On Mon, Feb 12, 2018 at 3:28 PM, Warren Kumari <war...@kumari.net> wrote:

> <author hat only>
>
> Hi all,
>
> Sorry it has taken so long to get a new version of this document
> posted - you deserve better.
>
> Anyway, we've finally posted an updated version -
> https://datatracker.ietf.org/doc/draft-ietf-dnsop-kskroll-sentinel/
>
> This version includes a (hopefully easily understood) description of
> how this would actually be used, and not just "here's a protocol, k,
> thnx, bye!". I've tried to layout what each party does, and how it all
> fits together - please let me know if it isn't clear. This section is
> towards the top of the document - we will likely make it an Appendix
> before publication.
>
> I've also updated it to use the kskroll-sentinel-is-ta-<id> format. It
> is easy to change again in the future, but this seemed to be what the
> working group liked. I also updated my demo implementation
> (http://www.ksk-test.net) to use this naming scheme.
>
> This version also clarifies that the test is "Is the Key ID a DNSSEC
> root KSK?" Originally my view was that it should be "Is there *any*
> key in the trust store with this keyID?", but after running some
> numbers I decided that there is a significant chance of false
> positives.
>
> As I mentioned, it took an embarrassingly long time to post the update
> - please let us know if we missed your comments.
>
> W
>
>
Thanks, the explanation helped.  I finally realized that it only works if
all (or most) validating resolvers are updated to support this new feature,
otherwise we have a bunch of responses that are uncertain.

Thinking out loud...
If an entry could be put in the root zone, that is signed only with the new
key, then could users query that and always get a yes/no answer to whether
they will be affected?
I realize it might be difficult politically or process-wise.
Not sure if it could be just a pair of A and AAAA records, or a delegation
to a temporary TLD.
And is it even possible, since it probably needs a new ZSK based on the new
KSK?

-- 
Bob Harold
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