# Proof

#### Uniqueness

If $$(F_1, + , \cdot)$$ and $$(F_2,\ast, \circ)$$ where two different fields fulfilling both properties (1) and (2), then because of the property (2), $$(F_1,+, \cdot)$$ would be a subfield of $$(F_2,\ast, \circ)$$ and vice versa. Therefore, both fields must be the same.

#### Existence

To improve the readability, in our proof we write $$ab$$ instead of $$a\cdot b$$. Our proof will be comprised of 4 steps:

### Step 1: Construct $$(F, \ast , \circ)$$ as a non-empty set with two binary operations: addition "$$\ast$$" and multiplication "$$\circ$$"

By assumption, $$(R, +, \cdot)$$ is an integral domain. As a first step, we define an equivalence relation on the cartesian product $$R\times (R\setminus\{0\})=\{(a,b)~|~a\in R,b\in R\setminus\{0\}\}$$ as follows:

$$(a,b)\sim (c,d)\Leftrightarrow ad=bc,~~~~~~(a,c\in R;~b,d\in R\setminus\{0\}).$$

Please note that this is indeed an equivalence relation, since it is reflexive, symmetric and transitive: * The reflexivity of "$$\sim$$" follows from the commutativity of multiplication in the integral domain $$(R, + , \cdot)$$: $$(a,b)\sim(a,b)\Leftrightarrow ab=ba,~~~~~~(a\in R;~b\in R\setminus\{0\}).$$ * Also the symmetry of "$$\sim$$" follows from the commutativity of multiplication in the integral domain $$(R, + , \cdot)$$: $$\begin{array}{ccl} (a,b)\sim (c,d) &\Leftrightarrow& ad=bc\\ &\Leftrightarrow &cb=da\\ &\Leftrightarrow& (c,d)\sim (a,b),~~~~~~(a,c\in R;~b,d\in R\setminus\{0\}). \end{array}$$ * The transitivity of "$$\sim$$" follows from $$R$$ being an integral domain: * Assume for $$a,c,e\in R;~b,d,f\in R\setminus\{0\}$$: \begin{align}\label{eq:889a}\begin{array}{ccc} (a,b)\sim (c,d)&\Leftrightarrow&ad=bc\ (c,d)\sim (e,f)&\Leftrightarrow&cf=de.\ \end{array}\end{align} * After multiplying both sides of both equations, we can first rearrange the variables, because the multiplication is commutative in any integral domain, and get \begin{align}\label{eq:889b}\begin{array}{ccc} (ad)(cf)&=&(bc)(de)\ adcf&=&bcde\ (af)(dc)&=&(be)(dc)\ \end{array}\end{align} * Now, if $$c\neq 0$$, then $$d\in R\setminus\{0\}$$ means that also $$d\neq 0$$. Therefore we have $$dc\neq 0$$, because in $$R$$ the element $$0$$ is the only zero divisor. This means that we can cancel $$(dc)$$ in the equation \eqref{eq:889b}, resulting in $$\begin{array}{ccc} (af)\cancel{(dc)}&=&(be)\cancel{(dc)}\\ &(a,b)\sim(e,f).& \end{array}$$ * On the other hand, if $$c=0$$, then it follows from \eqref{eq:889a} and from $$b,d,f\in R\setminus\{0\}$$, and from the fact that $$F$$ has no zero divisors except of $$0$$ that we must have $$a=e=0$$, which means that $$(a,b)\sim(c,d)$$ degrades to $$(0,b)\sim(0,d)\Leftrightarrow 0\cdot d=b\cdot 0\Leftrightarrow 0=0$$, and analogously $$(c,d)\sim(e,f)$$ degrades to $$(0,d)\sim(0,f)\Leftrightarrow 0\cdot f=d\cdot 0\Leftrightarrow 0=0$$, which again means $$(a,b)\sim(e,f)$$. * Therefore, "$$\sim$$" is transitive.

We have just shown that the relation "$$\sim$$" constitutes an equivalence relation. Therefore, each ordered pair $$(a,b)$$ represents its own equivalence class $$\lbrack a,b\rbrack \subseteq R\times (R\setminus\{0\})$$ and we can set $$F:=(R\times (R\setminus\{0\}))/\sim$$ as the set of all these equivalence classes. However, because the relation $$\sim$$ is symmetric and transitive, each equivalence class $$\lbrack a,b\rbrack$$ does not(!) depend on the special choice of its representative $$(a,b)$$. Therefore, we can define two new binary operations $$\ast$$, $$\circ$$ on $$F$$ using any representatives, e.g. the representatives $$(a,b)$$ and $$(c,d)$$ as we do in the following: $$\begin{array}{ccl} \lbrack a,b\rbrack \ast\lbrack c,d\rbrack &:=&\lbrack ad + cb,~bd\rbrack ,\\ \lbrack a,b\rbrack \circ\lbrack c,d\rbrack &:=&\lbrack ac,~bd\rbrack .\\ \end{array}$$

Please note that $$F$$ does contain equivalence classes suitable to define the above binary operations $$\ast$$, $$\circ$$. Because $$(R, + ,\cdot)$$ is an integral domain, it contains the element $$0$$, but also at least one other element $$x\in R$$ with $$x\neq 0$$. Therefore, $$F$$ is also not empty (i.e. it contains at least the two equivalence classes $$\lbrack 0,x\rbrack$$ and $$\lbrack x,x\rbrack$$ represented by $$(0,x)$$ and $$(x,x)$$). Thus, we have succeeded to define a non-empty set with two binary operations $$(F,\ast,\circ)$$.

