# Is the kernel of any ring homomorphism a subring, according to this definition?

This is an exercise taken verbatim from Birkhoff and MacLane, A Survey of Modern Algebra:

Show that if $\phi: R \rightarrow R'$ is any homomorphism of rings, then the set $K$ of those elements in $R$ which are mapped onto $0 \in R'$ is a subring.

Let me recall from this section of the text the definition of a subring:

... define a subring of a commutative ring $A$ as a subset of $A$ which contains, with any two elements $f$ and $g$, also $f \pm g$ and $fg$, and which also contains the unity of $A$.

It seems to me that the implication in the exercise is false in general. I agree that $a,b \in K$ implies $a \pm b, ab \in K$, but it seems $1 \notin K$ in general. For example, take $\phi$ to be the identity map from $\mathbb{Z}$ to itself, then $K = \{0\}$ and hence $K$ does not contain the unity element. But the definition of subring given here requires that it contain the unity element.

I wonder if I am missing something here, and in case I am correct that the implication doesn't hold, is there some variation of the statement which is true?

• Right. $K$ is an ideal, so $1 \in K \Rightarrow K = R$. Often, a subring is defined to not necessarily include $1$, probably the authors just mixed their definitions up and thought of the one without including the $1$ when posing the exercise. – Daniel Fischer Jul 7 '13 at 1:41
• @DanielFischer: ok, thanks. – AG. Jul 7 '13 at 2:34