In Lindsay Childs' Algebra text (3ed pg 141), it statements this proposition regarding a ring homomorphism: $ Let f: R \rightarrow S$ be a homomorphism where R is a field and $1 \ne 0$ in $S$. Then $f$ is one-to-one.
My confusion is about $1 \ne 0$. Is that possible? I can only see that in congruence where $[1]_{1} = [0]_{1}$, but again, congruence class of integers mod 1 is not really interesting... btw, the book's errata says nothing about this page.
Any help would be appreciated! Thanks!
There is exactly one ring in which $1 = 0$. It is the trivial ring where every element is equal to zero.
Whether this extra condition is necessary depends on your definition of a ring homomorphism. (I assume at least that your rings have multiplicative identity.) If ring homomorphisms are required to preserve the multiplicative identity then there are no homomorphisms from the trivial ring to any nontrivial ring, so this situation never occurs. (This addressed a different version of the question.)
The case $1 = 0$ occurs when you are talking about the trivial ring $R = \{0\}$. It turns out $R$ is the trivial ring if and only if $1 = 0$, so stating $1 \neq 0$ is arguably the easiest way to say $R$ is a nontrivial ring.
Based on the context I can guess the following: Your book includes as part of the definition of field that $1\neq 0$ (which is standard). Rings in your book have $1$ and ring homomorphisms send $1$ to $1$. If $1=0$ in $S$, then $f(1)=f(0)=0$, so $f$ is not one-to-one. That is why the condition is needed in this case. (As indicated in the other answers, this would only be the trivial case $S=\{0\}$. In that case $f$ would have to send everything to a single point.)
