The solution (still sans spaces):
Yes I missed the third i in division and an e from seen (buried somewhere in there). Congratulations to all those who solved it. Apparently the letters t and e were bang on predicted frequency which if you used those for a starting point (being the most used letters it makes sense) meant the problem wasn't too hard.
Interesting as a side note is frequency analysis in other languages. In English, e typically exceeds 10% of all letters (and is the only letter to do so) and 5 likely have frequencies below 1% (j, k, q, x and z). In Italian 3 letters exceed 10% and 9 are less than 1%. In German nearly 20% of all letters are e's. Note, thought that these are averages and a non-average passage may not conform to these patterns (see: The Void by Georges Perec)
Also a theoretical (if extremely ineffient) way to determine whether an attempt at decoding a mono-substitution cipher is gibberish would be to check for a series of common words. First you discard any attempt that doesn't contain the string 'the' somewhere in it, then discard all remaining ones which don't contain 'and' and so on through common words. This would only work with a long enough passage that you could be confident it would contain these common words.
Right next up is the Vigenère cipher (after mentioning it in the second post I expect you to solve this one RC). Basically it's a Caesar shift cypher except each position uses a different shift amount. It's kind of wordy to explain so I'll direct to an excellent and very useful Wikipedia page.
There you can find the exact details of the cipher, examples of it in use, the history of it and tips on how to crack it.
Ho qd uswqgn trm xwkf lkhmkbtk eadx bp dpp ljmnm jhkzea sywgn trm BJk pmlwidm zn zua ehcu wq mpbmkpexkp. Pw ua t mobupz bgvbvr mwwtwsm myaxaqmj fpta roiwtq avef ckup qffw aps yey ik m axuiyz lvv til zhyey ljmutaim qxxjadxtexb tv 1998. Zq ptz gbmlb htglpckt dsaxtl pnmtflazo lpzo, aamwp & IT hnn pp idew ahs k jlawnies bkkvojacgk, hkdtvy eqzuen etbz fpx Widbdjmdoa Wibiemk acm vf rqrp kophvl. Rm sik m amyoxo lze, m vbje actkc dmelacm lvv szxht zidaazo lrivtd, jmf px ps cbttd ewflwrie qfowgzicbpvl. Tm glena ew ayxkvvo wy pae wolrktw rmpoxtexb lvv pmvpsswy uswqgn, beb tb ae xkvbkjwg eazx vf k klaw an ehcu wq mpbmkpexkp. Ik tm vvndqycwe bh sekzy bg dmtk donpvkqa al wstw ljmutaimiwtq uuiyofm la s bttfeb. Pp usk ptce dpp paspxzt exdqvq wy ahsa rzggx, ubt sa azgniusy dpp ugeb "khw" yn epw qtbae aclzlqzuhcua pvlqzbug sv epw pztmt. Sn epw ZNE aeku epsf lkhfda sqe ua ihtsmyb, ltmr divt mm jqetydol tv luux didp lv wjkxslove xdmgxy.
I've left spaces and punctuation in this time to aid in the decoding process. Also I can tell you that the key isn't randomly generated (ie consists of real words) and it isn't the length of the code (that would be a one-time pad cipher which is uncrackable as long as each key is only used once) or even close to it.
Once again, PM me with either the key (which would show you've done the hard yards) or the plaintext if you solve it. Any questions can be asked either that way (if you feel you could be revealling something important) or in this thread.
Remember, your first step should be trying to find the length of the key.