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I am revising a little bit some Galois Theory using the book of P. Morandi, "Field and Galois Theory". I stack on the proof of separability polynomials in characteristic $0$, where he writes

If $f(X) \in K[X]$ is irreducible over the field $K$, then the only possibility for $\gcd(f(X),h(X))$ is $1$ or $f(X)$.

Well, say $g(X)=\gcd(f(X),h(X))$. So, $g(X)|f(X)$, thus $f(X)=k(X)g(X)$ for some $k(X)\in K[X]$. Since $f(X)$ is irreducible over $K$, it must be either $k(X)$ invertible, or $g(X)$ invertible. In other words, either $k(X)=c\in K\setminus \{0_K\}$, or $g(X)=1_K$ (since the gcd is monic). So, $g(X)=c^{-1}f(X)$ or $g(X)=1_K$.

Why does the author claim that $g(X)=f(X)$ in the first case?

Thanks!

Chris
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  • Well you need to start saying that $f$ is monic too, so you obtain $c=1$ by looking at the coefficient of the highest degree – Aitor Iribar Lopez Jan 09 '22 at 00:36
  • I would clearly agree, but the author doesn't assume that $f$ is monic too. But I can see that this is necessary. Do I miss something? – Chris Jan 09 '22 at 00:37
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    Are you sure that your deifnition and the author's of gdc agree? Because by the way you have written it it says either gcd is a unit or gcd is $f$. If gcd is always monic then $f$ is monic too – Aitor Iribar Lopez Jan 09 '22 at 00:40
  • I searched this, but it is nowhere stated clearly in the text... – Chris Jan 09 '22 at 00:51
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    They're either using the widespread convention that gcds in $K[x]$ are normalized to be monic, or else the common abuse of language with gcds where $,\gcd(a,b) = c,$ (or the gcd "is" $c),$ means that the gcd is associate to $c$, i.e. $,\gcd(a,b) \approx c,,$ esp. used in domains where there is no natural choice for unit-normalization, cf. this answer in the linked dupe, and other related answers. – Bill Dubuque Jan 09 '22 at 09:27
  • @AitorIribarLopez Thanks for both the answer and the comments. – Chris Jan 09 '22 at 14:19
  • @BillDubuque Actually the answer I was looking for was only for these lines, that is whether the author had a mistake/typo here or whether I was missing something. Anyhow, thanks for extending my question and for the useful suggestions. – Chris Jan 09 '22 at 14:23

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Usually in arbitrary rings there is no canonical choice of the $\gcd$. In the first known rings that have $\gcd$ such as $\mathbb Z$ or $K[X]$ one usually says "gcd is positive" or "gcd is monic". However, $-2$ as (from the perspective of divisibility) $\gcd$ of $4$ and $6$ as $2$ is. An example of a ring in which there is not a natural way to pick a $\gcd$ is the ring of Gaussian Integers $\mathbb Z[i]$.

For this reason, usually one says that $\gcd(a,b)$ (when it exists) is the set of all $c$ such that $c \mid a$, $c \mid b$ and $c$ is maximal with this property. In particular (when your ring is an integral domain), two $\gcd$ differ by multiplication by a unit. In this context, your text would be better written as something like this: if $f$ is irreducible then for all polynomials $h$ either $f$ is a $\gcd$ for $f$ and $h$ or $1$ is a $\gcd$ for $f$ and $h$, and written like this your proof is correct.

user26857
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Aitor Iribar Lopez
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