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Gaussian integers and Fermat’s Theorem

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 $(\Zn{p})^{\times}$ has an element of order $4$, so $4\divides (p-1)$. Notice that $(\Zn{p})^{\times}$ 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 $(\Zn{p})^{\times}$. 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.

Proposition: The ring $\Z[\sqrt{-5}]$ is not a Euclidean ring. In fact, the ideal $(3,1+\sqrt{-5})$ is not principal. It is a proper ideal, because the quotient of $\Z[\sqrt{-5}]$ by this ideal is $\Zn{3}$. If $\pi$ were a generator of this ideal, then $3=x\pi$ means that either $N(\pi)=3$ or $N(\pi)=9$. Also $(1+5i)=y\pi$ means that $N(\pi)$ divides $6$. Since $\pi$ is not a unit, $N(\pi)=3$. But the equation $x^2+5y^2=3$ has no integer solutions, so there is no element of norm 3 in this ring.

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: 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)

Note: The converse is not true, but the question of existence of Euclidean algorithms is subtle. See Conrad’s notes on the euclidean domains for a discussion. DF prove that $\Z[(1+\sqrt{-19})/2]$ is a PID but is not Euclidean with respect to any norm (see page 277).

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

Unique factorization

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

  1. A non-unit element $x\in R$ is called irreducible if whenever $x=ab$ in $R$, either $a$ or $b$ is a unit.
  2. 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.
  3. Two elements $a$ and $b$ are called associates in $R$ if there is a unit in $R$ such that $a=bu$.

Example: In a polynomial ring $F[x]$ over a field $F$, the irreducible elements are the irreducible polynomials. Every irreducible element is prime (by the Euclidean algorithm). In the ring $\Z[\sqrt{-5}]$ the element $2$ is irreducible by not prime, since $2$ divides $(1+\sqrt{-5})(1-\sqrt{-5})=6$ but does not divide either of the factors.

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

Proof: If $p$ is a prime element, and $p=xy$, then either $p|x$ or $p|y$. Assume $x=pu$. Then $p=puy$ so $p(1-uy)=0$ and therefore $uy=1$ so $y$ is a unit and $p$ and $x$ are associates. Similarly if $pR$ is a prime ideal then $R/pR$ is an integral domain, so $xy=0$ in $R/pR$ implies either $x\in pR$ or $y\in pR$.

If $R$ is a PID, and $q$ is irreducible, suppose $q$ divides $xy$. Let $d$ generate the ideal $(q,x)$. If $d$ is a unit then we can write $qa+xb=1$ so $qay+xby=y$ and therefore $q$ divides $y$. If $d$ is not a unit, then $q=du$ and $x=dv$ and since $q$ is irreducible and $d$ is not a unit, $u$ must be a unit. Then $d$ and $q$ are associated and therefore $q$ divides $x$.

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.

– This follows from uniquess of the factorization.

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

There are two features of the UFD property. One is that every nonzero element is a finite product of irreducibles; and the other is that this is unique.

Theorem: A principal ideal domain is a UFD.

  • Every element of a PID $R$ that is not a unit is a finite product of irreducible elements.

Proof: Choose a non-unit $x$ in $R$. Suppose $x$ does not have a finite factorization into irreducibles. Write $x=a_1b_1$ where $a_1$ and $b_1$ are non-units. Then one of $a_1$ or $b_1$ does not have a finite factorization into irreducibles; suppose it’s $a_1$. Notice that $xR\subset a_1R$ and the inclusion is strict since $b$ is a non-unit. Repeat this argument to construct an increasing sequence of proper ideals \(xR\subset a_1R\subset a_2R\subset\cdots\) Let $I$ be the union of all of these ideals inside $R$. This ideal must be principal, so $I=yR$ for some $y$. Now $y\in a_{j}R$ for some $j$, which means that at some point the increasing sequence stabilizes; $a_{k}R=yR$ for all $k\ge j$. This contradicts the assumption that $x$ did not have a finite factorization.

For the uniqueness, we know that every element of $R$ is a finite product of irreducible elements, and that irreducible elements in $R$ are prime. We proceed by induction on $n$, the minimal number of irreducible elements needed to write $x$ as a product. Suppose $n=1$. Then $x$ is irreducible and hence prime. Suppose that whenever $x$ is a product of up to $n$ irreducibles, that expression is unique. Suppose $y$ is a product of $n+1$ irreducibles and it has two factorizations \(y=p_1 p_2\cdots p_{n+1}=q_{1} q_{2}\cdots q_{s}\) where $s\ge n+1$. Since $p_1$ divides the product of the $q’s$, it must equal one of the $q’s$ up to a unit, so we can cancel $p_1$ from both sides of the equation. Now $y/p_1$ has a shorter expression as a product of irreducibles, so it’s expression is unique, and therefore $s=n+1$ and the $q’s$ are a rearrangement of the $p’s$.