About comment by Jacobson on proving that a morphism in $\mathbf{Ring}$ is monic iff it is injective

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.

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.