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Given a commutative ring $R$, we say that $x$ is a gcd of $(y,z)$ iff the following conditions hold:

  1. $x \mid y,z$
  2. For all $x' \in R$, if $x' \mid y,z$, then $x' \mid x$.

This gives a ternary relation on any commutative ring, and we may therefore ask which ring homomorphisms preserve it. Now recall that most monotone maps preserve neither meets nor joins. Hence, it is natural to suspect that most ring homomorphisms don't preserve gcd's. Actually, what is an example of this phenomenon?

Question. What is an example of a homomorphism of rings $f : R_0 \rightarrow R_1$ having the following properties?

  1. $R_0$ and $R_1$ are commutative rings satisfying: "every pair of elements has at least one gcd."
  2. $f$ fails to preserve the aforementioned gcd ternary relation.
goblin GONE
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    Interesting. Unless I made a mistake $R_0$ cannot be a PID. For if it were, then Bezout's identity would exist and a gcd of two elements is their $R_0$-linear combination. But a ring homomorphism necessarily respects such a relation, so... – Jyrki Lahtonen Apr 28 '15 at 17:11
  • Take two primes a, b and choose (f(a), f(b)) > 1 for example with f(x) = kx in N – Piquito Apr 28 '15 at 18:05

2 Answers2

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Let $R_0=R_1=\Bbb Z[x]$, and let $f$ be the map determined by $x\mapsto 2x$. Then $1$ is a gcd for $(2,x)$, but $f(1)$ is not a gcd for $(f(2),f(x))$.

PVAL-inactive
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  • @user26857 No, it is a unitary ring morphism. It is the same example as the other answer (which I didnt see before I answered.) A unitary ring endomorphism on $\Bbb Z[x]$ is determined by where $x$ goes (which is what I described). – PVAL-inactive Apr 28 '15 at 18:34
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$\,\gcd(x,2) = 1\,$ in $\,\Bbb Z[x]\,$ but it is a nonunit $ 2$ in $\,\Bbb Z[x/2]$

Remark $\ $ This cannot occur in Bezout domains (see here for further discussion).

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Bill Dubuque
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