# In a principal ideal domain, "prime" and "irreducible" are the same

Let $$R$$ be a ring which is a PID, and let $$r \not = 0$$ be an el­e­ment of $$R$$. Then $$r$$ is ir­re­ducible if and only if $$r$$ is prime.

In fact, it is eas­ier to prove a stronger state­ment: that the fol­low­ing are equiv­a­lent. noteEvery proof known to the au­thor is of this shape ei­ther im­plic­itly or ex­plic­itly, but when it’s ex­plicit, it should be clearer what is go­ing on.

1. $$r$$ is ir­re­ducible.

2. $$r$$ is prime.

3. The gen­er­ated ideal $$\langle r \rangle$$ is max­i­mal in $$R$$.

mo­ti­vate the third bul­let point

# Proof

## $$2 \Rightarrow 1$$

A proof that “prime im­plies ir­re­ducible” ap­pears on the page for ir­re­ducibil­ity.

## $$3 \Rightarrow 2$$

We wish to show that if $$\langle r \rangle$$ is max­i­mal, then it is prime. (In­deed, $$r$$ is prime if and only if its gen­er­ated ideal is prime.)

An ideal $$I$$ is max­i­mal if and only if the quo­tient $$R/I$$ is a field. (Proof.)

An ideal $$I$$ is prime if and only if the quo­tient $$R/I$$ is an in­te­gral do­main. (Proof.)

All fields are in­te­gral do­mains. (A proof of this ap­pears on the page on in­te­gral do­mains.)

Hence max­i­mal ideals are prime.

## $$1 \Rightarrow 3$$

Let $$r$$ be ir­re­ducible; then in par­tic­u­lar it is not in­vert­ible, so $$\langle r \rangle$$ isn’t sim­ply the whole ring.

To show that $$\langle r \rangle$$ is max­i­mal, we need to show that if it is con­tained in any larger ideal then that ideal is the whole ring.

Sup­pose $$\langle r \rangle$$ is con­tained in the larger ideal $$J$$, then. Be­cause we are in a prin­ci­pal ideal do­main, $$J = \langle a \rangle$$, say, for some $$a$$, and so $$r = a c$$ for some $$c$$. It will be enough to show that $$a$$ is in­vert­ible, be­cause then $$\langle a \rangle$$ would be the en­tire ring.

But $$r$$ is ir­re­ducible, so one of $$a$$ and $$c$$ is in­vert­ible; if $$a$$ is in­vert­ible then we are done, so sup­pose $$c$$ is in­vert­ible.

Then $$a = r c^{-1}$$. We have sup­posed that $$J$$ is in­deed larger than $$\langle r \rangle$$: that there is $$j \in J$$ which is not in $$\langle r \rangle$$. Since $$j \in J = \langle a \rangle$$, we can find $$d$$ (say) such that $$j = a d$$; so $$j = r c^{-1} d$$ and hence $$j \in \langle r \rangle$$, which is a con­tra­dic­tion.

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