P_2 (x_7, y_7) - P_3(x_1, y_1, y_2) $ . As in (4.7), the derivatives of $ h_3 = P_3 (x_7,y_7,y_6) - P_2(x_5, y_5) $ with respect to $ x_7$ , $ y_7 $ , and $ x_6 $ cannot vanish at the same time, i.e., $ V (h_3 ) $ is a regular hypersurface, which is transversal to $ V( z_4 ) $ . Observe that (using $ \mathrm{char}(K)=2 $ )

\begin{equation*} \begin{array}{l} h_1 = x_1 y_1 y_2 y_3 + x_1 y_1 + x_1 y_3 + y_2 y_3 + x_5 y_5 = x_1 z_1 + x_5 y_5 + y_2 y_3 \ . \end{array} \end{equation*}

This implies the remaining parts of the proposition.

5.1. $\textbf{B}_{\textbf{n}}$ cluster algebras

A possible exchange matrix B for type $B_n$ , $ n \geq 2 $ , is given as:

\begin{equation*} B=\begin{pmatrix} 0 & 1 & \cdots & 0 & 0 & 0 & 0 \\\ -1 & 0 & \ddots & 0 & 0 & 0 & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots & \vdots & \vdots \\ 0 & 0 & \ddots & 0 &1 & 0 & 0 \\ 0 & 0 & \cdots & -1 & 0 &1 & 0 \\ 0 & 0 & \cdots & 0 & -1 & 0 & 1 \\ 0 & 0 & \cdots & 0 &0 &-2 & 0 \\\end{pmatrix}\end{equation*}

(cf. [Reference Fomin, Williams and Zelevinsky16, Section 5.5, (5.31)]) and the corresponding Dynkin diagram is of type $ B_n $ (where n is the number of vertices):

Lemma 5.1. Let K be any field and $ n \geq 2 $ . The cluster algebra $ \mathcal{A}(B_n) $ is isomorphic to

\begin{equation*} {K[z_1, \ldots, z_{n-1},u_1,u_2, u_3]/\langle g_n, h_n \rangle} \ , \end{equation*}

where $ g_n \;:\!=\; (u_1 u_2 - 1 ) u_3 - u_1^2 - P_{n-2}(z_1,\ldots, z_{n-2}) $ and $ h_n \;:\!=\; u_1 u_2 - 1 - P_{n-1}(z_1,\ldots, z_{n-1}) $ . In particular, the variety $ \mathrm{Spec}(\mathcal{A}(B_n)) $ is isomorphic to a codimension 2 subvariety of $ \mathbb{A}_K^{n+2} $ .

Proof. Since the underlying diagram $\Gamma(B) $ is acyclic of type $B_n$ , we get the presentation:

\begin{align*}\mathcal{A}(B_n) \cong K[x_1, \ldots, x_n, y_1, \ldots, y_n]/ & \langle x_1y_1-x_2-1, x_2y_2 - x_1 - x_3, \ldots, x_{n-2}y_{n-2}-x_{n-3}-x_{n-1}, \\ &{x_{n-1}y_{n-1} - x_n^2 - x_{n-2}}, x_n y_n -x_{n-1}-1 \rangle \ .\end{align*}

As in the proof of Lemma 4.1 (type $ A_n $ ), one can express $x_k$ in terms of $x_1, y_1 ,\ldots, y_{k-1}$ :

\begin{equation*} x_k=P_{k}(x_1, \ldots, y_{k-1}), \ \ \mbox{ for } 2 \leq k \leq n-1 \ .\end{equation*}

If we plug this into the remaining generators $x_{n-1}y_{n-1} - x_n^2 - x_{n-2} $ and $ x_n y_n -x_{n-1}-1 $ , we obtain

\begin{equation*} \begin{array}{lll} \widetilde g_n \;:\!=\; P_{n-1}(x_1,y_1,\ldots, y_{n-2}) y_{n-1} - x_n^2 - P_{n-2}(x_1, y_1, \ldots, y_{n-3}) = 0 \ , \\[5pt] h_n \;:\!=\; x_n y_n - P_{n-1}(x_1,y_1,\ldots, y_{n-2}) - 1 = 0 \ . \end{array}\end{equation*}

We replace $ \widetilde g_n $ by $ g_n \;:\!=\; \widetilde g_n + y_{n-1} h_n =(x_n y_n - 1 ) y_{n-1} - x_n^2 - P_{n-2}(x_1,y_1,\ldots, y_{n-3}) $ . Therefore, $\mathcal{A}(B_n)$ is isomorphic to $K[x_1, x_n, y_1, \ldots, y_n]/\langle g_n , h_n\rangle$ , which yields the assertion after renaming the variables.

