On Jul 26, 2012, at 4:21 PM, Tero Kivinen wrote:
> If that is correct how does the PKIX solve this? I.e. when I have
> certificate signed by the some other certificate using DSA? If my
> reading of RFC5280 is correct there is this signatureAlgorithm ASN.1
> blob in front of the signature itself and that gives all that
> information (including the domain parameters and hash functions etc).
>
> Is my understanding correct?
Yes. If you print out a certificate signed with ECDSA with OpenSSL, it looks
like this:
MBA:tmp ynir$ openssl x509 -in the_ca_cert.crt -text
Certificate:
Data:
Version: 3 (0x2)
Serial Number: 1 (0x1)
Signature Algorithm: ecdsa-with-SHA1
Issuer:
Validity
Not Before: Jul 26 17:56:00 2012 GMT
Not After : Jul 26 17:56:00 2022 GMT
Subject:
Subject Public Key Info:
Public Key Algorithm: id-ecPublicKey
Public-Key: (256 bit)
pub:
04:b9:36:b6:65:ee:65:27:70:f0:f9:16:67:78:53:
b8:be:14:29:c5:36:09:a7:3b:0a:f0:0d:59:4d:31:
6d:9a:f3:be:fd:bf:e3:6e:0e:39:69:96:c9:d8:ae:
74:79:3d:f8:af:b5:5a:65:44:fe:76:c1:8c:52:18:
f3:6e:49:43:23
ASN1 OID: prime256v1
X509v3 extensions:
X509v3 Basic Constraints: critical
CA:TRUE
X509v3 Subject Key Identifier:
9B:D6:1D:1A:9B:B8:CE:2D:46:FD:B1:A5:6E:7D:35:E6:05:04:F0:36
X509v3 Key Usage:
Certificate Sign, CRL Sign
Netscape Cert Type:
SSL CA, S/MIME CA, Object Signing CA
Netscape Comment:
xca certificate
Signature Algorithm: ecdsa-with-SHA1
30:45:02:20:61:68:e4:5e:9c:47:2d:d6:49:f6:6b:24:cb:43:
cd:20:2e:c4:5d:bb:1a:45:d9:95:09:6b:89:93:5b:00:d2:cb:
02:21:00:b9:c4:d3:d4:4a:98:e0:d6:20:45:b2:95:8b:4a:06:
d5:6b:3d:90:f6:a7:81:be:1f:d0:c5:f1:a8:b5:6a:d0:b9
See the "Signature Algorithm" field before the signature itself? If we dump
the ASN.1 the signature is a sequence of an OID and a bitstring:
MBA:tmp ynir$ openssl asn1parse -in the_ca_cert.crt -i -dump
0:d=0 hl=4 l= 351 cons: SEQUENCE
…
270:d=1 hl=2 l= 9 cons: SEQUENCE
272:d=2 hl=2 l= 7 prim: OBJECT :ecdsa-with-SHA1
281:d=1 hl=2 l= 72 prim: BIT STRING
0000 - 00 30 45 02 20 61 68 e4-5e 9c 47 2d d6 49 f6 6b .0E. ah.^.G-.I.k
0010 - 24 cb 43 cd 20 2e c4 5d-bb 1a 45 d9 95 09 6b 89 $.C. ..]..E...k.
0020 - 93 5b 00 d2 cb 02 21 00-b9 c4 d3 d4 4a 98 e0 d6 .[....!.....J...
0030 - 20 45 b2 95 8b 4a 06 d5-6b 3d 90 f6 a7 81 be 1f E...J..k=......
0040 - d0 c5 f1 a8 b5 6a d0 b9- .....j..
In IKE we only have the bitstring, so we must infer the OID from something else.
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