LEC ALG3 7

Irreducible and prime elements

Definition 112.1(Definition).

Let $a,b\in R$. We say $a|b$ if there is an element $r\in R$ such that $b=ar$. If $a|b$ and $b|a$, we say $a$ and $b$ are associates.

Proposition 112.2(Proposition).

Let $a,b\in R$.

1. $a|b⟺(b)\subseteq (a)$.
2. $a$ and $b$ are associates $⟺$ $(a)=(b)$.
3. $u$ is a unit $⟺$ $u|a$ for all $a\in R$.
4. Association is an equivalence relation.

Definition 112.3(Definition).

An element $a$ is irreducible if $a$ is not a unit and $a=a_1{a}_2$ implies $a_1$ is a unit or $a_2$ is a unit.
An element $p$ is prime if $p|ab$ implies $p|a$ or $p|b$.

Every prime is irreducible in an integral domain. The converse is true for unique factorization domains, and in particular for PIDs, as we will show below.

Example 112.4(Example).

1. $(1+i)$ and $(1-i)$ are associates in the ring of Gaussian integers.
2. $(\overline{2})$ and $(\overline{3})$ are associates in $\mathbb{Z}/8\mathbb{Z}$.
3. In $\mathbb{Z}[\sqrt{-5}]$, $⟨6⟩=⟨2⟩⟨3⟩=⟨1-\sqrt{-5}⟩⟨1+\sqrt{-5}⟩$, where $2$, $3$, and $1\pm \sqrt{-5}$ are irreducible - factorization is not unique!

Proposition 112.5(Proposition).

Let $R$ be an integral domain, $a,b\in R$, $a,b\ne 0$. Then,

1. $a$ is prime $⟺$ $⟨a⟩$ is a nonzero prime ideal. (ID hypothesis not required.)

2. If $R$ is not a field, then $b$ is irreducible $⟺$ $⟨b⟩$ is maximal in the set of all proper principal ideals.

3. Every prime element is irreducible.

4. If $R$ is a PID, $p$ is prime $⟺$ $p$ is irreducible.

5. Every associate of an irreducible/prime element is irreducible/prime.

Proof.

$(1)$ Suppose $a$ is prime. Let $\alpha \beta \in ⟨a⟩$. $\alpha \beta =ar$. $a|ar=\alpha \beta$, so $a|\alpha \beta$, so $a|\alpha$ or $a|\beta$, whence $\alpha \in ⟨a⟩$ or $\beta \in ⟨a⟩$.

Conversely, suppose $⟨a⟩$ is prime and $a|\alpha \beta$ for some $\alpha ,\beta \in R$. This implies $\alpha \beta \in ⟨a⟩$, so $\alpha \in ⟨a⟩$ or $\beta \in ⟨a⟩$, whence $a|\alpha$ or $a|\beta$.

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$(2)$ Suppose $b$ is irreducible and $⟨b⟩\subseteq ⟨c⟩$ for $c\in R$. Since $b\in ⟨c⟩$, $b=cr$, and either $c$ or $r$ is a unit. If $c$ is a unit, $⟨c⟩=R$. If $r$ is a unit, we can write $c=b{r}^{-1}$, so $⟨c⟩\subseteq ⟨b⟩$ and $⟨b⟩=⟨c⟩$.

Conversely, suppose $⟨b⟩$ is maximal in the set of all principle ideals. If $b=b_1{b}_2$, then $⟨b⟩\subseteq ⟨b_1⟩$. $⟨b_1⟩$ is forced to be $R$ (in which case $b_1$ is a unit) or $⟨b⟩$. In the latter case, $b_1=br$, and $b=br{b}_2$. Since $R$ is an integral domain, we have $r{b}_2=1$, so $b_2$ is a unit.

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$(3)$. Let $p$ be prime and $p=ab$. Then, $p|a$ or $p|b$. WLOG, consider the first case. Then, $pr=a$. Then, $p=ab=prb$, so $p(1-rb)=0$. Since $R$ is an integral domain, we have $1=rb$, so $b$ is a unit.

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$(4)$ Suppose $R$ is a PID. Let $p\in R$ be irreducible. Since we are in a PID, $(2)$ implies that $⟨\alpha ⟩$ is a maximal ideal. So, $⟨\alpha ⟩$ is prime, so $\alpha$ is prime by $(1)$.

Note.

PIDs are not the only class of rings with this property; consider $\mathbb{Z}[x]$ as an example. It is not a PID, but every irreducible element is prime (recall our characterization of the prime ideals of $\mathbb{Z}[x]$!)

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$(5)$ Let $a$ be irreducible and $a$ and $b$ be associates. $a=bu$ for some unit $u$. If $b$ is not irreducible then $b=b_1{b}_2$, so $a=b_1{b}_{2}u=(b_1)(b_{2}u)$, so $a$ is not irreducible.□

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Example of an irreducible element which is not prime: Consider $2$ in $\mathbb{Z}[\sqrt{-5}]$. $2\cdot 3=(1+\sqrt{5})(1-\sqrt{5})$, and $2$ does not divide either $1+\sqrt{5}$ or $1-\sqrt{5}$. It is easily seen that $2$ is irreducible. This also shows that $6$ does not have a unique factorization!

Example 112.6(Example).

An irreducible element may become reducible in a quotient space. For example, $2$ is irreducible in $\mathbb{Z}[x]$, but is reducible in the Gaussian integers $\mathbb{Z}[x]/(x^2+1)$: $\overline{2}=\overline{(1+x)}\cdot \overline{(1-x)}$.