BUG #17211: Partitioned index partition does not inherit table partition tablespace

The following bug has been logged on the website:

Bug reference:      17211
Logged by:          Andy S
Email address:      gatekeeper.mail@gmail.com
PostgreSQL version: 14.0
Operating system:   Debian Bullseye
Description:

Suppose we have 2 tablespaces:
pg_default (which is slow)
fast

We also have a partitioned table:
create table a (id serial not null, b int not null primary key, c int not
null, d int) partition by hash (b);

and 2 partitions:
create table a_1 partition of a for values with (modulus 2, remainder 0);
create table a_2 partition of a for values with (modulus 2, remainder 1)
tablespace fast;

This successfully allocates table a_2 pages within the 'fast' tablespace;
but this fails to allocate a_2_pkey primary key index there too and the
result is the index is allocated on tablespace oid(0) which means pg_default
I guess, at least the corresponding relfilenode is allocated within the
pg_default tablespace.

There's also a CREATE TABLE hack for index/constraint declaration within the
command that allows USING INDEX TABLESPACE, though there's an unresolved
error loop in partitioned tables use-case for such a syntax:
create table a_2 partition of a (primary key using index tablespace fast)
for values with (modulus 2, remainder 1) tablespace fast;
fails since there's no primary key definition provided

create table a_2 partition of a (primary key (b) using index tablespace
fast) for values with (modulus 2, remainder 1) tablespace fast;
fails with 'ERROR:  multiple primary keys for table "a_2" are not allowed'

The only possibility for a user to then allocate both index(es) and a table
on a non-default tablespace is to create a partition, scan pg_class/pg_index
to find all the names of indexes' partitions allocated for the new table,
traverse them programmatically and ALTER INDEX SET TABLESPACE them. Which
does not seem convenient since there's a simpler syntax/approach for a
regular table.