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I am reading Herstein and it makes the following claim.

The sentence followed by the definition is what I don't get.

enter image description here

A prime element $\pi \in R$ has no non-trivial factorisation in $R$.

By definition, I see that if $\pi = ab$ for some $a,b\in R$, then exactly one of $a$ or $b$ must be a unit. (But not both, because otherwise their product $ab$ would be a unit and so would $\pi$, which is not allowed by the definition of prime elements)

By this I take it to mean that the only factorisation of $\pi$ is where one of $a$ or $b$ is a unit and the other must be $\pi$, but for this I can't see why the other non-unit, say $b$, cannot be something that isn't $\pi$ but where $ab$ multiply up to $\pi$.

Any help appreciated thanks!

Edit: snapshot from book added to clarify that this is the paragraph I don't understand.

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    $\begingroup$ Remind me, what is your definition of prime element? $\endgroup$
    – Malady
    Commented Jul 5 at 20:39
  • $\begingroup$ What is $R$? Any (commutative, unital) ring? An integral domain? $\endgroup$ Commented Jul 5 at 20:41
  • $\begingroup$ Multiplication by a unit is considered a trivial factorization. If $u \in R$ is a unit, then it's always true that $u(u^{-1}\pi)=\pi$. $\endgroup$ Commented Jul 5 at 20:55
  • $\begingroup$ Sorry I've updated with a picture that should explain that the set up is a Euclidean domain. I haven't read anything about irreducible elements so I'm not too sure what the connection is just yet. Thank you! $\endgroup$
    – Jeff8770
    Commented Jul 5 at 21:50
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    $\begingroup$ See the linked dupes. Beware that Herstein's definition of "prime" is nonstandard. Rather, nonzero elements in a domain satisfying his definiton are normally called irreducibles (in non-domains this notion bifurcates and their is no standard terminology). However, irredcibles and primes are equivalent notions in gcd domains, e.g. Euclidean domains. $\ \ $ $\endgroup$ Commented Jul 5 at 21:55

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