[Math] Maximal Ideal in a commutative ring with unit element

abstract-algebramaximal-and-prime-ideals

Exercice 1.8 of Sohrab, Basic Real Analysis

Let $R$ be a commutative ring with unit element $1\neq 0$. Show that $R$ has at least one maximal ideal, i.e., an ideal $M\subset R$ such that there is no ideal $N\subset R$ satisfying the proper inclusions $M\subset N\subset R$. Hint: Consider the set $\mathcal{I}$ of all ideals $I\subsetneqq R$ and note that $\{0\}\in\mathcal{I}$. Partially order $\mathcal{I}$ by inclusion and show that, if $(I_\alpha)_{\alpha\in A}$ is a chain in $\mathcal{I}$, then $\bigcup_{\alpha\in A}I_\alpha\in\mathcal{I}$. Now use Zorn's Lemma.

Proof

$S$ is an ideal of $R$ if $S$ subgroup of $(R,+)$ and $\forall s\in S,r\in R$ we have $rs,sr\in S$. An ideal is maximal if it is not contained in any other ideal.

Let us define $\mathcal{I}:=\{I\in \mathcal{P}(R)\mid I\text{ ideal of }R\}\setminus \{R\}$. Note that $\{0\}$ is an ideal, because $(\{0\},+)$ is a group and $r0=0r=0\in\{0\}$, thus $\{0\}\in\mathcal{I}$. Consider the partial order $(\mathcal{I},\subset)$. If $(I_\alpha)_{\alpha\in A}$ is a chain in $\mathcal{I}$, we show that $I_A:=\bigcup_{\alpha\in A}I_\alpha$ is in $\mathcal{I}$. In other words we have to prove that $I_A$ is an ideal and $I_A\neq R$.

  • $I_A\subset R$ (since all $I_\alpha$ are) and $(I_A,+)$ is a group:
    • $(I_A,+)$ is closed: Consider $a,a'\in I_A$, then there are $\alpha,\alpha'\in A$ such that $a\in I_\alpha$ and $a'\in I_{\alpha'}$. Since $(I_\alpha)_{\alpha\in A}$ is a chain, either $I_\alpha\subset I_{\alpha'}$ or $I_{\alpha'}\subset I_{\alpha}$. Without loss of generality $I_{\alpha'}\subset I_{\alpha}$ $\implies$ $a,a'\in I_\alpha$, which is closed under addition (being an ideal).
    • Associativity: With the same reasoning as by the closeness, for any $a,a',a''\in I_A$, there is an $I_\alpha$, such that $a,a',a''\in I_\alpha$. The associativity of $(I_A,+)$ follows from the one of $(I_\alpha,+)$.
    • The identity element is $0$, which is also the identity of all $(I_\alpha,+)$.
    • Inverse element: For any $a\in I_A$, there is an $I_\alpha$, such that $a\in I_\alpha$. Thus the inverse of $a$ is in $I_\alpha$ and also in $a\in I_A$.

  • $\forall s\in I_A,r\in R$ we show that $rs,sr\in I_A$: As above $\forall s\in I_A$, $\exists \alpha\in A$ such that $s\in I_\alpha$. $I_\alpha$ is an ideal thus $\forall s\in I_\alpha,r\in R$ we have $rs,sr\in I_\alpha\subset I_A$.

  • $I_A\neq R$ To be completed

Now we can apply Zorn's Lemma: Every chain in the partial ordered set $\mathcal{I}$ has an upper bound, then $\mathcal{I}$ contains a maximal element. That is, an ideal that contains all other ideal, that is an ideal that is not contained by any other ideal, that is a maximal ideal.

My questions

  • Main question. How do I prove that $I_A\neq R$?
  • I hope the rest is correct, but is something superfluous?

Best Answer

If $I_A=R$, $1\in I_A$, hence $1\in I_\alpha$ for some $\alpha\in A$. This cannot happen, as $I_\alpha\varsubsetneq R$.

I don't see the point of proving associativity for an ideal. A priori, there's no inverse, and I don't see what it's here for. Stability by multiplication by elements of the ring requires checking one side only in a commutative ring. Last point: the $A_\alpha$ seem to be $I_\alpha$.

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