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Intuitively, we may conclude that $3$ is not a unit in $\mathbb{Z}$ simply by observing that $1/3 \notin \mathbb{Z}$. However, what does this reasoning actually appeal to??? Is it true that:

Conjecture. For all rings $R$ and all $r \in R$, if there exists a ring $S$ and a homomorphism $\varphi : R \rightarrow S$ such that $\varphi(r)$ has a multiplicative inverse $s' \in S$ such that $s' \notin \mathrm{im}\, \varphi$, then $r$ is not a unit in $R$.

If not, what is the correct statement of the theorem?

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    $\begingroup$ How intuitively? What about $\frac{1}{3} \in \mathbb{Z}_8$? Is $3$ a unit there? $\endgroup$ – John May 31 '13 at 0:17
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I have often seen the proof "$\mathbb{Z}$ is not a field: $2$ is not invertible, since $1/2 \notin \mathbb{Z}$" (of course the same can be said with $3$ instead of $2$). But this is not a proof. It is only a reformulation of the claim: We use the embedding $\mathbb{Z} \to \mathbb{Q}$ and say that $1/2 \in \mathbb{Q}$ does not lie in the image of $\mathbb{Z}$. Ok but one has to argue why this is the case. Why is there no integer among $0,1,-1,2,-2,\dotsc$ which becomes $1$ when doubled? Of course we learn this in school, but in school one doesn't learn a proof for this either. $^{(1)}$

Anyway, here is a correct proof, in fact for a stronger statement: $\pm 1$ are the only units of $\mathbb{Z}$. For that, let $n \in \mathbb{Z}$. Assume $n \geq 2$. Then for every $m \in \mathbb{Z}$ we have either $m \leq 0$, hence $mn \leq 0$, or $m \geq 1$, hence $mn \geq n > 1$, in particular $mn \neq 1$ (these properties of inequalities can be derived from any formal definition of the set of natural numbers, using induction). This shows $mn \neq 1$. Hence, $n$ is not a unit in $\mathbb{Z}$. Assuming $n \leq -2$, then $-n \geq 2$, hence $-n$ is not a unit, and therefore $n$ is not a unit. $\square$

$^{(1)}$ Here is another example which hopefully convinces you that this proof is incomplete: Consider the ring of formal power series with, say, rational coefficients $\mathbb{Q}[[x]]$. Then one might say "this is not a field, since $\frac{1}{1-x} \notin \mathbb{Q}[[x]]$". Well, it is true that a priori this fraction is not represented as a formal power series. And this fake proof could be be able to convince students who have just learned the definition of formal power series, and remember that $1-x$ is not a unit in $\mathbb{Q}[x]$. But actually $1-x$ is a unit, the inverse is the well-known geometric series $1+x+x^2+\dotsc$. Actually this is the key step to the more general result $R[[x]]^* = \{f \in R[[x]] : f(0) \in R^*\}$ for commutative rings $R$. From this we can build more complicated examples, such as

$$\frac{1}{x^2+x+1} = \sum_{k=0}^{\infty} x^{3k} - \sum_{k=0}^{\infty} x^{3k+1},$$ $$\frac{1}{x^2+x-2}=\sum_{k=0}^{\infty} \bigl(-\frac{1}{3} - \frac{(-1/2)^k}{6}\bigr) x^k.$$

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    $\begingroup$ PS: If one already knows how to manipulate inequalities of fractions, we have $|n|>1 \Rightarrow 0 < |1/n| < 1$, hence $1/n \notin \mathbb{Z}$. But this basically comes down to the proof above. $\endgroup$ – Martin Brandenburg May 31 '13 at 0:36
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    $\begingroup$ I guess many of us are guilty of just assuming that the integer multiples of 2 are all even (irony) :) $\endgroup$ – rschwieb May 31 '13 at 20:44
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Hint: Think about the following true statement and how it relates to your conjecture.

Suppose $r\in R$ is a unit. Then, for all ring homomorphisms $\varphi:R\to S$, $\varphi(r)$ is a unit in $S$, with inverse $\varphi(r^{-1})$.

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Your conjecture is true! Suppose to the contrary that $r$ is a unit in $R$, say $rs=1$, then $\varphi(rs)=\varphi(r)\varphi(s)=1$, contradicting $\varphi(r)s'=1$.

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  • $\begingroup$ @PeterTamaroff Thanks! $\endgroup$ – Ma Ming May 30 '13 at 22:03

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