linear algebra – q (T) = 0 implies that q is a multiple of the minimum polynomial


I read Sheldon Axler Linear Algebra Done Right Third Edition. There is evidence on page 264. I don't understand any of the equality.

Sentence: Accept $ T in mathcal {L} (V) $ and $ q in mathcal {P} ( mathbb {F}) $. Then $ q (T) = 0 $ then and only if $ q $ is a polynomial multiple of the minimal polynomial of $ T $.

Proof:
To let $ p $ denote the minimal polynomial of $ T $.
Accept $ q (T) = 0 $.
According to the division algorithm for polynomials, there are polynomials $ s, r in mathcal {P} ( mathbb {F}) $ so that $ q = ps + r $ and degrees $ r $ <Degrees $ s $.

We have $ 0 = q (T) = p (T) s (T) + r (T) = r (T) $.

The above equation implies that r = 0 (otherwise divided $ r $ its highest degree coefficient would produce a monic polynomial which, when applied to $ T $ returns 0; This polynomial would be less than the minimal polynomial, which would be a contradiction.

My question is:
How do we know? $ p (T) s (T) + r (T) = r (T) $?