Proof. We use the presentation $\mathcal{A}(B_n) \cong K[z_1, \ldots, z_{n-1},u_1,u_2,u_3]/\langle g_n, h_n \rangle $ of Lemma 5.1, where $ g_n = (u_1 u_2 - 1 ) u_3 - u_1^2 - P_{n-2}(z_1,\ldots, z_{n-2}) $ and $ h_n \;:\!=\; u_1 u_2 - 1 - P_{n-1}(z_1,\ldots, z_{n-1}) $ . We consider the $ 2 \times 2 $ minors of the Jacobian matrix. The columns corresponding to $ u_1, u_2 , u_3 $ yield that the following equations hold for the singular locus:

\begin{equation*} u_1 (u_1 u_2 - 1) = u_2 (u_1 u_2 - 1 ) = 2 u_1^2 = 0. \end{equation*}

First, assume $ \operatorname{char}(K) \neq 2 $ . Then, we have $ u_1 = 0 $ , which implies $ u_2 = 0 $ and the $ u_3 $ column of the Jacobian matrix is $ ( -1 , 0 )^T $ . This implies that the singular locus of $ \mathrm{Spec}(\mathcal{A}(B_n)) $ is contained in the singular locus of $ P_{n-1}(z_1,\ldots, z_{n-1}) + 1 = 0 $ . The latter is empty if $ n - 1 \equiv 1 \mod 2 $ (Lemma 3.7(2)) or if $ n -1 = 2m $ for some $ m \in {\mathbb{Z}}_+ $ and $ m \equiv 0 \mod 2 $ (Proposition 3.8). Therefore, $ \mathrm{Spec}(\mathcal{A}(B_n)) $ is regular if $ n \not\equiv 3 \mod 4 $ .

Consider the case $ n \equiv 3 \mod 4 $ . Then the singular locus of $ P_{n-1}(z_1,\ldots, z_{n-1}) + 1 $ is $ V ( z_1, \ldots, z_{n-1}) $ (Proposition 3.8). In particular, $ P_{n-2}(z_1, \ldots, z_{n-2}) = 0 $ , by Lemma 3.6. Moreover, $ u_1 = 0 $ implies that $ g_n = 0 $ is equivalent to $ u_3 = 0 $ . Hence, $ \mathrm{Spec}(\mathcal{A}(B_n)) $ has an isolated singularity at the origin. Locally at the origin, $ u_1 u_2 - 1 $ is invertible and thus the generator $ g_n = (u_1 u_2 - 1 ) u_3 - u_1^2 - P_{n-2}(z_1,\ldots, z_{n-2}) $ can be eliminated. This has no effect on $ h_n $ since $ u_3 $ does not appear in it. We obtain that, locally at the singular point, $ \mathrm{Spec}(\mathcal{A}(B_n)) $ is isomorphic to an $ A_1 $ -hypersurface singularity, where the type of the singularity can be seen by applying the same coordinate transformation as in the proof of Proposition 3.8.

Suppose $ \operatorname{char}(K) = 2 $ . We make a case distinction for $ u_1 (u_1 u_2 - 1) = 0 $ . If $ u_1 = 0 $ , then the same arguments as for $ \operatorname{char}(K) \neq 2 $ apply, which provides an isolated $ A_1 $ -singularity at the origin if and only if $ n \equiv 1 \mod 2 $ .

