## Maximal Ideals

**Proposition:** (Krull) Every ideal in a ring with unity is contained in a maximal ideal.

## Ring factorization theorem

Let $R$ be a commutative ring with unity.

**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 $I_{1}\cdots I_{k}$.

## A first look at unique factorization: Euclidean domains and PID

Let $R$ be an integral domain.

**Definition:** Let $\mathbb{N}$ be the natural numbers *starting at zero.* A function $N:R\to \mathbb{N}$ with $N(0)=0$ is called a norm. If $N(a)=0\implies a=0$ then $N$ is calledf a positive norm.

**Definition:** $R$ is called a *Euclidean domain* if there is a norm on $R$ such that, given $a,b\in R$, with $b\not=0$, there are elements $q$ and $r$ in $R$ such that \(a=qb+r\) and either $N(r)=0$ or $N(r)<N(b)$.

Euclidean domains have a euclidean algorithm.

**Key Examples:** $F[x]$ when $F$ is a field; $Z$; $Z[i]$; $Z[\sqrt{-2}]$.

**Proposition:** Every ideal in a Euclidean Domain $R$ is principal. More precisely, if $I$ is a nonzero ideal in $R$, then $I=aI=(a)$ where $a$ is any nonzero element of $I$ of minimal norm.

## Divisibility and ideals

**Definition:** Let $R$ be a commutative ring, with $a,b\in R$ and $b\not=0$.

- We say $a$ divides $b$ ($a\divides b$) if there is $x\in R$ with $b=ax$.
- A greatest common divisor of $a$ and $b$ is an element $d\in R$ with $d\divides a$ and $d\divides b$, and such that, if $x\divides a$ and $x\divides b$ then $x\divides d$. (In general the gcd need not be unique)

Translations.

- $a\divides b$ if and only if $bR\subset aR=(a)$. (
*to contain is to divide*) - Let $I$ be the ideal of $R$ generated by $a$ and $b$: $I=(a,b)=aR+bR$. Then $d=\gcd(a,b)$ if and only if $I\subset dR$ and if $aR$ is any principal ideal containing $I$ then $dR\subset aR$.

**Proposition:** Let the ideal $I=(a,b)$. If $I=dR$ (so that $I$ is principal) then $d$ is the greatest common divisor of $a$ and $b$.

**Proposition:** Two principal ideals $aR$ and $bR$ are equal if and only if $a=bu$ for some unit $u\in R$.

In a Euclidean domain, the ideal $I=(a,b)$ is principal and generated by the “last remainder” obtained from Euclid’s algorithm.

## Principal Ideal domains

**Definition:** An integral domain in which every ideal is principal is called a Principal Ideal Domain.

Principal ideal domains satisfy the conclusions of the Euclidean algorithm (but maybe without the algorithm).

That is, given $a,b\in R$ if $R$ is a PID, then the ideal $(a,b)=(d)$ where $d$ is a greatest common divisor of $R$, and there are $x$ and $y$ in $R$ such that $ax+by=d$. The gcd $d$ is unique up to multiplication by a unit.

**Proposition:** In a principal ideal domain, every nonzero prime ideal is maximal.

Proof: Suppose $(p)$ is a prime ideal and $(m)$ is an ideal with $(p)\subset (m)$. Then $p=mx$ for some $x\in R$. Since (p) is prime, either $m\in P$ or $x\in P$. If $m\in P$, then $(m)=(p)$. If $x\in P$, then $x=pr$ and so $p=mpr$ or $p(1-mr)=0$, meaning $mr=1$ and so $m$ is a unit. Then $(m)=R$. So the only ideals of $R$ containing $(p)$ are $(p)$ and $R$, and $(p)$ is maximal. (Note: this is the ideal theoretic version of the statement that, if $p|xm$, then either $p|x$ or $p|m$.)

**Proposition:** A Euclidean ring is a PID. (DF p. 281 contains a strengthening of this result, proving that an integral domain $R$ is a PID if and only if it has a “Dedekind-Hasse” norm, which is a slightly more general type of norm that isn’t necessarily positive)

## Unique factorization

**Key Terminology:** Let $R$ be an integral domain.

- A non-unit element $x\in R$ is called irreducible if whenever $x=ab$ in $R$, either $a$ or $b$ is a unit.
- A non-unit element $x\in R$ is called prime if, whenever $p$ divides $ab$, either $p$ divides $a$ or $p$ divides $b$. Equivalently, $p$ is prime if the ideal $pR$ is a prime ideal.
- Two elements $a$ and $b$ are called associates in $R$ if there is a unit in $R$ such that $a=bu$.

**Lemma:** If $R$ is an integral domain, then every prime is irreducible. If $R$ is a principal ideal domain, then the converse is true.

**Definition:** A unique factorization domain (UFD) is an integral domain such that every nonzero element $r\in R$ which is not a unit is a product \(r=p_1p_2\cdots p_n\) where the $p_{i}$ are (not necessarily distinct) irreducible elements of $R$ and, if $r=q_1q_2\cdots q_k$ is another such factorization, then there is a rearrangement of the $q_{i}$ so that $q_{i}$ and $p_{i}$ are associates.

**Lemma:** in a UFD, $p$ is prime if and only if it is irreducible.

**Lemma:** A UFD has greatest common divisors (computed using the factorization into primes as in $\Z$).

**Theorem:** A principal ideal domain is a UFD.