The degree of a vertex in a hypergraph is defined as the number of hyperedges in which it occurs. The degree of the hypergraph is the maximum of the degrees of the vertices.
Given a hypergraph $\mathcal{H}$, is there a way to decompose it into two hypergraphs $\mathcal{H'}$ and $\mathcal{H}\setminus\mathcal{H'}$ such that both have strictly smaller degree?
I tried taking $\mathcal{H'}$ to be a minimal vertex cover (of hyperedges) of minimum degree. Then clearly $\mathcal{H}\setminus\mathcal{H'}$ has strictly smaller degree. But I'm finding it hard to argue (by contradiction) that $\mathcal{H'}$ also has strictly smaller degree (Disclaimer: I'm not sure about the truth of this statement.)
Breaking a hypergraph into two hypergraphs of smaller degree
combinatoricshypergraphs
Best Answer
Example: a hypergraph $\mathcal H$ of degree $d$ which can't be decomposed into $k$ hypergraphs of degree less than $d$.
Choose an integer $n\gt k(d-1)$ and let $[n]=\{1,2,\dots,n\}$.
The vertex set of $\mathcal H$ is the set $$V=\binom{[n]}d=\{v\subseteq[n]:|v|=d\}$$ of all $d$-element subsets of $[n]$.
The edge set of $\mathcal H$ is the set $$E=\{e_1,e_2,\dots,e_n\}$$ where $$e_i=\{v\in V:i\in V\}.$$
Plainly, the hypergraph $\mathcal H=(V,E)$ is regular of degree $d$.
Consider any decomposition $E=E_1\cup\dots\cup E_k$; this corresponds to a partition $[n]=N_1\cup\dots\cup N_k$ where $E_j=\{e_i:i\in N_j\}$.
Since $n\gt k(d-1)$, there is some $j\in[k]$ with $|N_j|\ge d$; so we can choose $v\in E$ with $v\subseteq N_j$, and then $v$ will be a vertex of degree $d$ in the subhypergraph $\mathcal H_j=(V,E_j)$.
Intuition. In terms of the dual hypergraph (interchanging vertices and edges), you're asking to bound the chromatic number of a hypergraph of rank $d$ (edges of size $d$), and of course there is no such bound.