Number 28 In class proof J.O.

28. (Preliminary for Th, April 5; finalized Sat, April 14) Prove that an ideal I of a ring R is maximal if and only if the quotient ring R/I is a field.

Assume r/I is a field. Let J be an ideal such that $I \subset J$. Then $\exists x\in J x\notin I$. Therefore, $I+x\notin I+0$.
Since R/I is a field, $\exists (I+y) \in R/I \longrightarrow (I+x)(I+y)=I+1$.
Thus $xy-1\in I$.
Since $I\subset J, (xy-1)\in J$.
Then consider $1= xy-(xy-1)$.
All element of the equation are in J, so $1\in J$.
Therefore, J=R and I is maximal.

Now assume that I is maximal.
Let $I+a \in R/I \longrightarrow (I+a)\neq (I+0) \wedge a\notin I$
Consier $J=I+Ra :=\{i+ra:i\in I, r\in R\}.$
Then $\forall i\in I, i=i+0a \longrightarrow I\subset J$.
Since I is maximal, J=I or J=R. However, $a=0+1a$. so $a\in J \wedge a\notin I$. so $J=R$.
Since I+Ra=R and $1\in R$,
1=i+ra for some $i\in I \wedge r\in R$.
$\Longrightarrow 1-ra=i\in I$, so (I+1)=(I+ra) implies (I+1)=(I+r)(I+a).
I+a is the inverse of I+r.
$\forall a\in R$, exists r such that ar=1.
All elements have an inverse, so R/I is afield.

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