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As above, I am trying to answer the following question:

"Does every commutative ring with $1 \neq 0$ and a non-zero, non-unit element contain a prime element?"

Since in some books prime elements are only defined for integral domains, one can reformulate the following question in the following way:

""Does every integral domain containing a non-zero, non-unit element contain a prime element?"

Of course if we get rid of the second assumption, the answer would be no, since any field would be an example of a ring with no prime elements. However, what would happen in a non-field case?

Also, I understand that the answer to the second question may be different than the answer to the first question. However, any help regarding any of the above would be greatly appreciated.

EDIT $1$: My definition of a prime element is the following: $p$ is prime if $p$ is a non-zero non-unit and whenever $p|ab$ then $p|a$ or $p|b$.

EDIT $2$: By integral domain I mean a commutative ring with $1 \neq 0$ and no zero divisors.

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    $0$ is prime in any integral domain if by "prime element" you mean "generates a prime ideal." – 2017-01-24
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    No, e.g. in the ring of all algebraic integers there are no irreducibles (so no primes), since $\, a =\sqrt a \sqrt a\ \ \ $ – 2017-01-24
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    so the ring is commutative? – 2017-01-25
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    @JorgeFernándezHidalgo In the first question not-necessarily, in the second question yes. – 2017-01-25
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    how do you define divisibility in the non-commutative case? – 2017-01-25
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    @JorgeFernándezHidalgo You are right. It needs to be commutative. Thank you for your comment. I'll edit the question. – 2017-01-25
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    @AdamHughes are you sure? As far as I know, prime elements are taken to be non-zero and non-unit as the OP wrote in his post. – 2017-01-25
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    @Xam he said the ring contains non-zero elements, not that the prime he's looking for needs to be non-zero. And yes, since $R$ is an integral domain, $R/(0)$ is an integral domain, so by definition $(0)$ is a prime ideal, and if a prime element means "generates a prime ideal" $0$ fits the bill. It seems he's edited to add in new conditions, so that's no longer the case, but it was when I wrote the comment. – 2017-01-25

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