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?

  • 7
    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jul 7 '13 at 1:41
  • $\begingroup$ @DanielFischer: ok, thanks. $\endgroup$ – AG. Jul 7 '13 at 2:34

This is a community wiki answer intended to remove this question from the unanswered queue.

As Daniel Fischer pointed out in the comments, the OP's reasoning is correct, and this was probably just a momentary lapse in consistency with definitions.


For a ring to be a subring it does not need to have the unit element. It only has to have the zero element, closure under addition and multiplication and an additive inverse, meaning a+(some element)=(zero element). If k={0} it has the zero element of the integers, it is closed under Addition and multiplication and 0+(-0)=0.

  • $\begingroup$ Not if a subring has the unit element it just means it is a ring with identity, but rings don't have to have identity or a multiplicative inverse at that $\endgroup$ – Helper Oct 7 '15 at 17:13
  • 1
    $\begingroup$ The definition in the book used by the OP requires a subring to contain the unity of the ambient ring. $\endgroup$ – Daniel Fischer Oct 7 '15 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.