Search Results

For $k=4$ the sets $\{1,2,3\}$, $\{1,2,4\}$, $\{1,3,4\}$ and $\{2,3,4\}$ are linearly independent as elements of $\mathbb Z_2^4$. However, the chain of $Y_i$ can not be strictly increasing. In the ot …

You can think of this ring as the semigroup ring of the semigroup $S$ generated by $$(1,0,0,1),(-1,0,0,1),(0,1,0,1),(0,-1,0,1),(0,0,1,1),(0,0,-1,1).$$ The above semigroup elements correspond to $f_1,f …

One can prove this statement along the following lines. Prove that ${\rm Trace}(M(l,n)) = 1+l +\cdots + l^{n-1}$. Prove that $M(l,n)^p = M(l^p,n)$. Clearly, these statements 1 and 2 together imply …

In general, $g(n)$ is strictly less than $f(n)$, although they have the same leading term. One sufficient condition to get $g(n)=f(n)$ is to require smoothness of the corresponding toric variety. Comb …

Since you have changed the formulation of the question slightly to now require identity for all $n$, not just for large ones (or maybe it was just me misreading the original post), let me offer a suff …

Consider the generating functions $$ R_k(u) = \sum_{j_1,\ldots,j_k\geq 0} u^{j_1 + 2 j_2 + \cdots + k j_k} \prod_{t=1}^k \frac 1{j_t! t^{j_t}} f_t(j_1,\ldots,j_t). $$ I will prove by induction on $k$ …

This is a minor curiosity that came up in a joint project recently. Consider the sequence $a_n=3\frac {(2n)!}{(n+2)!(n-1)!}$ (A000245 in OEIS). It has multiple combinatorial descriptions. One can w …

Bernd Sturmfels and others have studied varieties defined by binomial ideals. This is what you are looking for.

A "cube" $[-1,1]^n$ has $2^n$ vertices and is reflexive.

Sorry, don't know a reference, but here is a quick argument. If $M=p^ab+c$ with $0\leq c\leq p^a-1$, then $$(1+x)^M=(1+x)^{p^ab}(1+x)^c =(1+x^{p^a})^b(1+x)^c \mod p. $$ In turn, this equals $$ (1+bx …