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Two Useful TheoremsIn this section we prove two extremely useful results concerning rings, both of which make use of the notion of embedding. Definition 1: Let R and S be rings. We say that R is embedded in S if there exists an isomorphism If R is embedded in S via the embedding f, then In the concept of rings section, we saw that 2Z is a ring without an identity element. Note, however, that 2Z can be embedded in Z, which is a ring with identity. Our first result shows that this is quite a general phenomenon. Theorem 2: Let R be a ring. Then R can be embedded in a ring with identity. Before proceeding with the proof of Theorem 2, let us define a general notion which will come up rather frequently in our studies. Let R be a ring and let
if n > 0
= 0 if n = 0
if n < 0
Note that
n · (r_{1} + r_{2}) = n · r_{1} + r_{2},
(2)
n(r) = (n · r),
(3)
(n_{1} + n_{2}) · r = n_{1} · r + n_{2} · r,
(4)
(n_{1} · n_{2})· r = n_{1} · (n_{2} · r),
(5)
1 · r = r.
It should be noted however that in general the product ab for a and b in R and the scalar product nr are two different concepts entirely. The concept of scalar multiplication is derived from a shorthand notation fro certain statements about addition. Proof of Theorem 2: Let us consider the Cartesian product
Let us define addition and multiplication on
(r_{1},n_{1}) + (r_{2},n_{2}) = (r_{1} + r_{2}, n_{1} + n_{2}),
(r_{1},n_{1})·(r_{2},n_{2}) = (r_{1}·r_{2} + n_{1}·r_{2} + n_{2}·r_{1}, n_{1}·n_{2}).
Then, with respect to these operations,
(r_{1},n)·(0,1) = (r_{1}· 0 + n · 0 + 1 · r_{1}, 1 · n) = (r_{1},n),
(0,1)·(r_{1},n) = (0 · r_{1} + 1 · r_{1} + n · 0, 1 · n) = (r_{1},n).
Define the mapping
f :RR Z,
f(r) = (r,0).
Then
f(r_{1} + r_{2}) = (r_{1} + r_{2}, 0) = f(r_{1}) + f(r_{2}),
f(r_{1}·r_{2}) = (r_{1}·r_{2},0) = f(r_{1})·f(r_{2}).
Therefore, f is a ring homomorphism. Moreover, Let us now pose the following problem: Can an integral domain D always be embedded in a field? In order to get a good feel for the problem as well as its solution, let us consider the special case
a/b = (a/1)(b/1)^{1}.
Our goal in the second theorem is to generalize these facts to the case of an arbitrary integral domain in place of Z. Our main result is the following theorem.
Theorem 3: Let R be an integral domain. There exists a field F_{R} and an embedding Proof: Let us mimic the situation in the case of the rational numbers. Let
S = {(a,b) R R  b0},
and let ~ denote the relation defined on S defined by
(a,b) ~ (c,d) ad  bc = 0.
Then ~ is an equivalence relation on S. Let F_{R} denote the set of equivalence classes of S with respect to ~, and let us denote the equivalence class containing
a/b + c/d = (ad + bc)/bd,
(a/b) · (c/d) = (a · c)/(b · d).
Note that Let us prove that these operations are consistent. By way of example, let us supply the details only in the case of addition. Suppose that
a/b + c/d = a'/b' + c'/d'.
Indeed,
(ad + bc)b'd' = (a'd' + b'c')bd.
But, The mapping
f :RF_{R},
f(r) = r/1
is an embedding. Moreover, if
a/b = (a/1)·(1/b) = (a/1)·(b/1)^{1}
= f(a) · f(b)^{1}.
Thus, F_{R} satisfies the conditions of the theorem. Usually, we will dispense with the embedding f and will merely identify Example 1: If Example 2:
F_{R} = {f/g  f, g F[X], g0}.
Two elements, f/g and f'/g', of F_{R} are equal if and only if
fg' = f'g.
The two elements of F_{R} are called rational functions in X. 
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