# Every homomorphism $\varphi : F \to R$ from a field $F$ to a nonzero ring $R$ is injective: proof details in Artin

In Artin's Algebra the statement in the title is introduced as a corollary of the following:

The only ideals of a field are the zero ideal and the unit ideal. $\star$

The proof of the statement in the title then goes as following:

The kernel of $\varphi$ is an ideal of $F$. So according to $\star$, the kernel is either $(0)$ or $(1)$. If $\ker \varphi$ were the unit ideal $(1)$, $\varphi$ would be the zero map. But the zero map isn't a homomorphism when $R$ isn't the zero ring. Therefore $\ker \varphi = (0),$ and $\varphi$ is injective.

Now $R$ sure isn't the zero ring, but the zero ring is still embedded in $R$. Suppose $\varphi$ is the zero map, then

$$\varphi(a + b) = 0 = 0 + 0 = \varphi(a) + \varphi(b)\\ \varphi(ab) = 0 = 0 \cdot 0 = \varphi(a)\varphi(b) \\ \varphi(1_R) = 0 = 1,$$

so $\varphi$ is a homomorphism from $F$ to $R$ (although it doesn't hit anything except $0$ in $R$).

So is the statement in bold technically incorrect, or am I missing something? Other questions on this topic indicate that other authors like to phrase the original statement as "$\varphi$ is injective or the zero map".

It is the last claim which fails: $$\varphi(1_R) = 0 \neq 1$$
Note that if $R$ is a ring and $0=1$, then $R$ is the zero ring.
Indeed, for all $x \in R$ we have $$x=x \cdot 1=x \cdot 0=0$$
• @AndreyPortnoy The last claim fails because in an unitary ring, unless the ring is trivial, $1 \neq 0$. And part of the definition of ring homomorphism is $\phi(1)=1$. – N. S. Feb 13 '18 at 5:22