# Abstract Nonsense

## Maximal Ideals (Pt. II)

Point of Post: This is a continuation of this post.

Just as was the case for prime ideals there is a theorem for maximal ideals that says there are ‘enough of them’. More directly, this will say something to the effect that every ideal is contained in a maximal ideal (for unital rings). This theorem is famous enough that it warrants a name-attachment, in particular to Wolfgang Krull who first stated the theorem in 1929. As we shall see (and it really should be apparent to you since we are trying to prove the existence of a maximal element of a poset) that we are going to use Zorn’s lemma. More surprising perhaps is the fact that Krull’ls lemma is equivalent to the Axiom of Choice (for more information one can see the great book Axiom of Choiceby the renowned set theorist Jech).

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Theorem (Krull): Let $R$ be a unital ring. Then, given any ideal $\mathfrak{a}$ of $R$, there exists a maximal ideal $\mathfrak{m}$ containing $\mathfrak{a}$.

Proof: We are evidently trying to show that the poset $(P,\subseteq)$ of all proper ideals containing $\mathfrak{a}$ has a maximal element. To do this, let $\mathcal{C}$ be a chain in $P$ and let $\mathfrak{u}$ be it’s union. As was checked in the similar argument for prime ideals, we have that $\mathfrak{u}$ is an ideal, and it evidently contains $\mathfrak{a}$. Thus, to see that $\mathfrak{u}\in P$ it suffices to check that it’s proper. But, from first principles we know that $\mathfrak{u}=R$ if and only if $1\in\mathfrak{u}$, but since $1$ is not contained in any element of $\mathcal{C}$ it evidently isn’t contained in their union. Thus, $\mathfrak{u}\ne R$, so $\mathfrak{u}\in P$. Since $\mathcal{C}$ was arbitrary we may conclude from Zorn’s lemma that $P$ has a maximal element $\mathfrak{m}$, which is evidently the desired maximal ideal. $\blacksquare$

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Applying this lemma to $(0)$ gives:

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Corollary: Every non-zero unital ring contains a maximal ideal.

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Remark: There do exist rings that don’t have maximal ideals, but they are fairly ugly, and go too far off into a tangent to pursue.

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Another thing this theorem does is illuminate the connection between non-units of $R$ and $\text{MaxSpec}(R)$. Namely, the maximal ideals of $R$ form the entirety of the nonunits of $R$. Said more directly:

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Theorem: Let $R$ be a commutative unital ring, then

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$\displaystyle R=R^\times\sqcup\bigcup_{\mathfrak{m}\in\text{MaxSpec}(R)}\mathfrak{m}$

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Proof: Evidently all $R=R^\times\sqcup\text{non-units}$ and so it suffices to prove that the non-units are the union over $\text{MaxSpec}(R)$. To see this we merely note that every nonunit is contained in a maximal ideal since, by the previous theorem, we know that $Ra$ is contained in some maximal ideal $\mathfrak{m}$ and since $a\in Ra$ the conclusion follows. Conversely, we know that if $m$ is an element of the union over $\text{MaxSpec}(R)$ then it cannot be unit, otherwise the ideal $\mathfrak{m}$ it belongs to would be equal to $R$ contradictory to assumption. $\blacksquare$

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An obvious question one may ask is, we already have a set of ideals of $R$ which are, in some way, special–prime ideals. So, how do these maximal ideals relate to them? Well, the long and short of it is that:

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Theorem: Let $R$ be a commutative unital ring, then $\text{MaxSpec}(R)\subseteq\text{Spec}(R)$.

Proof: This follows immediately from the fact that if $\mathfrak{m}\in\text{MaxSpec}(R)$ then $R/\mathfrak{m}$ is a field, and so $R/\mathfrak{m}$ is, in particular, a prime ideal. It follows that $\mathfrak{m}\in\text{Spec}(R)$. $\blacksquare$

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So, now we’d like to see $\text{MaxSpec}(R)$ for some classes of, or specific rings. For example, it should not be hard to see, similar to the case for prime ideals, that:

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Theorem: Let $R_1,\cdots,R_n$ be finitely many rings, then

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$\text{MaxSpec}(R_1\times\cdots\times R_n)=\left\{A_1\times\cdots\times A_n:\text{There exists }j\in[n]\text{ s.t. }A_i=R,i\ne j\text{ and }A_j\in\text{MaxSpec}(R_j)\right\}$

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Perhaps more useful is the fact that the fourth isomorphism theorem clearly implies that:

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Theorem: Let $R$ be a ring and $\mathfrak{a}$ an ideal of $R$. Then, $\text{MaxSpec}(R/\mathfrak{a})=\left\{\mathfrak{m}/\mathfrak{a}:\mathfrak{m}\in\text{MaxSpec}(R)\text{ and }\mathfrak{m}\supseteq\mathfrak{a}\right\}$.

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Perhaps more fun is the following gem:

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Theorem: Let $C[a,b]$ be, as usual, the set of continuous maps $f:[a,b]\to\mathbb{R}$ and let, for $x\in[a,b]$, $\mathfrak{m}_x=\left\{f\in C[a,b]:f(x)=0\right\}$. Then, $\text{MaxSpec}(R)=\left\{\mathfrak{m}_x:x\in[a,b]\right\}$.

Proof: Evidently $\mathfrak{m}_x$ is maximal (as discussed before) since $C[a,b]/\mathfrak{m}_x\cong\mathbb{R}$. Conversely, let $\mathfrak{m}$ be a maximal ideal, and suppose that for each $x\in[a,b]$ there existed a $f_x\in C[a,b]$ such that $f_x(x)\ne 0$. Since $f_x$ is continuous there exists some open neighborhood $U_x$ of $x$ such that $f_x$ is non-vanishing on $U_x$. Since $[a,b]$ is evidently the union of the $U_x$ we may, by compactness, produce a finite subcover $U_{x_1},\cdots,U_{x_n}$. Considering then $f_{x_1}^2+\cdots+f_{x_n}^2$ gives a non-vanishing element of $\mathfrak{m}$, but a non-vanishing element of $C[a,b]$ is invertible, which creates an obvious contradiction. $\blacksquare$

August 31, 2011 -

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