In a UFD if $x$ is irreducible, then $(x)$ is prime

Question

I know how to show that if $(x)$ is prime ideal then $x$ irreducible element.

I know the converse is true if the ring is a UFD, but how to show it?

i.e. in a UFD, if $x$ is irreducible, then $(x)$ is a prime ideal.

Answer 1

Let $R$ be a UFD. Suppose $p$ is an irreducible element of $R$. In order to show that $p$ is a prime we will show that $(p)$ is a prime ideal of $R$. We know that $(p) \neq \{0\}$ because an irreducible element is nonzero. Also $(p) \neq R$ because an irreducible element is not a unit. Suppose the product $ab$ of two elements $a, b \in R$ is an element of $(p)$. Then $ab = mp$ for some $m \in R$.

First suppose $m = 0$. Then $ab = 0$. Since $R$ is an integral domain then either $a = 0$ or $b = 0$. If $a = 0$ then $a \in (p)$ and if $b = 0$ then $b \in (p)$. This means $(p)$ is a prime ideal and $p$ is prime.

Now suppose $m \neq 0$, but $m$ is a unit. This means that $mp$ is an associate of $p$. Then since $a, b \in R$, which is a UFD, then we have \begin{align*} a = y_1 \cdots y_r \\ b = x_1 \cdots x_m \end{align*} for some irreducibles $y_k, x_l$. So $$(y_1 \cdots y_r)(x_1 \cdots x_m) = mp$$ By the uniqueness of the product of irreducibles, $mp$ is equal to $y_k$ for some $k$ or $x_l$ for some $l$. If $mp$ is equal to $y_k$ for some $k$ then $mp\ |\ a$. If $mp$ is equal to $x_l$ for some $l$ then $mp\ |\ b$. It follows that $a \in (p)$ or $b \in (p)$ and $(p)$ is a prime ideal. In either case $p$ is prime in $R$.

Lastly suppose $m \neq 0$ and $m$ is not a unit. Then since $m,a, b \in R$, which is a UFD, then we have \begin{align*} m = z_1 \cdots z_n \\ a = y_1 \cdots y_r \\ b = x_1 \cdots x_m \end{align*} for some irreducibles $z_i, y_k, x_l$. So $$(y_1 \cdots y_r)(x_1 \cdots x_m) = (z_1 \cdots z_n)p$$ By the uniqueness of the product of irreducibles, $p$ or an associate of $p$ is equal to $y_k$ for some $k$ or $x_l$ for some $l$. If $p$ or an associate of $p$ is equal to $y_k$ for some $k$ then $p\ |\ a$. If $p$ or an associate of $p$ is equal to $x_l$ for some $l$ then $p\ |\ b$. It follows that $a \in (p)$ or $b \in (p)$ and $(p)$ is a prime ideal. In either case $p$ is prime in $R$.

Answer 2

Let $R$ be a unique factorization domain. Let's call two elements in $R$ associate if one can be multiplied by a unit to get the other. This defines an equivalence relation on $R$, and you can easily check that an associate of an irreducible element is irreducible. Let $P$ be a complete set of representatives of irreducible elements of $R$ under this relation.

Then to say that $R$ is a unique factorization domain is to say that every nonzero element $x$ in $R$ can be written as

$$u_x\prod\limits_{p \in P} p^{x_p}$$

for a unit $u$ and nonnegative integers $x_p$, and that $u_x$ and $x_p$ are uniquely determined. This is a finite product.

Let's say that an irreducible element $q \in P$ occurs in a given element $x$ if $x_q \geq 1$. Obviously, $q$ divides $x$ if and only if $q$ occurs in $x$. If $a, b$ are nonzero elements of $R$, we have

$$u_{ab} \prod\limits_{p \in P} p^{(ab)_p}= ab =(u_a \prod\limits_{p \in P} p^{a_p})(u_b \prod\limits_{p \in P} p^{b_p}) = u_au_b \prod\limits_{p \in P} p^{a_p + b_p}$$

so by uniqueness, $u_{ab} = u_au_b$ and $a_p + b_p = (ab)_p$ for all $p$. It follows from here that if an irreducible element in $P$ occurs in a product $ab$, it must occur in either $a$ or $b$.

This shows the elements of $P$ are prime. But every irreducible element of $R$ is associate to a unique element of $P$, so every irreducible element of $R$ is prime.

What's also interesting is the converse: if in an integral domain, every nonzero element is a product of irreducibles, and if every irreducible is prime, then you are in a unique factorization domain.

Answer 3

Hint: Take any $ab\in (x)$ and consider the factorisations of $a$, $b$ and $ab$ into irreducible elements.