By Irena Swanson

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**Example text**

The full configuration of the graph is the loading of each vertex by |v| − 1 grains of sand. An overfull configuration is a loading the full configuration by a finite number of extra grains of sand. Let G be the set of all stable configurations (for this graph/sink) obtained from toppling each overfull configuration. This will be the sandpile group. ) Clearly G is a finite set, with at most Πv |v| elements, as v varies over the vertices of the graph. This is a very rough estimate, however. Note that the constant zero loading is not allowed in G if the graph has at least one edge, in fact, the configuration having an edge between two vertices with zero grains of sand on them is not a possible stable configuration.

From any loading of the graph with grains of sand, the toppling produces a unique stable configuration in finitely many steps. The stable configuration and the number of topplings do not depend on the order of the topplings. Proof. , which means, by the assumption on how the sink is connected, that the sink would receive infinitely many grains of sand. But we only allow finitely many grains of sand, so a stable configuration is always reached in finitely many steps. We start with some initial configuration.

Sk ). If k = 1, we also write s1 R for (s1 ). The empty set generates the ideal (0) = {0}. Note that the ideal (s1 , . . , sk ) consists of all elements of the form over elements of R. 4 Let R = Z. Prove that (2, 3) = (1), or more generally, that (n1 , . . , nk ) = (gcd(n1 , . . , nk )). 5 Prove that every ideal in Z is generated by (at most) one element. 6 Let X, Y, Z, T be variables over Q. 2, there exists a unique ring homomorphism ψ : Q[X, Y, Z] → Q[T ] such that ψ|Q is identity, ψ(X) = T , ψ(Y ) = T 2 , ψ(Z) = T 3 .