LEC ALG3 6

Existence of maximal ideals

Theorem 111.1(Krull's theorem).

Let $R$ be a non-zero ring. Maximal ideals exist in $R$ and every proper ideal is contained in some maximal ideal.

Proof.

Let $\Sigma$ be the set of all proper ideals in $R$. Order the elements of $\Sigma$ by inclusion. Let $C=\{I_i{\}}_{i\in I}$ be a chain of ideals. Let $I={\bigcup }_{i\in I}{I}_i$. It is easy to verify that $I$ is a proper ideal in $R$. By Zorn’s lemma, there exists a maximal ideal $M$ in $\Sigma$.

A similar construction works to show that every proper ideal $I$ is contained in some maximal ideal. Let $P=\{A\supset I:A\text{ is a proper ideal in }R\}$. Again, by Zorn’s lemma, $P$ has a maximal element $N$. It is easy to see that $N$ must be a maximal ideal in $R$. By definition of $P$, $N$ must contain $I$.□

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Chinese remainder theorem, reprise

The ideals $A$ and $B$ of a commutative ring $R$ are said to be comaximal if $A+B=R$.

Theorem 111.2(Chinese remainder theorem).

Let $A_1,\dots ,A_k$ be ideals in $R$. The map $R\to R/A_1\times \cdots \times R/A_k$ defined by

$$r\mapsto (r+A_1,\dots ,r+A_k)$$

is a ring homomorphism with kernel $A_1\cap \cdots \cap A_k$. If $A_1,\dots ,A_k$ are pairwise comaximal, then this map is surjective and $A_1\cap \cdots \cap A_k=A_1\dots A_k$, so

$$\frac{R}{A_1\dots A_k}=\frac{R}{A_1\cap \cdots \cap A_k}\cong \frac{R}{A_1}\times \cdots \times \frac{R}{A_k}.$$

Proof.

We first prove this for$k=2$; the general case will follow by induction. Consider the map $\phi :R\to R/A_1\times R/A_2$ defined by $\phi (r)=(r+A_1,r+A_2)$. This map is clearly a ring homomorphism with kernel $A_1\cap A_2$.

It remains to show that if $A_1$ and $A_2$ are comaximal, then $\phi$ is surjective and $A_1\cap A_2=A_1{A}_2$. Since $A_1+A_2=R$, there are elements $x\in A_1$ and $y\in A_2$ such that $x+y=1$. This equation shows that $\phi (x)=(0,1)$ and $\phi (y)=(1,0)$. If now $(r_1+A_1,r_2+A_2)$ is an arbitrary element in $R/A_1\times R/A_2$, then the element $r_{2}x+r_{1}y$ maps to this element since

$$\begin{array}{rl}\phi (r_{2}x+r_{1}y)& =\phi (r_2)\phi (x)+\phi (r_1)\phi (y)\\ & =(r_2+A_1,r_2+A_2)(0,1)+(r_1+A_1,r_1+A_2)(1,0)\\ & =(r_1+A_1,r_2+A_2).\end{array}$$

Thus, $\phi$ is surjective.

Finally, the ideal $A_1{A}_2$ is always contained in $A_1\cap A_2$. If $A_1$ and $A_2$ are comaximal and $x$ and $y$ are as above, then for any $c\in A_1\cap A_2$, $c=c1=c(x+y)=cx+cy\in A_1{A}_2$.

The general case follows by induction from the case of two ideals using $A=A_1$ and $B=A_2\dots A_k$ once we show that $A_1$ and $A_2\dots A_k$ are comaximal. By hypothesis, for each $i\in \{2,3,\dots ,k\}$, there are elements $x_i\in A_1$ and $y_i\in A_i$ such that $x_i+y_i=1$. It follows that $1=(x_2+y_2)\dots (x_k+y_k)$ is an element in $A_1+(A_2\dots A_k)$.□

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Example 111.3(Example).

Consider $n\in \mathbb{Z}$. $\mathbb{Z}$ is a UFD, so we can write $n=p_1^{a_1}\dots p_r^{a_r}$ for primes $p_1,\dots ,p_r$. Since $\mathbb{Z}$ is a PID, prime ideals are maximal ideals, so $⟨p_1⟩,\dots ,⟨p_r⟩$ are maximal ideals, and hence pairwise comaximal. It follows¹ that the ideals $⟨p_1{⟩}^{a_1},\dots ,⟨p_r{⟩}^{a_r}$ are pairwise comaximal. By Theorem 2, we have² $⟨p_1^{a_1}⟩\cap \cdots \cap ⟨p_r^{a_r}⟩=⟨p_1^{a_1}⟩\dots ⟨p_r^{a_r}⟩=⟨p_1^{a_1}\dots p_r^{a_r}⟩=⟨n⟩$, and

$$\frac{\mathbb{Z}}{n\mathbb{Z}}\cong \frac{\mathbb{Z}}{p_1^{a_1}\mathbb{Z}}\times \cdots \times \frac{\mathbb{Z}}{p_r^{a_r}\mathbb{Z}}.$$

Additionally,

$$\frac{\mathbb{Z}}{n\mathbb{Z}}[x]\cong \frac{\mathbb{Z}}{p_1^{a_1}\mathbb{Z}}[x]\times \cdots \times \frac{\mathbb{Z}}{p_r^{a_r}\mathbb{Z}}[x].$$

Footnotes

1. In any commutative ring, $I+J=R$ implies $I^m+J^k=R$ for all ideals $I,J\subseteq R$ and positive integers $m,k$ (just raise the identity $i+j=1$ to the power $m+k$). ↩

2. $⟨a⟩⟨b⟩=⟨ab⟩$ in commutative rings. ↩