To encrypt something using RSA algorithm you need modulus and encryption (public) exponent pair (n, e). That's your public key. To decrypt something using RSA algorithm you need modulus and decryption (private) exponent pair (n, d). That's your private key.
To encrypt something using RSA public key you treat your plaintext as a number and raise it to the power of e modulus n:
ciphertext = ( plaintext^e ) mod n
To decrypt something using RSA private key you treat your ciphertext as a number and raise it to the power of d modulus n:
plaintext = ( ciphertext^d ) mod n
To generate private (d,n) key using openssl you can use the following command:
openssl genrsa -out private.pem 1024
To generate public (e,n) key from the private key using openssl you can use the following command:
openssl rsa -in private.pem -out public.pem -pubout
To dissect the contents of the private.pem private RSA key generated by the openssl command above run the following (output truncated to labels here):
openssl rsa -in private.pem -text -noout | less
modulus - n
privateExponent - d
publicExponent - e
prime1 - p
prime2 - q
exponent1 - d mod (p-1)
exponent2 - d mod (q-1)
coefficient - (q^-1) mod p
Shouldn't private key consist of (n, d) pair only? Why are there 6 extra components? It contains e (public exponent) so that public RSA key can be generated/extracted/derived from the private.pem private RSA key. The rest 5 components are there to speed up the decryption process. It turns out that by pre-computing and storing those 5 values it is possible to speed the RSA decryption by the factor of 4. Decryption will work without those 5 components, but it can be done faster if you have them handy. The speeding up algorithm is based on the Chinese Remainder Theorem.
Yes, private.pem RSA private key actually contains all of those 8 values; none of them are generated on the fly when you run the previous command. Try running the following commands and compare output:
# Convert the key from PEM to DER (binary) format
openssl rsa -in private.pem -outform der -out private.der
# Print private.der private key contents as binary stream
xxd -p private.der
# Now compare the output of the above command with output
# of the earlier openssl command that outputs private key
# components. If you stare at both outputs long enough
# you should be able to confirm that all components are
# indeed lurking somewhere in the binary stream
openssl rsa -in private.pem -text -noout | less
This structure of the RSA private key is recommended by the PKCS#1 v1.5 as an alternative (second) representation. PKCS#1 v2.0 standard excludes e and d exponents from the alternative representation altogether. PKCS#1 v2.1 and v2.2 propose further changes to the alternative representation, by optionally including more CRT-related components.
To see the contents of the public.pem public RSA key run the following (output truncated to labels here):
openssl rsa -in public.pem -text -pubin -noout
Modulus - n
Exponent (public) - e
No surprises here. It's just (n, e) pair, as promised.
Now finally answering the initial question: As was shown above private RSA key generated using openssl contains components of both public and private keys and some more. When you generate/extract/derive public key from the private key, openssl copies two of those components (e,n) into a separate file which becomes your public key.