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I am currently studying irreducibility of polynomials, and the notes I am studying from defines irreducible polynomials as those non constant polynomials whose only lower degree divisors are constant polynomials.

Later the notes claim that $3x-3$ is irreducible over $\mathbb{Q}$, which is valid since $3(x-1)$ is non constant and the only lower degree divisor of $3(x-1)$ is $3$, which is constant. But then it claims that $3(x-1)$ is reducible over $\mathbb{Z}$. since $3$ is not a unit in $\mathbb{Z}$

My question is, according to the definition of irreducible polynomial, the only two conditions are that the polynomial should be non zero, and any lower degree divisor of the polynomial must be constant. Now $3x-3$ satisfy both conditions, then why is it reducible over $\mathbb{Z}$?

Devansh Kamra
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    It's the product of two non-unit elements in $\mathbb Z[x]$, that's what reducible means. – lulu Jun 21 '22 at 11:09
  • @lulu So should the definition be, "Irreducible polynomials over a ring are those non constant polynomials whose only lower degree divisors are constant polynomials which are UNITS in the mentioned ring"? Or am I still missing something? – Devansh Kamra Jun 21 '22 at 11:12
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    See also this duplicate. We only need to see how an irreducible element is defined in a domain: "an irreducible element of a domain is a non-zero element that is not invertible (that is, is not a unit), and is not the product of two non-invertible elements." – Dietrich Burde Jun 21 '22 at 11:12
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    Factoring with units is never considered as a "true" factoring. So irreducible just means that the element can't be written as the product of two non-units. – lulu Jun 21 '22 at 11:19

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