### Step 2: Demonstrate that $$(F,\ast,\circ)$$ is a field

We will show the defining properties of a field, namely: 1. $$(F,\ast)$$ is a commutative group with $$\lbrack 0,x\rbrack$$ as neutral element (of addition "$$\ast$$") 1. $$F^*:=(F \setminus \{\lbrack 0,x\rbrack \},\circ)$$ is a commutative group with $$\lbrack x,x\rbrack$$ as neutral element (identity of multiplication "$$\circ$$") 1. The distributivity law holds for all $$\lbrack a,b\rbrack ,\lbrack c,d\rbrack ,\lbrack e,f\rbrack \in F$$.

Ad 1): As mentioned in the last note on Step 1, $$\lbrack 0,x\rbrack \in F$$, so $$(F,\ast)$$ is not empty.

a) The operation $$\ast$$ is associative, because for all $$\lbrack a,b\rbrack ,\lbrack c,d\rbrack ,\lbrack e,f\rbrack \in F$$ we have $$\begin{array}{ccl} (\lbrack a,b\rbrack \ast\lbrack c,d\rbrack )\ast \lbrack e,f\rbrack &=&\lbrack ad + cb,bd\rbrack \ast\lbrack e,f\rbrack \\ &=&\lbrack (ad + cb)f + ebd,bdf\rbrack \\ &=&\lbrack adf + cfb + edb,bdf\rbrack \\ &=&\lbrack adf + (cf + ed)b,bdf\rbrack \\ &=&\lbrack a,b\rbrack \ast\lbrack cf + ed,df\rbrack =\lbrack a,b\rbrack \ast(\lbrack c,d\rbrack \ast\lbrack e,f\rbrack ).\\ \end{array}$$

b) The operation $$\ast$$ is commutative, because for all $$\lbrack a,b\rbrack ,\lbrack c,d\rbrack \in F$$ we have $$\begin{array}{ccl} \lbrack a,b\rbrack \ast\lbrack c,d\rbrack &=&\lbrack ad + cb,bd\rbrack \\ &=&\lbrack cb + ad,db\rbrack =\lbrack c,d\rbrack \ast\lbrack a,b\rbrack .\\ \end{array}$$

c) The equivalence class $$\lbrack 0,x\rbrack$$, represented by $$(0,x)$$ with $$0\in R;~x\in R\setminus\{0\}$$ is the neutral element of the operation $$\ast$$. $$\begin{array}{ccl} \lbrack a,b\rbrack \ast\lbrack 0,x\rbrack &=&\lbrack ax + 0\cdot b,bx\rbrack =\lbrack ax,bx\rbrack \\ &\Leftrightarrow&ax=xb\\ &\Leftrightarrow&a\cancel x=\cancel xb~~~~~~~~~\text{ note that we can cancel by }x\neq 0!\\ &\Leftrightarrow&a\cdot 1=1\cdot b\\ &\Leftrightarrow&\lbrack a\cdot 1,b\cdot 1\rbrack =\lbrack a,b\rbrack \\ \end{array}$$

d) Each equivalence class $$\lbrack a,b\rbrack \in(F,\ast)$$ has an (additive) inverse element $$\lbrack - a,b\rbrack \in(F,\ast)$$. $$\lbrack a,b\rbrack \ast\lbrack - a,b\rbrack =\lbrack ab - ba,bb\rbrack =\lbrack 0,bb\rbrack =\lbrack 0,x\rbrack ,$$ because $$(0,bb)\sim(0,x)$$.

Ad 2): As mentioned in the last note on Step 1, $$\lbrack x,x\rbrack \in F \setminus \{\lbrack 0,x\rbrack \}$$, so $$(F \setminus \{\lbrack 0,x\rbrack \},\circ)$$ is not empty.