Thus, let $ u_1 u_2 - 1 = 0 $ . Then $ g_n = h_n = 0 $ is equivalent to $ u_1^2 + P_{n-2} (z_1, \ldots, z_{n-2}) = P_{n-1} (z_1, \ldots, z_{n-1}) = 0 $ . The minor of the Jacobian matrix of $ g_n $ and $ h_n $ corresponding to $ ( u_2, z_{n-1} ) $ provides that we have to have $ u_1 u_3 P_{n-2} (z_1, \ldots, z_{n-2} ) = 0 $ . Since $ u_1 u_2 - 1 = 0 $ and $ P_{n-2} (z_1, \ldots, z_{n-2}) = - u_1^2 $ , we obtain $ u_3 = 0 $ . The column of the Jacobian matrix with respect to $ u_2 $ is $ (0,u_1)^T $ and $ u_1 \neq 0 $ provides that all derivatives of $ P_{n-2} (z_1, \ldots, z_{n-2} ) $ have to vanish. If $ n \equiv 1 \mod 2 $ , then $ V ( P_{n-2} + u_1^2 ) $ is regular by Lemma 3.7(2) (where $ u_1^2 \neq 0 $ takes the role of $ \lambda $ ). On the other hand, if $ n \equiv 0 \mod 2 $ , then Proposition 3.8 implies that $ V( P_{n-2} + u_1^2 ) $ has a singularity of type $ A_1 $ at $ V (z_1, \ldots, z_{n-2}) $ if and only if $ u_1 = 1 $ . From this, we obtain assertion (2^′).

5.2. $\textbf{C}_{\textbf{n}}$ cluster algebras

We choose the exchange matrix B for type $C_n$ , $n \geq 3$ , as:

\begin{equation*}B=\begin{pmatrix} 0\;\;\;\; & 1\;\;\;\; & \cdots\;\;\;\; & 0\;\;\;\; & 0\;\;\;\; & 0\;\;\;\; & 0 \\ -1\;\;\;\; & 0\;\;\;\; & \ddots\;\;\;\; & 0\;\;\;\; & 0\;\;\;\; & 0\;\;\;\; & 0 \\\vdots\;\;\;\; & \ddots\;\;\;\; & \ddots\;\;\;\; & \ddots\;\;\;\; & \vdots\;\;\;\; & \vdots\;\;\;\; & \vdots \\0\;\;\;\; & 0\;\;\;\; & \ddots\;\;\;\; & 0\;\;\;\; &1\;\;\;\; & 0\;\;\;\; & 0 \\0\;\;\;\; & 0\;\;\;\; & \cdots\;\;\;\; & -1\;\;\;\; & 0\;\;\;\; &1\;\;\;\; & 0 \\0\;\;\;\; & 0\;\;\;\; & \cdots\;\;\;\; & 0\;\;\;\; & -1\;\;\;\; & 0\;\;\;\; & 2 \\0\;\;\;\; & 0\;\;\;\; & \cdots\;\;\;\; & 0\;\;\;\; &0\;\;\;\; &-1\;\;\;\; & 0 \\\end{pmatrix}\end{equation*}

(cf. [Reference Fomin, Williams and Zelevinsky16, Section 5.5, (5.32)]). The corresponding Dynkin diagram is of type $ C_n $ (where n is the number of vertices):

For this cluster algebra, the characteristic of the field makes a significant difference. First, we provide a suitable presentation of $ \mathcal{A}(C_n) $ .

Lemma 5.3. Let K be any field and $ n \geq 3 $ . The cluster algebra $ \mathcal{A}(C_n) $ is isomorphic to

\begin{equation*} K[z_1, \ldots, z_{n+1}]/\langle P_n(z_1, \ldots, z_n)z_{n+1} - P_{n-1}(z_1, \ldots, z_{n-1})^2 -1 \rangle \ , \end{equation*}

where $ P_{n} $ is the continuant polynomial defined in Section 3. In particular, the variety $ \mathrm{Spec}(\mathcal{A}(C_n)) $ is isomorphic to a hypersurface in $ \mathbb{A}_K^{n+1} $ .

