Length of fields in an RSA CRT Key In a 2048 bit RSA key. Are $P, Q$, exponent $P$, exponent $Q$, and the coefficient guaranteed to be half of the length of the modulus (128 bytes)?
No, it is not guaranteed that in a 2048-bit RSA key, the prime factors $p$ and $q$ of the modulus are exactly half of the length of the modulus or 128 octets; much less so for $d_p$, $d_q$, and $q_\text{inv}$ . 

In particular, PKCS#1v2.2 allows almost total freedom on size of $p$ and $q$ (and even allows more than two prime factors for the public modulus). $p$ and $q$ being of exactly equal size however follows from requirements mandated by some common standards, in particular FIPS&nbsp;186-4, which requires $p$ and $q$ to be in the interval $(2^{1023.5},2^{1024})$ for a 2048-bit modulus.  This can be traced to its ancestor on RSA matters, ANSI&nbsp;X9.31 (1998).

While having $p$ and $q$ of exactly half the modulus is definitely common, there are implementations around which generate $p$ and $q$ of different size, and even standards like ETSI&nbsp;TS&nbsp;102 176-1 with recommendation that clearly allow that, and have been interpreted as favoring it:


  $p$ and $q$ should have roughly the same length, e.g. set a range such as $0,1 &lt; | \log_2p - \log_2q | &lt; 30$;




Even if $p$ and $q$ are exactly 1024-bit, this does not insure that $d_p$, $d_q$, and $q_\text{inv}$ are exactly 1024-bit or 128-octet. They are often slightly smaller: the high order bits of $q_\text{inv}$ behave about as if $q_\text{inv}$ was random in interval $(0,p)$, thus $q_\text{inv}&lt;2^{1016}$ with probability I guess at least $1/256$ (I welcome a proof or refutation of the estimate, or/and a better one).