What ring homomorphisms do induce well-defined mapping between principal ideals?

Question

The original question I was going to ask was an attempt to define the notion of a "GCD domain homomorphism", but ultimately it boiled down to another question, so here I go.

Given a ring $R$, let the set of principal ideals of $R$ be denoted by $\langle R \rangle$. Given a ring homomorphism $\phi : R \to S$, a natural question to follow is whether $\phi$ induces a well-defined mapping from $\langle R \rangle$ to $\langle S \rangle$, denoted by $\langle \phi \rangle$. That is, whether the map $\langle \phi \rangle$ defined as $\langle \phi \rangle (\langle x \rangle) = \langle \phi(x) \rangle$ is well-defined.

As long as $R$ is a GCD domain, since $\langle R \rangle$ can be endowed a lattice order structure where $\gcd$ is the join and $\mathrm{lcm}$ is the meet, the notion of a GCD domain homomorphism should follow. But I'm concerned whether the requirement of well-definedness of $\langle \phi \rangle$ is too restrictive.

I tested some ring homomorphisms, such as inclusion map $i : \mathbb{Z} \to \mathbb{Q}$, the quotient map $q : \mathbb{Z} \to \mathbb{Z}_2$, and the evaluation map $\epsilon_0 : \mathbb{Z}[x] \to \mathbb{Z}$. And surprisingly, they are all successful examples. So what is a non-example?

Answer 1

You don't specify if your rings are commutative or have unities, so let me do this as generally as possible for as long as possible...

Let $R$ and $S$ be ring, and let $\phi\colon R\to S$ be a ring homomorphism.

If $I$ is an ideal of $R$, then $\phi(I)$ need not be an ideal of $S$, but we can form the ideal of $S$ generated by $\phi(I)$, $(\phi(I))$. In the commutative ring context this is usually called the "extension of the ideal $I$", and for rings with unity it corresponds to $S\phi(I)S$.

This defines a map from the set of ideals of $R$ to the set of ideals of $S$, mapping $I$ to $(\phi(I))$. The map is monotone: if $I\subseteq J$, then $\phi(I)\subseteq \phi(J)$, hence $(\phi(I))\subseteq (\phi(J))$.

In addition, the map sends principal ideals to principal ideals: suppose $I=(a)$. I claim that $(\phi(I))=(\phi(a))$. Indeed: since $\phi(a)\in \phi(I)$, then $(\phi(a))\subseteq(\phi(I))$. Conversely, if $x\in I$, then $x=na+ra+as + \sum_{j=1}^m r_jas_j$ with $n\in\mathbb{Z}$, $m\geq 0$, $r,s,r_j,s_j\in R$; so $\phi(x) = n\phi(a) + \phi(r)\phi(a)+\phi(a)\phi(s) + \sum_{j=1}^m \phi(r_j)\phi(a)\phi(s_j)\in (\phi(a))$, so $\phi(I)\subseteq (\phi(a))$, and hence $(\phi(I))\subseteq (\phi(a))$.

So we certainly get a map from the set of principal ideals of $R$ to the set of principal ideals of $S$; if we partially order both under inclusion, the map is monotone.

So now we move to the case where $R$ and $S$ are GCD domains, so that they are integral domains in which any two elements have a gcd and an lcm, which make the poset of principal ideals into a lattice. Now we have a map from the lattice of principal ideals of $R$ to the lattice of principal ideals of $S$, and the map is monotone. Is it a lattice homomorphism?

The main difficulty is that in general it is not true that a monotone map between two lattices is a lattice homomorphism. That is because if $L_1$ and $L_2$ are lattices, then $f\colon L_1\to L_2$ is a lattice homomorphism if and only if $f(a\wedge b) = f(a)\wedge f(b)$ and $f(a\vee b) = f(a)\vee f(b)$. This implies that the function is monotone, but the converse is not true; for a counterexample, consider the map between the four element diamond lattice $\{0,a,b,1\}$ and the three-element chain $\{0,1,2\}$ that sends $0$ to $0$, $a$ and $b$ to $1$, and $1$ to $2$. This map is (strictly) monotone: if $x\lt y$ then $f(x)\lt f(y)$. But $f(a\wedge b) = f(0) = 0$, and $f(a)\wedge f(b) = 1\neq 0 = f(a\wedge b)$, and $f(a)\vee f(b) = 1\neq 2 = f(a\vee b)$.

So the question here is: if $a,b\in R$ will $(\phi(\gcd(a,b))) = (\gcd(\phi(a),\phi(b)))$?

Since $\gcd(a,b)$ divides both $a$ and $b$, then $\phi(\gcd(a,b))$ divides both $\phi(a)$ and $\phi(b)$, so we always have that $(\gcd(\phi(a),\phi(b)))\subseteq (\phi(\gcd(a,b)))$. But the converse inclusion need not hold:

Let $R=\mathbb{Z}[x,y]$ and $S=\mathbb{Z}[z]$. They are both GCD-domains (in fact, UFDs). Note that $\gcd(x,y)=1$. Now let $\phi\colon R\to S$ be the map that is the identity on the coefficients and sends $x$ and $y$ both to $z$. Then we have that $\gcd(\phi(x),\phi(y)) = \gcd(z,z) = z$, but $\phi(\gcd(x,y))=1$. So here $(\gcd(\phi(x),\phi(y))=(z)\subsetneq (1) = (\phi(\gcd(x,y)))$.

If $R$ is a Bézout domain, though, then it will work. From the above observations we just need to show that $(\phi(\gcd(a,b)))\subseteq (\gcd(\phi(a),\phi(b))$,, or equivalently, that $\gcd(\phi(a),\phi(b))$ divides $\phi(\gcd(a,b))$.

In this situation, we have $(a,b)=(\gcd(a,b))$, so $\gcd(a,b) = ar+bt$ for some $r,t\in R$. Therefore, $\phi(\gcd(a,b)) = \phi(a)\phi(r)+\phi(b)\phi(t)$ which is clearly a multiple of $\gcd(\phi(a),\phi(b))$. So in this situation, we do have equality and we get a lattice morphism between the principal ideals of $R$ and the principal ideals of $S$.