§ Handy characterization of adding an element into an ideal, proof that maximal ideal is prime

§ The characterization

Let II be an ideal. The ideal generated by adding (aR)(a \in R) to II is defined as A(I{a})A \equiv (I \cup \{ a\}). We prove that A=I+aRA = I + aR.
(I{a})={αi+βaiI,α,βR}={i+βaiI,α,βR}(I is closed under multiplication by R)=I+aR \begin{aligned} &(I \cup \{a \}) \\ &= \quad \{ \alpha i + \beta a | i \in I, \alpha, \beta \in R \} \\ &= \quad \{ i' + \beta a | i' \in I, \alpha, \beta \in R \} \qquad \text{($I$ is closed under multiplication by $R$)} \\ &= I + aR \end{aligned}

§ Quotient based proof that maximal ideal is prime

An ideal PP is prime iff the quotient ring R/PR/P is an integral domain. An ideal MM is maximal R/MR/M is a field. Every field is an integral domain, hence: M is maximal     R/M is a field     R/Mis an integral domain    M is primeM \text{ is maximal } \implies R/M \text{ is a field } \implies R/M \text {is an integral domain} \implies M \text{ is prime}. I was dissatisfied with this proof, since it is not ideal theoretic: It argues about the behaviour of the quotients. I then found this proof that argues purly using ideals:

§ Ideal theoretic proof that maximal ideal is prime

§ Sketch

Let II be a maximal ideal. Let a,bRa, b \in R such that abIab \in I. We need to prove that aIbIa \in I \lor b \in I. If aIa \in I, the problem is done. So, let aIa \notin I. Build ideal A=(Ia)A = (I \cup {a}). IAI \subsetneq A. Since II is maximal, A=RA = R. Hence, there are solutions for 1RA    1rI+aR    iI,rR,1R=i+ar1_R \in A \implies 1_r \in I + aR \implies \exists i \in I, r \in R, 1_R = i + ar. Now, b=b1R=b(i+ar)=bi+(ba)rI+IR=Ib = b \cdot 1_R = b(i + ar) = bi + (ba)r \in I + IR = I. ( baIba \in I by assumption). Hence, bIb \in I.

§ Details

let ii be a maximal ideal. let a,bra, b \in r such that abiab \in i. we need to prove that aibia \in i \lor b \in i. if aia \in i, then the problem is done. so, let aia \notin i. consider the ideal AA generated by adding aa into II. A(I{a})A \equiv (I \cup \{a\}). We have shown that A=I+aRA = I + aR. Hence, I+a0=IAI + a0 = I \subset A. Also, 0+ac1˙=aA0 + ac \dot 1 = a \in A, aIa \neq I \implies AIA \neq I. Therefore, IAI \subsetneq A. Since II is maximal, this means that A=RA = R Therefore, I+aR=RI + aR = R. Hence, there exists some iI,rRi \in I, r \in R such that i+ar=1Ri + ar = 1_R. Now, b=b1R=b(i+ar)=bi+(ba)rI+IR=Ib = b \cdot 1_R = b \cdot (i + ar) = bi + (ba) r \in I + IR = I Hence, bIb \in I.