How to understand the definition of prime subring in GTM 167? In the Appendix of the book GTM 167 'Field and Galois Theory', the author defines something called 'prime subring'. Also, he assumes that every ring in this book has an identity. I have two questions:
1, How to prove the prime subring is the unique minimal subring of R? How to understand the uniqueness here: the subring itself is unique or unique in the meaning of ring-isomorphism? And how to understand the term 'minimal'? 
2, Does the author mean there can be a subring of R whose identity is not the one of the R? For example, an ideal containing the identity of R is exactly the R itself. So proper ideal must contain an identity different from the one of R.

 A: I'm going to answer your questions in the reverse order.
(2) Not everyone assumes that rings have a multiplicative identity, so there is no consensus on the way to define a ring. On the other hand, I do think there is a consensus on the way to define a subring. Mathematicians define a subring $S$ of a ring $R$ to be a subset of $R$, such that when you restrict the operations of $R$ to $S$, $S$ becomes a ring. For mathematicians who do not assume that rings have a multiplicative identity, this just means that $S$ contains $0$, $S$ contains additive inverses, and $S$ is closed under addition and multiplication. For mathematicians who do assume that rings have a multiplicative identity, this means that $S$ contains $0$, $S$ contains additive inverses, $S$ is closed under addition and multiplication, and that $S$ contains the multiplicative identity of $R$ (i.e. $1_R\in S$).
Since your book is assuming that rings have multiplicative identities, your book is also assuming that if $S$ is a subring of $R$ then $S$ contains the multiplicative identity of $R$ (i.e. $1_R\in S$).
Also, please note that subrings and ideals are slightly different. Let $S\subseteq R$. Then $S$ is an ideal of $R$ iff the following hold:
$$0\in S.$$
$$\forall x,y\in S,\;x+y\in S.$$
$$\forall r\in R,\;\forall x\in S,\;r\cdot x\in S\text{ and }x\cdot r\in S.$$
And as we discussed a moment ago, $S$ is a subring of $R$ iff the following hold:
$$0\in S.$$
$$\forall x\in S,\;-x\in S.$$
$$\forall x,y\in S,\;x+y\in S.$$
$$\forall x,y\in S,\;x\cdot y\in S.$$
$$1_R\in S.$$
Notice that these definitions are slightly different. One consequence of this difference is the following: every subring of $R$ contains $1_R$, but the only ideal of $R$ that contains $1_R$ is $R$ itself.
(1) Let $P$ be the prime subring of $R$. The prime subring of $R$ is the unique minimal subring of $R$. What this means is (A) that $P$ is a subring of $R$, (B) that $P\subseteq S$ for every subring $S$ of $R$, and (C) that $P$ is the only subring of $R$ that satisfies properties (A) and (B). To prove these, use the fact that $P=\phi(\Bbb{Z})$ where the $\phi:\Bbb{Z}\to R$ is the ring homomorphism which sends $n\mapsto n\cdot1_R$ for all $n\in\Bbb{Z}$.
(A) In general, if $\psi:R_1\to R_2$ is a ring homomorphism, then $\psi(R_1)$ is a subring of $R_2$. Since $P$ is defined to be $\phi(\Bbb{Z})$, where $\phi:\Bbb{Z}\to R$ is the ring homomorphism which sends $n\mapsto n\cdot1_R$ for all $n\in\Bbb{Z}$, it follows that $P$ is a subring of $R$.
(B) Let $S$ be a subring of $R$. Then $1_R\in S$. Since $S$ contains additive inverses, we also have that $(-1)\cdot1_R=-1_R\in S$. Also, since $S$ is closed under addition, it can be shown that $n\cdot1_R\in S$ for all $n\in\Bbb{Z}$. Hence $P=\phi(\Bbb{Z})\subseteq S$.
(C) Suppose $P_1$, $P_2$ are subrings of $R$ such that $P_1\subseteq S$ for every subring $S$ of $R$ and $P_2\subseteq S$ for every subring $S$ of $R$. It follows that $P_1\subseteq P_2$ and $P_2\subseteq P_1$. Hence $P_1=P_2$.
