Can a semi-decidable problem be also decidable? As far as I understand, a semi-decidable (recursively enumerable) problem could be: 


  
  decidable (recursive) or 
  undecidable (nonrecursively enumerable)
  


This post made me wonder if this is not conventionally followed. This is my answer to it and as far as I understand it is correct:


  A semidecidable problem (or equivalently a recursively enumerable
  problem) could be:
  
  Decidable: If the problem and its complement are both semidecidable
  (or recursively enumerable), then the problem is decidable
  (recursive).
  
  Undecidable: If the problem is semidecidable and its complement is not
  semidecidable (that is, is not recursively enumerable).
  
  Important note: Remember that a decidable (recursive) problem is also
  semidecidable (recursively enumerable). Conversely, if a problem is
  not recursively enumerable (semidecidable), then is not recursive
  (decidable).


What the Wikipedia entry says is that:


  Partially decidable problems that are not decidable are called
  undecidable.
  
  In general, a semidecidable problem (recursively enumerable) could be
  decidable (recursive) or undecidable (nonrecursively enumerable).
  
  Also note that a problem and its complement could both (or just one of
  them) be not even semi-decidable (nonrecursively enumerable). Also
  note that, if a problem is recursive, its complement is also
  recursive.


Is it conventionally (always) understood this way? Is there some literature that presents semi-decidability (partially decidable, recursively enumerable) problem as an equivalent of undecidability?
Yes, a recursively enumerable language may be either decidable or undecidable. To see this, you ust look at the definitions of the terms.

A language $L$ is recursive (aka decidable) if there is a Turing machine that halts for all inputs, accepting every word in&nbsp;$L$ and rejecting every word not in&nbsp;$L$. $L$ is recursively enumerable (aka semi-decidable) if there is a Turing machine that halts and accepts any input in&nbsp;$L$ and, for any input not in&nbsp;$L$, it either halts and rejects or it does not halt.

Therefore, every recursive language is recursively enumerable. The machine that decides the recursive language is a special case of the machine required for a recursively enumerable language: specifically, it is allowed to either loop forever or reject for words not in&nbsp;$L$; in fact, it always rejects and never uses the option of looping forever.