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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!

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    $1=0$ when the additive identity and multiplicative identity are the same element. This occurs when your set is $\{0\}$, but I don't believe this is considered a field.2011-04-01
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    @yunone: no, it's not a field. It's not even an integral domain.2011-04-01
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    @Qiaochu, ok, good, thank you.2011-04-01
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    @mathcat Whenever you edit a question in a way that invalidates prior answers (as below) you should, as a courtesy, add a note to the question saying so. Otherwise readers may waste valuable time trying to comprehend answers that no longer make sense.2011-04-01
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    @bill actually I only edited it to fix some latex issues, not the content of the question. but thanks anyway for letting me know :)2011-04-01
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    @mathcat Your original question had the map in the opposite direction - which is a different, simpler question (which was answered by Qiaochu)2011-04-01

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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.)

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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.)

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    Why do you think the above has anything to do with the "situation" in the stated theorem? It has nothing to do with maps *from* the trivial ring.2011-04-01
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    @Bill: the question was originally stated with the arrow in the other direction. (In hindsight, the statement is incorrect in this direction, but I was tired.)2011-04-01
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    Ah, that explains the mixup. The OP should've mentioned the drastic change.2011-04-01
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    @bill ok, now I got your point. sorry, it was a typo; I put the wrong arrow there...2011-04-01
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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.

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    Why wouldn't they just say "R is not trivial" ?2011-04-01
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    @The By definition $1\ne 0\:$ in a field so the field $\rm\:R\:$ cannot be a trivial ring.2011-04-01
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    @B: I see (more or less!)2011-04-01
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    @The point of enforcing the convention that $1\ne0\:$ in fields and domains is to eliminate uninteresting and bothersome extremal cases like those in OP's theorem. By the way, you need to put at least 3 characters of the username after the @ for it to notify the user, e.g. see above!2011-04-01
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    @Bill: My name has never been so functional! Thanks again.2011-04-01