I know that in an integral domain, prime implies irreducible. Moreover, in a principal integral domain, these notions are equivalent. The ring of polynomials $\mathbb{Z}[x]$ is not a principal integral domain (if it were, an irreducible ideal $I$ would be maximal, and the quotient $\mathbb{Z}[x]/I$ would be a field, but the quotient by the irreducible ideal $(x)$ is $\mathbb{Z}$ which is not a field). Is there a polynomial in $\mathbb{Z}[x]$ which is irreducible but not prime?
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3$\mathbb Z[x]$ is a unique factorization domain, hence all irreducible elements are prime. – Randy Marsh Nov 12 '20 at 11:56
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As said in comment, you cannot find such an example in $\mathbb{Z}[X]$. In other rings, this could be possible : for example, $2 \in \mathbb{Z}[2i]$ is irreducible but not prime.

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