P

is prime when

A B \subseteq P

A \subseteq P

B \subseteq P

k e_{21}

lies in every prime of

T_{2} (k)

, and the quotient is

k \times k

. Hence the two primes are

P_{a} = {(a, c, d) : a = 0}, P_{d} = {(a, c, d) : d = 0} .

Here

C (P_{a}) = {a \neq = 0}

and

C (P_{d}) = {d \neq = 0}

Direct multiplication of triples

(a, c, d) (a^{'}, c^{'}, d^{'}) = (a a^{'}, c a^{'} + d c^{'}, d d^{'})

shows that all of

C (P_{a})

satisfies the left Ore equations. For

P_{d}

, the largest left Ore subset is

{(a, c, d) : a \neq = 0, d \neq = 0} = T_{2} (k)^{*} .

Indeed units always form an Ore set. If

d \neq = 0

but

a = 0

, applying the Ore equation successively to

e_{21}

and

e_{22}

gives incompatible equations, so no left Ore subset can contain that element.

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