Every ring morphism from an integral domain to a field is injective?

Question

I know that every morphism from a field to an integral domain must be injective since fields contain no proper ideals.

I'd like to see a counterexample for the inverse statement. Or even better, can someone please tell me what's wrong with the following proof. I can't find the mistake and it bugs me a lot.

($x/y$ is a shorthand for $xy^{-1}$). For every integral domain $D$ there is a field $Q(D)$, the field of quotients, together with an injective ring morphism $j: D \to Q(D)$ such that every element in $Q(D)$ can be written as a quotient $ja/jb$ for some elements $a$ and $b\neq 0$ from $D$ (see MacLane and Birkhoff, Algebra, Chapter III,5).

Given any ring morphism $\mu: D\to F$ from $D$ to a field $F$, define $\nu: Q(D) \to F$ by $ja/jb \mapsto \mu a/\mu b$.

First, check that $\nu$ is well-defined. If $ja/jb = ja'/jb'$, then $(ja)(jb') = (ja')(jb)$. Then $j(ab') = j(a'b)$ since $j$ is a ring morphism and $ab' = a'b$ since $j$ is injective. Since $\mu$ is a ring morphism, we get $(\mu a)(\mu b') = (\mu a')(\mu b)$ and in consequence $\mu a/ \mu b = \mu a' / \mu b'$. $\nu$ is a well-defined function. $\nu$ is also a morphism of fields (proof below), and thus injective. We have that \begin{align*} \nu(ja) = \nu(ja/1) = \nu(ja/j1) = \mu a/ \mu 1 = \mu a/ 1 = \mu a \end{align*} for all $a\in D$. Thus, $\mu = \nu \circ j$ is the composition of two injective functions and itself injective.

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$\nu$ is a morphism of fields... $\nu$ preserves multiplication: \begin{align*}\nu((ja/jb)(jc/jd)) = \nu((ja\, jc)/(jb\, jd)) = \nu(j(ac)/j(bd)) = \mu(ac)/\mu(bd) = ... = (\mu a / \mu b) (\mu c/ \mu d) = \nu(ja/jb)\nu(jc/jd)\end{align*} Addition: \begin{align*} \nu(ja/jb + jc/jd) = \nu((ja\,jd)/(jb \, jd) + (jc\, jb) / (jd\, jb) ) = \nu((ja\, jd + jc\, jb) / (jb\, jd)) = ...= \nu(j(ad + cb)/j(bd)) = \mu(ad+cb) /\mu(bd) = ... =\mu a / \mu b + \mu c/\mu d = \nu(ja/jb) + \nu(jc/jd) \end{align*} And finally unit $\nu(1) = \nu(1/1) = \nu(j1/j1) = \mu 1/ \mu 1 = 1/1 = 1$.

Answer 1

There is nothing preventing $\mu b$ from being $0$. For example, $R=\mathbb{Q}[x]$ is an integral domain with field of fractions $Q(R)=\mathbb{Q}(x)$. There is a homomorphism $R\to\mathbb{Q}$ sending $x\mapsto 0$. There is no corresponding homomorphism $Q(R)\to \mathbb{Q}$ as you have constructed it, because you may be dividing by a multiple of $x$ which is sent to $0$.

For a simpler example, you could take $R=\mathbb{Z}$ and $Q(R)=\mathbb{Q}$. There is a homomorphism $\mathbb{Z}\to\mathbb{Z}_2$, but none from $Q(R)$ to $\mathbb{Z}_2$.

Answer 2

The issue is that by defining $ja/jb\mapsto \mu a/\mu b$ you have implicitly assumed that $\mu b\neq0$, which is about the same as assuming $\mu$ is injective to start with; indeed you can get a counter-example by taking $k[x]\to k$ to be the homomorphism mapping a polynomial $a_0+a_1x+\cdots+a_nx^n$ to the element $a_0\in k$, for any field $k$.