About comment by Jacobson on proving that a morphism in $\mathbf{Ring}$ is monic iff it is injective I am reading Nathan Jacobson's Basic Algebra II, Chapter 1 Categories, and in $\S$2 Some Basic Categorical Concepts, he introduces the notion of a morphism being monic or epic. He asks the reader to observe that in the category $\mathbf{Set}$, a morphism is monic (epic) iff it is injective (surjective). Then, he states and proves the following proposition (on page 16):

PROPOSITION 1.1. A morphism $f$ in $R$-$\mathbf{mod}$ or in $\mathbf{Grp}$ is monic (epic) iff the map of the underlying set is injective (surjective).

I reproduce the part of the proof that is relevant for my question below:

Proof: Let $f : A \to B$ in $R$-$\mathbf{mod}$ or $\mathbf{Grp}$. If $f$ is injective (surjective) as a map of sets, then it is left (right) cancellable in $\mathbf{Set}$. It follows that $f$ is monic (epic) in $R$-$\mathbf{mod}$ or $\mathbf{Grp}$. Now suppose the set map $f$ is not injective. Then $C = \ker(f) \neq 0$ in the case of $R$-$\mathbf{mod}$ and $\neq 1$ in the case of $\mathbf{Grp}$. Let $g$ be the injective homomorphism of $C$ into $A$ (denoted by $C \hookrightarrow A$), so $g(x) = x$ for every $x \in C$. Then $fg$ is the homomorphism of $C$ into $B$, sending every $x \in C$ into the identity element of $B$. Next let $h$ be the homomorphism of $C$ into $A$, sending every element of $C$ into the identity element of $A$. Then $h \neq g$ since $C \neq 0$ (or $\neq 1$), but $fg = fh$. Hence $f$ is not monic.

Now, after completing the proof Jacobson asks what happens in the category $\mathbf{Ring}$, which is the category with objects as unital (associative) rings and morphisms as ring homomorphisms (mapping $1$ to $1$). He writes (on page 17):

In the first place, the proof of Proposition 1.1 shows that injective homomorphisms are monic and surjective ones are epic. The next step of the argument showing that monics are injective breaks down totally, since the kernel of a ring homomorphism is an ideal and this may not be a ring (with unit). Moreover, even if it were, the injection map of the kernel is most likely not a ring homomorphism.

My confusion is regarding the sentence that I've bolded. The kernel of a ring homomorphism is always an ideal, and if it is also a unital ring, then it must be the full ring. In this case, the injection is just the identity map on the ring which is always a ring homomorphism. So, is this comment by Jacobson an error on his part?
 A: In the category of rings with unity (with a $1$), morphisms are required to take the unity to the unity. In the case at hand, unless the morphism is trivial, the kernel will not be a ring with unity that embeds as a ring with unity.
Consider for example the case of $R=\mathbb{Z}\times\mathbb{Z}$, and the morphism $R\to\mathbb{Z}$ obtained by mapping $(a,b)$ to $a$. This is a ring-with-unit morphism.
The kernel of the map is the ideal $I=\{(0,b)\mid b\in\mathbb{Z}\}$. Now, as an abstract ring, $I$ is a ring with unity: the element $(0,1)$ is a unity for $I$. (In fact, $I$ is isomorphic as a ring-with-unity to $\mathbb{Z}$).
Thus, in this case, the kernel is in fact in the category: it is a ring with unity. However, the embedding $I\hookrightarrow R$ is not a morphism in the category of rings-with-unity, because the unit of $I$, $(0,1)$, is not mapped to the unit of $R$. Thus, the embedding is not a morphism in this category.
That’s Jacobson’s point: even if the kernel happens to, abstractly and by happenstance, be a ring with unity, you will almost never have that the embedding of the kernel into the ring is a morphism in the category. The slight error in your argument is that it is possible for the kernel to be a ring-with-unity, abstractly, even if it is not a subring-with-unity (which would require the unity of the ideal to be the same as the unity of the whole ring, in which case you are correct that the ideal would have to be the whole ring).
Note that this issue does not show up in the category of rings/rngs, where you do not require rings to have a unity and you do not require morphisms to map $1$ to $1$ even when the two rings do. In that category, the same proof shows that a morphism is monic if and only if it is one-to-one. 
A: It may be possible for an ideal of a ring to be a ring with a different one. For instance, we can make a ring out of finite sums of the variables $\{X_n | n \in \mathbb{N} \}$ where multiplication is given by $X_n X_m = X_{max(n,m)}$ and extending linearly. The one of this ring is $X_0$.
Then, consider the ideal given by sums of variables $\{X_n | n > 0 \}$. This by itself is a ring (in fact, it's isomorphic to the original ring) but it has a different one given by $X_1$.
Therefore, even in the very special case where the ideal is a unital ring, like this one, the inclusion need not be a unital ring homomorphism. 