Proof. Since the underlying diagram $ \Gamma(B) $ is acyclic, the cluster algebra $ \mathcal{A}(C_n)$ has a presentation as:

\begin{equation*} K[x_1, \ldots, x_n, y_1, \ldots, y_n]/\langle h_1, \ldots, h_n \rangle, \end{equation*}

where we define $ h_1 \;:\!=\; x_1 y_1-x_2-1 $ , $ h_n \;:\!=\; x_n y_n -x_{n-1}^2-1 $ , and $ h_k \;:\!=\; x_k y_k - x_{k-1} - x_{k+1} $ , for $ k \in \{ 2, \ldots, n- 1 \} $ . Similar as for type $A_n$ (see Lemma 4.1) one can for $2 \leq k \leq n$ stepwise express $x_k$ in terms of $x_1, y_1,\ldots, y_{k-1}$ :

\begin{equation*}x_k=P_{k}(x_1, y_1, \ldots, y_{k-1}) \ . \end{equation*}

The only difference is in the last generator $ h_n $ , which becomes

\begin{equation*}h_n(x_1, y_1, \ldots, y_n)= P_n(x_1, y_1, \ldots, y_{n-1})y_n - P_{n-1}(x_1, y_1, \ldots, y_{n-2})^2 - 1 \ .\end{equation*}

After renaming the variables, we obtain the assertion.

Proof. The proof follows the steps of the proof of Proposition 3.8. We apply the Jacobian criterion to the presentation $\mathcal{A}(C_n) \cong K[z_1, \ldots, z_{n+1}]/\langle h_n(z_1, \ldots, z_{n+1}) \rangle $ of Lemma 5.3. Since $ \dfrac{\partial h_n}{\partial z_{n+1}} = P_n (z_1, \ldots, z_n) = 0 $ , the equation $ h_n = 0 $ for the singular locus is equivalent to $ P_{n-1}(z_1, \ldots, z_{n-1})^2 = -1 $ . If $ -1 \notin K^2 $ is not a square in K, the last equality cannot hold and we have shown that the claim.

Assume $ -1 \in K^2 $ and let $ \lambda \in K \setminus \{ 0 \} $ be such that $ \lambda^2 = - 1 $ and $ P_{n-1}(z_1, \ldots, z_{n-1}) = \lambda $ . Using Lemma 3.5, we get

\begin{equation*} 0 = \frac{\partial h_n}{\partial z_n} = P_{n-1} (z_1, \ldots, z_{n-1}) z_{n+1} = \lambda z_{n+1} \end{equation*}

and thus $ z_{n+1} = 0 $ .

We prove by induction on k that $ h_n = \dfrac{\partial h_n}{\partial z_{n+1}} \ldots = \dfrac{\partial h_n}{\partial z_{n-2k}} = 0 $ imply

1. (a _k ) $ z_{n+1} = \cdots = z_{n-2k+1} = 0 $ ,

2. (b _k ) $ P_{n-2k-1}(z_1, \ldots, z_{n-2k-1}) = (-1)^{k} \lambda $ , and

3. (c _k ) $ P_{n-2k}(z_{1}, \ldots, z_{n-2k}) = 0 $ .

As we have seen above, the statements are true for $ k = 0 $ . Let us deduce (a $_{k+1}$ ), (b $_{k+1}$ ), (c $_{k+1}$ ) from (a $_{k}$ ), (b $_{k}$ ), (c $_{k}$ ). First, (a $_k$ ) and Lemma 3.6 imply $ P_{2k}(z_{n-2k}, \ldots, z_{n-1}) = \pm 1 $ . Using $ z_{n+1} = 0 $ , (b $_0$ ), and Lemma 3.5, we have

\begin{equation*} 0 = \frac{\partial h_n}{\partial z_{n-2k-1}} = - 2 P_{n-1} (z_1, \ldots, z_{n-1}) \frac{\partial P_{n-1} (z_1, \ldots, z_{n-1}) }{\partial z_{n-2k-1}} = \end{equation*}

\begin{equation*} = (-2) \lambda P_{n-2k-2} (z_1, \ldots, z_{n-2k-2}) P_{2k}(z_{n-2k}, \ldots, z_{n-1}) = ( \mp 2) \lambda P_{n-2k-2} (z_1, \ldots, z_{n-2k-2}) \ , \end{equation*}

