§ Prime numbers as maximal among principal ideals

I learnt of this characterization from benedict gross's lectures, lecture 31. We usually define a number pRp \in R as prime iff the ideal generated by pp, (p)(p) is prime. Formally, for all a,bRa, b \in R, if ab(p)ab \in (p) then a(p)a \in (p) or b(p)b \in (p). This can be thought of as saying that among all principal ideals, the ideal (p)(p) is maximal: no other principal ideal (a)(a) contains it.

§ Element based proof

  • So we are saying that if (p)(a)(p) \subseteq (a) then either (p)=(a)(p) = (a)
  • Since (p)(a)(p) \subseteq (a) we can write p=arp = ar. Since (p)(p) is prime,and ar=p(p)ar = p \in (p), we have that either a(p)r(p)a \in (p) \lor r \in (p).
  • Case 1: If a(p)a \in (p) then we get (a)(p)(a) \subseteq (p). This gives (a)(p)(a)(a) \subseteq (p) \subseteq (a),or (a)=(p)(a) = (p).
  • Case 2: Hence, we assume a∉(p)a \not \in (p), and r(p)r \in (p).Since r(p)r \in (p), we can write r=rpr = r'p for some rRr' \in R.This gives us p=arp = ar and p=a(rp)p = a(r'p). Hence ar=1ar' = 1. Thus aais a unit, therefore (a)=R(a) = R.