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Zorn’s Lemma

Definition: A partial order on a set $A$ is a relation $\le$ on $A$ such that is reflexive (so $a\le a$ for all $a\in A$), antisymmetric (so $a\le b$ and $b\le a$ implies $a=b$) and transitive (so $a\le b$ and $b\le c$ implies $a\le c$).

Definition: A total order on $A$ is a partial order with the additional property that, given $a,b\in A$, either $a\le b$ or $b\le a$.

Definition: A chain in $A$ is a subset of $A$ which is totally ordered by $\le$.

Definition: An upper bound for a subset $B$ of a partially ordered set $A$ is an element $a\in A$ such that, for all $b\in B$, $b\le a$.

Definition: A maximal element of a partially ordered set $A$ is an element $m\in A$ such that $m\le x\implies m=x$ for all $x\in A$.


  • integers under divisiblity are partially ordered; powers of a prime $p$ are chains.
  • subsets of a set $X$ under inclusion are partially ordered; a chain is a nested sequence of sets. The union of elements in a chain is an upper bound for the chain. The whole set $X$ is a maximal element.
  • Let $A$ be the set of pairs $(X,f)$ where $X\subset \R$ is open and $f:X\to \R$ is continuous (or differentiable, …). The relation $(X,f)\le (Y,g)$ if $X\subset Y$ and $g$ restricted to $X$ is $f$.

Zorn’s Lemma: If $A$ is a nonempty partially ordered set in which every chain has an upper bound then $A$ has a maximal element.

Not a lemma – really an axiom.

If $R$ is a ring with unity, let $J$ be a proper ideal of $R$ and let $A$ be the set of proper ideals of $R$ containing $J$. Then $A$ satisfies the conditions of Zorn’s lemma – a chain is an increasing system of proper ideals; the union of proper ideals is a proper ideal (if the union weren’t proper, it would contain $1$, so $1$ would belong to one of the elements in the sequence, which can’t happen); that union is the upper bound for that chain. So $A$ has a maximal element which is a proper ideal containing $J$.

Decomposition of rings

Suppose $R$ is a commutative ring with unity.

Definition: Two ideals $I$ and $J$ of a ring $R$ are called coprime or comaximal if $I+J=R$.

Lemma: If $I+J=R$ then $IJ=I\cap J$.

Proof: We know $IJ\subset I\cap J$. Choose $x\in I\cap J$ and also write $1=u+v$ with $u\in I$ and $v\in J$. Then $x=xu+xv$. But both $xu$ and $xv$ are in $IJ$, so $x\in IJ$.

This is a (pretty big) generalization of the statement that if $a$ and $b$ are relatively prime integers then their least common multiple is their product.

Proposition: Let $I_{1},\ldots, I_{k}$ are ideals of $R$, then there is a ring homomorphism \(R\to R/I_{1}\times \cdots\times R/I_{k}.\) Its kernel is the intersection $\bigcap_{i=1}^{k} I_{i}$. If, for every pair, $I_{j}+I_{k}=R$, the map is surjective and its kernel is $I_{1}\cdots I_{k}$.

Key examples: Polynomials and integers.

Ideals and divisibility

Euclidean Domains

Three notable examples:

  • $\Z$
  • $F[x]$ where $F$ is a field
  • $\Z[i]$

Proposition: Every ideal in a Euclidean domain is principal.

Proposition: (Fermat) A prime number is the sum of two squares if and only if it is $2$ or is congruent to $1$ mod $4$.

Lemma: The congruence $x^2\equiv -1\pmod{p}$ has a solution modulo a prime $p$ if and only if $p=2$ or $p\equiv 1\pmod{4}$.

Proof: If $p=2$, $1$ is a solution. If $p$ is odd, and $x^2=-1$ has a solution, then $(\Zn{p})^{\ast}$ has an element of order $4$, so $4\divides (p-1)$. Notice that $(\Zn{p})^{*}$ has only two elements of order dividing $2$, because of $x^2\equiv 1\pmod{p}$ then $p\divides (x^2-1)$, so $p\divides (x+1)(x-1)$, so either $x\equiv 1\pmod{p}$ or $x\equiv -1\pmod{p}$. If $4\divides (p-1)$ then let $H$ be the Sylow $2$-subgroup of $(\Zn{p})^{\ast}$. If $H$ were not cyclic, then there would be too many elements of order $2$ in $H$. So $H$ must be cyclic and therefore there is an element of order $4$.

Now suppose that $p\equiv 1\pmod{4}$. Let $u$ be a solution to $x^2+1\equiv 0\pmod{p}$. Consider the ideal $I=(p,u+i)\subset \Z[i]$. This is a maximal ideal. If $\pi=a+bi$ is a generator of this ideal, then $p=x\pi$. If $x$ were a unit, then $u+i$ would have to be a multiple of $p$, which it visibly isn’t. Therefore $N(\pi)$ must be $p$.
But $N(\pi)=a^2+b^2$, so we’ve found our representation.