which provides (c $_{k+1} $ ) as $ \operatorname{char}(K) \neq 2 $ . If we apply the recursion of Lemma 3.4 for $ P_{n-2k}(z_{1}, \ldots, z_{n-2k}) $ and use (b $_k$ ) and (c $_{k+1} $ ), we obtain $ z_{n-2k} = 0 $ . On the other hand, Lemma 3.4 applied for $ P_{n-2k-1} (z_1, \ldots, z_{n-2k-1}) $ , (b $_{k} $ ), and (c $_{k+1} $ ) provide (b $_{k+1} $ ). It remains to prove $ z_{n-2k-1} = 0 $ . For this, we consider the derivative of $ h_n $ by $ z_{n-2k-2} $ (using Lemma 3.5) and apply (b $_0$ ), (b $_{k+1}$ ), as well as $ P_{2k+1} (z_{n-2k-1}, 0, \ldots, 0 ) = \pm z_{n-2k-1} $ (by Lemma 3.6).

Next, we distinguish two cases: if $ n = 2m $ for some $ m \in {\mathbb{Z}}_+$ , then we have not used the derivative by $ z_1 $ in the induction. Using $ z_{n+1} = 0 $ and (b $_0$ ), we obtain the contradiction:

\begin{equation*} 0 = \frac{\partial h_n}{\partial z_1} = -2 \lambda P_{2m-2} (z_2, z_3,\ldots, z_{n-1}) = \mp \lambda \neq 0 \ , \end{equation*}

where the last equality holds by applying (a $_{m-1}$ ) and Lemma 3.6.

Assume $ n = 2m +1 $ , for some $ m \in {\mathbb{Z}}_+ $ . Statement (a $_{m}$ ) implies $ z_2 = 0 $ and (b $_{m-1}$ ) states $ P_2 (z_1,z_2) = \pm \lambda $ . This leads to the equality $ \lambda = \mp 1 $ , which contradicts the property $ \lambda^2 = - 1 $ (see the definition of $ \lambda $ above).

Let us discuss the case $ \mathrm{char}(K) = 2 $ , which turns out to be more complicated.

Proof of Proposition 5.5. As we have seen above, $ \mathcal{A}(C_n) $ is isomorphic to the hypersurface determined by $ h_n (z_1, \ldots, z_{n+1}) = P_n(z_1, \ldots, z_n)z_{n+1} - P_{n-1}(z_1, \ldots, z_{n-1})^2 -1 $ . Since $ \mathrm{char}(K) = 2 $ , we can rewrite this as:

\begin{equation*} h_n (z_1, \ldots, z_{n+1}) = P_n(z_1, \ldots, z_n)z_{n+1} + \big( P_{n-1}(z_1, \ldots, z_{n-1}) + 1 \big)^2 \ . \end{equation*}

The partial derivatives are

\begin{equation*} \begin{array}{rcl} \displaystyle \frac{\partial h_n}{\partial z_{n+1}} & = & P_n(z_1, \ldots, z_n) \ , \\[12pt] \displaystyle \frac{\partial h_n}{\partial z_{k}} & = & \displaystyle z_{n+1} \frac{\partial P_n(z_1, \ldots, z_n) }{\partial z_k} \ , \ \ \ \mbox{ for } k \in \{ 1, \ldots, n \} \ . \end{array} \end{equation*}

This implies that we can replace the condition $ h_n(z_1, \ldots, z_{n+1}) = 0 $ by:

\begin{equation*} f_{n-2}(z_1, \ldots, z_{n-1}) = P_{n-1}(z_1, \ldots, z_{n-1}) + 1 = 0 \end{equation*}

when determining the singular locus, where $ f_{n-2}(z_1, \ldots, z_{n-1}) $ is the polynomial providing a hypersurface presentation for the variety $ \mathrm{Spec}(\mathcal{A}(A_{n-2})) $ (see Lemma 4.1 and use $ \mathrm{char}(K) = 2 $ ).

Since the second factor of $ \dfrac{\partial h_n}{\partial z_{n}} = z_{n+1} P_{n-1} (z_1, \ldots, z_n) $ cannot vanish if $ f_{n-2} = 0 $ and since $ P_n (z_1, \ldots, z_n) = z_n P_{n-1} (z_1, \ldots, z_{n-1}) + P_{n-2} (z_1, \ldots, z_{n-2}) $ , we obtain

\begin{equation*} \mathrm{Sing} (\mathcal{A}(C_n)) = V (u_n, z_{n+1}, f_{n-2}(z_1, \ldots, z_{n-1})) \;=\!:\;D, \end{equation*}

where we introduce $ u_n \;:\!=\; z_n + P_{n-2} (z_1, \ldots, z_{n-2}) $ . Observe that

\begin{equation*} h_n = f_{n-1}^2 + z_{n+1} \Big( u_n (1 + f_{n-1}) + f_{n-1} P_{n-2} \Big) \ . \end{equation*}

Corollary 4.3(1^′) implies that $ \mathrm{Sing} (\mathcal{A}(C_n)) $ is singular if and only if $ n \equiv 1 \mod 2 $ .

