Direct Product, and Subdirect Product in Herstein text I'm reading Noncommutative Rings by I. N. Herstein. And I find one lemma pretty strange. It's on page 52 of the book. I'm typing everything necessary all here, so everyone can have a look at it.
Definition of direct product

Given a family of rings $\{ R_\gamma \}$, for $\gamma \in I$, where I is some index set. We define the direct product, or completely direct sum of $\{ R_\gamma \}$ to be:
$$\prod_{\gamma \in I} R_\gamma = \left\{f:I \to \bigcup_{\gamma \in I} R_\gamma \middle|f(\gamma)\in R_\gamma, \forall \gamma \in I \right\}$$

Definition of subdirect product

R is said to be a subdirect product of the rings $\{R_\gamma\}_{\gamma \in I}$ if there exists a monomorphism $\phi: R \to \prod\limits_{\gamma \in I} R_\gamma$, such that $\pi_\gamma\phi(R) = R_\gamma, \forall \gamma \in I$, where $\pi_\gamma$ is the projection of $\prod\limits_{\gamma \in I} R_\gamma$ onto $R_\gamma$.


Here comes the lemma that I don't really get.

Lemma 2.2.1
Let $R$ be a ring, and let $\phi_\gamma: R \to R_\gamma$ be homomorphism of $R$ onto rings $R_\gamma$. Let $U_\gamma = \text{ker }\phi_\gamma$. Then $R$ is a subdirect product of the $R_\gamma$ if and only if $\bigcap\limits_\gamma U_\gamma = 0$.


Herstein didn't give a proof to it. But the $\Leftarrow$ should be easy. I'm having trouble with the $\Rightarrow$, since i don't think it's correct at all. But after reading herstein's book for a while, I come to realize that his wording is very dense, and there are times when something appears false on the first read, but when think about it for a while, it turns out to be so true.
Here's my counter-example of the lemma, so if someone can have a look at it, I would be very please.

Counter example
Consider the index set $I$ to have only 1 element. And consider the ring $R = \prod\limits_{i \in \mathbb{N}} \mathbb{Z}$, $R$ can be thought of as the direct product of itself (i.e $R = \prod\limits_{i \in I} R$, since $I$ has cardinality of 1), as well as subdirect product of itself (with the injection $\phi$ being the identity).
Now I let $\phi: R \to R$ sends each element $(a_1; a_2; a_3; ...) \in \prod\limits_{i \in \mathbb{N}} \mathbb{Z}$ to $(a_2; a_3; ...) \in \prod\limits_{i \in \mathbb{N}} \mathbb{Z}$. Of course $\phi$ is a ring epimorphism. But  $\text{ker }\phi \cong \mathbb{Z} \neq 0$.

What I'm trying to show above is that even though $R$ is the subdirect product of $R$, I can construct $\phi$ (stands for $\phi_\gamma$'s in the lemma), such that $\text{ker } \phi \neq 0$ (stands for $\bigcap_\gamma \text{ker } \phi_\gamma$).
Am I missing something here? Is there any way to prove the $\Rightarrow$ direction?
I can prove the $\Rightarrow$ direction if it states that there exists a set of $\phi_\gamma$'s instead of $\phi_\gamma$'s being chosen randomly. Or am I misunderstanding Herstein's ideas?
Thank you guys very much,
And have a good day.

Here's an online version of the book: http://books.google.com/books?id=hXa7mVyZCzYC&printsec=frontcover&hl=vi&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false

Edit
If the above theorem needs to be rephrased then, I don't really get the author's idea on page 53:

"On the other hand, suppose that R is a subdirect product of the rings $R_\phi = R / U_\phi$. Therefore, $\bigcap U_\phi = 0$."

I'm not sure, but how can he deduce this fact? It's illegal to use the canonical epimorphisms $\pi_\phi: R \to R / U_\phi$ to conclude that $\bigcap U_\phi = \bigcap \text{ker }\pi_\phi = 0$, right? Because, not EVERY set of epimorphisms $f_\phi: R \to R/U_\phi$ will satisfy the conditions that $\bigcap \text{ker }f_\phi = 0$. There exists some set that does, but not all.
 A: You did not take into account the condition
$$\pi_\gamma\phi(R) = R_\gamma, \forall \gamma \in I.$$
In your case $\pi_\gamma={\rm id}$, so $\phi(R) = R.$ But in the definition it is required that $\phi$ to be an monomorphism.
A: Let me see if I understood correctly your counterexample: you take the product of $\,\Bbb Z\;$ with itself just once, i.e. we get $\,\Bbb Z\,$ itself, both as a direct and as a semidirect product according to Herstein's definitions.
If so, I can't see how your defined $\,\phi\,$ works since an element of the direct product $\,R=\Bbb Z\,$ in this case is just a $\,1-$ton "vector" $\,(n)\,$ , i.e.: there are no $\,(a_1,a_2,...)\,$ as there is only one single coordinate!
Thus, for your function $\,\phi\,$ to work it must be the injection $\,(n)\mapsto n\;$ , which kernel indeed is zero, as promised by the theorem in Herstien's....
A: Yes, you're right.
But I'm pretty sure that what Herstein meant to say was that $R$ is a subdirect product of the $R_\gamma$ using the map $\phi:R\to\prod\limits_{\gamma\in I}R_\gamma$ induced by the $\phi_\gamma$ as the monomorphism in the definition of subdirect product if and only if the intersection of kernels is zero.
