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

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.

• – user26857 Jun 6 '19 at 9:56

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 $R$.

• Most of the answers make great sense to me. Just one point, why the uniqueness implies mp equals to only one element in the product? – JacobsonRadical Nov 29 '17 at 17:54
• I think you meant “divide” – JacobsonRadical Nov 29 '17 at 18:05
• Yes I think divide is correct. I guess it is not guaranteed that $mp$ will actually equal one of the individual irreducibles. – user475040 Nov 29 '17 at 19:55

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.

• This was really awesome proof! Will you please let me know why '$q$ divides $x$ then $q$ occurs in $x$' holds? – Silent Mar 21 '19 at 15:46
• If $q$ divides a given nonzero $x$, then $qa = x$ for some nonzero $a$. Writing $$a = u_a \prod\limits_{p \in P} p^{a_p}$$ in its unique form, you have $$x = qa = u_aq^{a_{q}+1}\prod\limits_{\substack{p \in P \\ p \neq q}}p^{a_p}$$ in its unique form. Since the exponent $x_{q} = a_q+1$ is positive, $q$ occurs in $x$. – D_S Mar 21 '19 at 16:39

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