First, assume $ n \equiv 0 \mod 2 $ . Then, $ V (f_{n-1}) $ is regular and we may take $ u_{n-1} \;:\!=\; f_{n-1} $ as a local variable locally at the singular locus. Furthermore, $ 1 + f_{n-1} $ is a unit and we may introduce the local variable $ w_n \;:\!=\; u_n (1 + f_{n-1}) + f_{n-1} P_{n-2} $ . Hence, $ h_n = u_{n-1}^2 + z_{n+1} w_{n} $ and claim (1) follows.

Suppose that $ n \equiv 1 \mod 2 $ . By Corollary 4.3(2^′), $ \mathrm{Sing} (\mathcal{A}(C_n)) $ has an isolated singularity of type $ A_1 $ at the origin $ V (z_1, \ldots, z_{n-1}, u_n, z_{n+1} ) $ . The statement about the the type of singularity away from origin follows with the same argument as in the case $ n \equiv 0 \mod 2 $ .

Let us study the situation at the origin. Write $ n - 2 = 2m - 1 $ , for $ m \in {\mathbb{Z}}_+ $ . As we have seen in the proof of Proposition 3.8, there is a coordinate transformation $ (z_1, \ldots, z_{2m}) \mapsto (t_1, \ldots, t_{2m}) $ such that $ f_{n-2} (t_1, \ldots, t_{2m}) = \sum_{i=1}^{2m} t_{2i-1} t_{2i} $ , locally around the origin. Furthermore, we can again introduce $ w_n $ above such that, locally at the origin, we obtain $ h_n = ( \sum_{i=1}^{m} t_{2i-1} t_{2i} )^2 + w_n z_{n+1} $ , as desired.

Let us discuss the desingularization of the variety $ \mathrm{Spec}(\mathcal{A}(C_n)) $ . First, we blow up with center the origin. In order to simplify the presentation of the charts, we abuse notation and write $ z_n \;:\!=\; u_n $ .

We fix the notation when considering explicit charts of a blowup: (we only discuss this for the blowup of the origin, but it can be adapted for any blowup with a smooth center.) Recall that the blowup in the origin of $ {\mathbb{A}}^{n+1}_K = \mathrm{Spec}(R) $ , where $ R \;:\!=\; K[z_1, \ldots, z_{n+1}] $ , is given by $ \operatorname{Bl}_0({\mathbb{A}}^{n+1}_K) \;:\!=\; \operatorname{Proj}(R[Z_1, \ldots, Z_{n+1}]/ \langle z_i Z_j - z_j Z_i \mid i, j \in \{ 1, \ldots, n +1 \} \rangle ) $ , where $ (Z_1, \ldots, Z_{n+1} ) $ are projective variables. In particular, $ \operatorname{Bl}_0({\mathbb{A}}^{n+1}_K) $ is covered by the open subsets $ D_i $ given by $ Z_i \neq 0 $ , for $ i \in \{ 1, \ldots, n + 1 \} $ . We also say that $ D_i $ is the $ Z_i $ -chart.

Fix $ i \in \{ 1, \ldots, n +1 \} $ . Since $ z_i Z_j - z_j Z_i = 0 $ , for $ j \neq i $ , we obtain that $ z_j = z_i \dfrac{Z_j}{Z_i} $ in the $ Z_i $ -chart. This provides that the $ Z_i $ -chart is isomorphic to $ \mathrm{Spec} (K[z'_{\!\!1}, \ldots, z'_{\!\!n+1}] ) $ , where we set $ z'_{\!\!i} \;:\!=\; z_i $ and $ z'_{\!\!j} \;:\!=\; \dfrac{Z_j}{Z_i} $ for $ j \neq i $ . In order to keep the notation light, we abuse it by using the same letter for the variables after the blowup as before, i.e., the transformation of the variables will be written as $ z_j = z_i z_j $ for $ j \neq i $ .

