Thanks the correction by TakuyaMurata! In general the following statement would not be true.
An arithmetic genus (of a smooth projective variety) is a birational invariant. --Enyokoyama (talk) 12:07, 20 July 2013 (UTC)
But apparently an arithmetic genus is a birational invariant for the most of the times; for the curves (since arithmetic genus is geometric genus and the latter is a birational invariant) or over C by Hodge theory. In any case, the matter should be discussed in detail in the body of the article. -- Taku (talk) 21:21, 20 July 2013 (UTC)