# In a PID without unit an ideal is maximal iff it is prime.

As the titles says, I need to show that in a PID $R$ an ideal is maximal iff it is prime. This is easy to do if $R$ has a multiplicative identity. I can not do it if $R$ does not have an identity. It would be great if someone could help me out.

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A PID is an integral domain, which has a multiplicative identity. – Alan Simonin Jan 19 '13 at 14:34
@AlanSimonin I checked my book's definition. It doesn't include this.I checked my book's definition. It doesn't include this. – Amr Jan 19 '13 at 14:34
Quote from Wikipedia : "It is usually assumed that commutative rings and integral domains have a multiplicative identity even though this is not always included in the definition of a ring" – Alan Simonin Jan 19 '13 at 14:39
It's an interesting question, even for commutative rings. Is it right to assume that you meant to ask "a non-zero ideal is maximal iff it is prime"? Since $(0)\subset\mathbb{Z}$ seems a prime ideal to me, while it most certainly isn't maximal. – HSN Jan 19 '13 at 17:01
@zaltan It's really unlikely Hungerford intended to ask about a domain without a unit. For instance, you can see that he assumes integral domains have a unit in the proof to theorem 3.4. If he did not intend for there to be units, he would have a really hard time defining irreducible elements and handling UFD's to begin with. Finally, most authors don't put unnecessarily complicated problems as problem #1 in a section. – rschwieb Jan 20 '13 at 12:50