$ Z_n $ -chart. We have $ z_i = z_n z_i $ for every $ i \neq n $ . The strict transform of $ h_n $ is

\begin{equation*} h{'}_{\!\!n} = f{'}_{\!\!n-1}^{2} z_n^{2} + z_{n+1} ( 1 + z_n^2 f{'}_{\!\!n-1} + z_n^{2} f{'}_{\!\!n-1} P{'}_{\!\!n-2} ) \ , \end{equation*}

where we denote by $ f'_{\!\!n-1} $ (resp. $ P'_{\!\!n-2} $ ) the strict transform of $ f_{n-2} $ (resp. $ P_{n-2} $ ). The strict transform D ^′ of D is empty in this chart. Hence, the only singularities which may appear have to be contained in the exceptional divisor $ V (z_n) $ . On the other hand, we have $ \dfrac{\partial h'_{\!\!n}}{\partial z_{n+1}'} = 1 + z_n^2 f'_{\!\!n-1} + z_n^2 f'_{\!\!n-1} P'_{\!\!n-2} $ , which implies that the strict transform of $ \mathrm{Spec} (\mathcal{A} (C_n)) $ is regular in this chart.

The analogous argument applies for the $ Z_{n+1} $ -chart.

$ Z_1 $ -chart. (All other charts remaining are analogous.) We get $ z_i = z_1 z_i $ for every $ i \neq 1 $ and

\begin{equation*} h{'}_{\!\!n} = f{'}_{\!\!n-1}^{2} z_1^{2} + z_{n+1} \Big( z_n ( 1 + z_1^2 f{'}_{\!\!n-1}) + z_1^2 f{'}_{\!\!n-1} P{'}_{\!\!n-2} \Big) \ . \end{equation*}

By Corollary 4.3(2^′) $ V(f'_{\!\!n-1}) $ is regular and thus the same is true for D ^′. Lemma 3.6 implies that $ f'_{\!\!n-1} = z_2 + z_4 + \cdots + z_{n-1} + H $ for some $ H \in \langle z_1, z_2, \ldots, z_{n-1} \rangle^2 $ . In particular, $ V(f'_{\!\!n-1}) $ is transversal to $ V(z_1 z_n z_{n+1} ) $ and we may introduce the variable $ u_2 \;:\!=\; f'_{\!\!n-1} $ locally at D ^′.

Since any newly created singularities have to be contained in the exceptional divisor $ V(z_1) $ , the element $ 1 + z_1^2 f'_{\!\!n-1} $ is invertible locally at the singular locus of $ V ( h'_{\!\!n} ) $ . We introduce the local variable $ w_n \;:\!=\; z_n ( 1 + z_1^2 f'_{\!\!n-1}) + z_1^2 f'_{\!\!n-1} P'_{\!\!n-2} $ and we get

\begin{equation*} h{'}_{\!\!n} = f{'}_{\!\!n-1}^{2} z_1^2 + w_n z_{n+1}. \end{equation*}

Obviously, we have $ \mathrm{Sing} (V(h'_{\!\!n})) = D' \cup V (z_1, w_n, z_{n+1} ) $ . Both components are regular and $ V(h'_{\!\!n}) $ is an $ A_1 $ -singularity at every point expect their intersection. Let us define $ E \;:\!=\; V (z_1, w_n, z_{n+1} ) $ .

Next, we blow up with center D ^′. Observe that this is a well-defined global center, which is seen in any $ Z_i $ -chart with $i \in \{ 1 ,\ldots, n-1 \} $ as it is the strict transform of $ \mathrm{Sing}(\mathrm{Spec}(\mathcal{A}(C_n))) $ . There are no new singularities contained in the exceptional divisor of the second blowup, and hence the singular locus of $ V(h''_{\!\!\!n}) $ has to be the strict transform E ^′′ of E (where $ h''_{\!\!\!n} $ is the strict transform of $ h'_{\!\!n} $ after the second blowup). Locally at E ^′′, the hypersurface is given by an equation of the form $ x^2 - yz = 0 $ . Therefore, after blowing up E ^′′ all singularities are resolved. Again observe that E ^′′ is a well-defined global center at this step of the resolution process, since it is the singular locus of the strict transform after the second blowup.

