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An ideal will be a ring with identity iff it is a two sided ideal of the form $eR$ where $e$ is an idempotent and $eR(1-e)=0$. Clearly $e$ is the identity element, and in fact $eR=eRe$, the "corner ring" of e. This is all straightforward to see: the identity element is an idempotent, and acts as a left identity on $eR$, so we want it to be a right identity ...
A direct sum of $FH$-modules is also a direct sum of vector spaces, so it is enough to consider a single term $M \otimes t_i$ in the decomposition. Now $M \otimes t_i = \{ m \otimes t_i : m \in M \}$. So, if $a_1,\ldots,a_n$ is a basis of $M$ as a vector space over $F$, then every element of $M$ can be written uniquely as $\sum_{i=1}^n f_i a_i$ with \$f_i ...