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Let $R$ be a UFD and $f(x)=a_0+a_1x+...+a_nx^n \in R[x]$. Then the content of $f(x)$, denoted by $C(f)=gcd(a_0,a_1,...,a_n)$.

Let $R$ be a UFD and $f(x) \in R[x]$ is a non-constant irreducible polynomial in $R[x]$.Then is it must that $C(f)=1$?

I am asking this because if I take $f(x)=4x^3+2x^2+2x+2 \in \mathbb{Z}[x]$. Then $f(x)$ is irreducible in $\mathbb{Z}[x]$, but it's $C(f)=2$

Alexander
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  • Doesn't $2\mid f$ ? More generally, doesn't $C(f)\mid f$? – Sassatelli Giulio Jun 07 '22 at 06:18
  • But $f(x)$ is reducible if it is product of non constant polynomials, here 2 is constant – Alexander Jun 07 '22 at 06:20
  • Read again the definitions. An element $x$ of a domain is irreducible if and only if it isn't $0$ or invertible and, for all $a,b$ such that $x=ab$, $a$ is invertible or $b$ is invertible. – Sassatelli Giulio Jun 07 '22 at 06:25
  • @SassatelliGiulio yeah I know that definition, according to that definition f(x) is not irreducible as 2 is not unit. – Alexander Jun 07 '22 at 06:33
  • I have read that definition in case of an element of $R$ that an element $q \in R$ is said to be irreducible in $R$ iff $q$ is non zero non unit and whenever $q=bc $ for $b,c \in R$ then $b$ or $c$ is a unit. – Alexander Jun 07 '22 at 06:35
  • @AlvinL The definition of reducible be as it may, the exercise asks for an irreducible polynomial, not for a polynomial that isn't reducible. – Sassatelli Giulio Jun 07 '22 at 06:44
  • @SassatelliGiulio But I am also confusing with that definition of irreduciblity of polynomials, that says $f(x) \in R[x]$ is irreducible if there doesn't exist $u(x),v(x)$ in $R[x]$ suchthat $f(x)=u(x)v(x)$ where $0 \lt deg u \lt deg f$ and $0 \lt deg v \lt deg f$ – Alexander Jun 07 '22 at 06:46
  • @Epsilon101 I have stated what I claim is the standard definition of irreducible polynomial in a domain, and uner that definition the theorem holds. You claim it's a different one, and under that definition the claim is false. I see no point in further discussion. – Sassatelli Giulio Jun 07 '22 at 06:49
  • @SassatelliGiulio is there a difference between not reducible and irreducible? – Alexander Jun 07 '22 at 06:50
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    With my definitions, no. With yours, no. With a mix of the two, yes, for the reason we've discussed. – Sassatelli Giulio Jun 07 '22 at 06:51
  • @SassatelliGiulio My mistake, got hung up on semantics. – AlvinL Jun 07 '22 at 07:11
  • @AlvinL I don't know if you did, but I certainly did. – Sassatelli Giulio Jun 07 '22 at 07:16
  • You seem to be attempting to use a definition of "irreducible" polynomial that works only for polynomials over a field - see the linked dupe. This is a common oversight. – Bill Dubuque Jun 07 '22 at 11:22
  • See also here on definition(s) of "irreducible" in integral domains. – Bill Dubuque Jun 07 '22 at 11:29

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