Case (3) is seen by explicit computation.

5.3. $\textbf{F}_{4}$ and $\textbf{G}_{2}$ cluster algebras

Let us discuss the remaining cases of $ \mathcal{A}(F_4) $ and $ \mathcal{A}(G_2 ) $ .

For the cluster algebra $ \mathcal{A}(F_4) $ , we pick the exchange matrix $ B = \begin{pmatrix} 0 & 1 & 0 &0 \\ -1 & 0 &1 & 0 \\ 0 & -2 & 0 & 1\\ 0 & 0 & -1 & 0 \end{pmatrix}$ (cf. [Reference Fomin, Williams and Zelevinsky16, Exercise 4.4.12]), whose corresponding Dynkin diagram is of type $F_4$ :

Proof. The underlying graph $ \Gamma(B) $ is acyclic and thus $ \mathcal{A}(F_4 ) $ is isomorphic to its lower bound cluster algebra. Similar as above, we obtain the presentation:

\begin{equation*} \mathcal{A}(F_4) \cong K[x_1, \ldots, x_4, y_1, \ldots, y_4]/ \langle x_1y_1 -x_2 -1, x_2y_2 -x_1 -x^2_3, x_3y_3 -x_2 -x_4, x_4y_4-x_3-1 \rangle \ . \end{equation*}

By eliminating $ x_1, x_2, x_3 $ , this can be simplified to

\begin{equation*} \mathcal{A}(F_4) \cong K[x,y,z,w,t]/\langle xyzwt-x^2yt^2-xyz-yzw+2xyt-xwt+x-y+w-1 \rangle \ . \end{equation*}

Hence, $\mathrm{Spec}(\mathcal{A}(F_4))$ is isomorphic to a hypersurface in ${\mathbb{A}}^5_K$ . Moreover, the Jacobian criterion shows that the latter is regular.

Next, let us come to $ \mathcal{A}(G_2 ) $ . A possible exchange matrix is $ B = \begin{pmatrix} 0\;\;\;\; & 1 \\ -3\;\;\;\; & 0 \end{pmatrix} $ (cf. [Reference Fomin, Williams and Zelevinsky16, Section 5.7]) and the corresponding Dynkin diagram is of type $G_2$ :

Proof. Since $ \Gamma(B) $ is acyclic, the cluster algebra $\mathcal{A}(G_2)$ is isomorphic to its lower bound cluster algebra by Theorem 2.12. We get the presentation:

\begin{equation*} \mathcal{A}(G_2) \cong K[x_1,x_2,y_1,y_2]/\langle x_1y_1-1-x_2^3,x_2y_2-1-x_1\rangle \ . \end{equation*}

Since $x_1$ can be expressed in term of the other variables, we find

\begin{equation*} \mathcal{A}(G_2)\cong K[x,y,z]/\langle z^3-xyz+y+1\rangle \ . \end{equation*}

By the Jacobian criterion, this algebra is regular for $\mathrm{char}(K) \neq 3$ , and for characteristic 3 the Jacobian criterion yields $\mathrm{Sing}(\mathcal{A}(G_2)) \cong V(x+1,y,z+1) $ , a closed point.

Assume $\mathrm{char}(K)=3$ and set $x'\;:\!=\;x+1$ , and $z'\;:\!=\;z+1$ . Then, we obtain $\mathcal{A}(G_2)\cong K[x',y,z']/\langle z'^3-x'yz'+yz'+x'y\rangle$ , where the singular point is the origin in the new coordinates. Applying the local coordinate change $x'=\dfrac{\tilde x -z'}{1-z'}$ yields

\begin{equation*} \mathcal{A}(G_2)_{\langle \tilde x, y, z' \rangle} \cong K[\tilde x, y, z']_{\langle \tilde x, y, z' \rangle} /\langle z'^3+\tilde x y \rangle \ , \end{equation*}

which is a singularity of type $ A_2 $ , see [Reference Greuel and Kröning27, Definition 1.2].