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Algebra-II: Companion notes

Abhay Soman

Section 3.3 Prime and maximal ideals

We prove the following.
Let \(A\) be a nonzero commutative ring. Let \(S\) be a set of all proper ideals in \(A\text{.}\) Since \((0)\in S\) the set \(S\neq\emptyset\text{.}\) We order \(S\) by inclusion. We will use Zorn’s lemma 1  to prove the statement. In that vein, let \((\mfa_i)_{i\in I}\) be a chain of ideals in \(S\text{,}\) i.e., for any \(i,j\) either \(\mfa_i\subseteq\mfa_j\) or \(\mfa_j\subseteq\mfa_i\text{.}\) Then \(\bigcup_i\mfa_i\) is a proper ideal in \(A\) (verify!). Thus \(\bigcup_i\mfa_i\in S\text{.}\) We have shown that every chain in a nonempty set \(S\) has an upper bound in \(S\text{.}\) Hence, by Zorn’s lemma, \(S\) has a maximal element.
More generally we have the following.
Let \(A\) be a commutative ring and \(\mfm\) be a maximal ideal in \(A\text{.}\) Let \(a,b\in A\) be such that \(ab\in\mfm\text{.}\) Suppose that \(a\not\in\mfm\text{.}\) Then the ideal generated by \(\mfm\) and \(a\text{,}\) \((\mfm,a)\) properly contains \(\mfm\text{.}\) Hence, \((\mfm,a)=A\text{.}\) In particular, \(1\in(\mfm,a)\text{,}\) say \(1=m+ax\) for some \(m\in\mfm\) and \(x\in A\text{.}\) Hence, \(b=b1=bm+bax\in(\mfm,a)\text{.}\) Since \(m,ba=ab\in\mfm\) we get that \(b\in\mfm\text{.}\)
  1. Let \(\mfp\) be a prime ideal, and let \(a,b\in A\) be such that \((a+\mfp)(b+\mfp)=0+\mfp\text{,}\) i.e., \(ab\equiv 0\mod\mfp\text{.}\) Thus, \(ab\in\mfp\) and either \(a\in\mfp\) or \(b\in\mfp\) because \(\mfp\) is a prime ideal. Hence, \(A/\mfp\) is an integral domain. Conversely, assume that \(A/\mfp\) is an integral domain for a proper ideal \(\mfp\text{.}\) Let \(ab\in\mfp\text{.}\) Thus \(ab\equiv 0\mod\mfp\text{.}\) Since \(A/\mfp\) is an integral domain we have either \(a\equiv 0\mod\mfp\) or \(b\equiv 0\mod\mfp\text{.}\) Hence either \(a\in\mfp\) or \(b\in\mfp\text{,}\) i.e., \(\mfp\) is a prime ideal.
  2. Suppose that \(A/\mfm\) is a field. By Theorem 3.2.2, ideals of \(A\) containing \(\mfm\) are \(\mfm\) and \(A\text{.}\) Thus \(\mfm\) is a maximal ideal. Conversely, assume that \(\mfm\) is a maximal ideal. By Lemma 3.3.3 and the first part of this theorem, \(A/\mfm\) is an integral domain. By Theorem 3.2.2 only ideals of \(A/\mfm\) are the trivial ideal and the whole ring. Thus for any nonzero \(x\in A/\mfm\) the ideal generated by \(x\) is \(A/\mfm\text{.}\) Therefore, there exists \(y\in A/\mfm\) such that \(xy=1\in A/\mfm\text{.}\) Thus, \(A/\mfm\) is a field.
In the following proposition we use criterion for prime/maximal ideal obtained in Proposition 3.3.4 without explicitly mentioning it.
Let \(\mfa\subseteq\mfp\) be a prime ideal in \(A\text{.}\) Consider the ideal \(\mfp/\mfa\) in \(A/\mfa\text{.}\) By Theorem 3.2.5, \((A/\mfp)\simeq(A/\mfa)\big/(\mfp/\mfa)\text{.}\) As \(A/\mfp\) is an integral domain, we have \(\mfp/\mfa\) is a prime ideal. Conversely assume that \(I\) is a prime ideal in \(A/\mfa\text{.}\) Thus there exists an ideal \(\mfp\) in \(A\) such that \(I=\mfp/\mfa\) (see Theorem 3.2.2). By Theorem 3.2.5,
\begin{equation*} A/\mfp\simeq(A/\mfa)\big/(\mfp/\mfa)\simeq(A/\mfa)\big/ I. \end{equation*}
Since \(A/I\) is an integral domain, \(A/\mfp\) is also an integral domain. Hence, \(\mfp\) is a prime ideal.
The mapping \(\mfp\mapsto\mfp/\mfa\) is one-one follows from Theorem 3.2.2.
The statement about maximal ideals follow by similar arguments as given above.
Let \((S,\leq)\) be a nonempty partially ordered set. A subset \(T\subset S\) is a chain of \(S\) if for any \(x,y\in T\) either \(x\leq y\) or \(y\leq x\text{.}\) The Zorn’s lemma asserts that if every chain \(T\) of \(S\) has an upper bound in \(S\text{,}\) i.e., there exists \(s\in S\) such that for every \(t\in T\) we have \(t\leq s\text{,}\) then \(S\) has at least one maximal element.
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