a) The operation $$\circ$$ is associative, because for all $$\lbrack a,b\rbrack ,\lbrack c,d\rbrack ,\lbrack e,f\rbrack \in F \setminus \{\lbrack 0,x\rbrack \}$$ we have $$\begin{array}{ccl} (\lbrack a,b\rbrack \circ\lbrack c,d\rbrack )\circ \lbrack e,f\rbrack &=&\lbrack ac,bd\rbrack \circ\lbrack e,f\rbrack \\ &=&\lbrack (ac)e,(bd)f\rbrack \\ &=&\lbrack (a(ce),b(df)\rbrack \\ &=&\lbrack a,b\rbrack \circ\lbrack ce,df\rbrack \\ &=&\lbrack a,b\rbrack \circ(\lbrack c,d\rbrack \circ\lbrack e,f\rbrack ).\\ \end{array}$$

b) The operation $$\circ$$ is commutative, because for all $$\lbrack a,b\rbrack ,\lbrack c,d\rbrack \in F \setminus \{\lbrack 0,x\rbrack \}$$ we have $$\begin{array}{ccl} \lbrack a,b\rbrack \circ\lbrack c,d\rbrack &=&\lbrack ac,bd\rbrack \\ &=&\lbrack ca,db\rbrack =\lbrack c,d\rbrack \circ\lbrack a,b\rbrack .\\ \end{array}$$

c) The equivalence class $$\lbrack x,x\rbrack$$, represented by $$(x,x)$$ with $$x\in R\setminus\{0\}$$ is the neutral element of the operation $$\circ$$. $$\begin{array}{ccl} \lbrack a,b\rbrack \circ\lbrack x,x\rbrack &=&\lbrack ax,bx\rbrack \\ &\Leftrightarrow&ax=xb\\ &\Leftrightarrow&a\cancel x=\cancel xb~~~~~~~~~\text{ note that we can cancel by }x\neq 0!\\ &\Leftrightarrow&a\cdot 1=1\cdot b\\ &\Leftrightarrow&\lbrack a\cdot 1,b\cdot 1\rbrack =\lbrack a,b\rbrack \\ \end{array}$$

d) Each equivalence class $$\lbrack a,b\rbrack \in(F \setminus \{\lbrack 0,x\rbrack \},\circ)$$ has a (multiplicative) inverse element $$\lbrack b,a\rbrack \in(F \setminus \{\lbrack 0,x\rbrack \},\circ)$$. $$\lbrack a,b\rbrack \circ\lbrack b,a\rbrack =\lbrack ab,ba\rbrack =\lbrack x,x\rbrack ,$$ because $$(ab,ba)\sim(x,x)$$.

Ad 3): In $$(F,\ast,\circ)$$, the distributivity law is fulfilled, because for all $$\lbrack a,b\rbrack ,\lbrack c,d\rbrack ,\lbrack e,f\rbrack \in F$$ we have $$\begin{array}{ccl} \lbrack a,b\rbrack \circ(\lbrack c,d\rbrack \ast\lbrack e,f\rbrack )&=&\lbrack a,b\rbrack \circ\lbrack cf + de,df\rbrack \\ &=&\lbrack a(cf + de),bdf\rbrack \\ &=&\lbrack acf + ade),bdf\rbrack \\ &=&\lbrack (acf + ade)b,(bdf)b\rbrack ~~~~~~~~~\text{ note that we can multiply by }b\neq 0!\\ &=&\lbrack (ac)(bf) + (ae)(bd),(bd)(bf)\rbrack \\ &=&\lbrack ac,bd\rbrack \ast\lbrack ae,bf\rbrack \\ &=&(\lbrack a,b\rbrack \circ\lbrack c,d\rbrack )\ast(\lbrack a,b\rbrack \circ\lbrack e,f\rbrack ). \end{array}$$

### Step 3: Demonstrate that $$(F,\ast,\circ)$$ fulfills property (1)

We shall show that there is a subset $$S\subseteq F$$, which is a ring isomorphic to $$R$$, i.e. where $$(S, \ast, \circ)\simeq (R, +, \cdot)$$. We will do so by constructing the subset $$S\subseteq F$$ and finding a bijective homomorphism $$f:R\mapsto S$$. * For the multiplicative identity element $$\lbrack x,x\rbrack \in F$$, consider the subset $$S:=\{\lbrack ax,x\rbrack ~|~a\in R\},$$ and set $$f:\begin{cases}R&\mapsto S\\a&\mapsto \lbrack ax,x\rbrack .\end{cases}$$ * Then $$f$$ surjective by definition and it is injective since we can conclude that $$\begin{array}{ccl} f(a)=f(b)&\Leftrightarrow&\lbrack ax,x\rbrack =\lbrack bx,x\rbrack \\ &\Rightarrow&(ax,x)\sim(bx,x)\\ &\Rightarrow&(ax)x=x(bx)\\ &\Rightarrow&a\cancel{x^2}=b\cancel{x^2}~~~~~~~~~\text{ note that we can cancel by }x\neq 0!\\ &\Rightarrow&a=b\\ \end{array}.$$ Since $$f$$ is both, surjective and injective, it is bijective. * $$f$$ is also a ring homomorphism, since we have $$\begin{array}{ccl} f(a+b)&=&\lbrack (a + b)x,x\rbrack \\ &=&\lbrack (ax + bx),x\rbrack \\ &=&\lbrack (ax)x + (bx)x,xx\rbrack ~~~~~~~~~\text{ note that we can multiply by }x\neq 0!\\ &=&\lbrack ax,x\rbrack \ast\lbrack bx,x\rbrack \\ &=&f(a)\ast f(b). \end{array}$$ On the other hand, we have also $$\begin{array}{ccl} f(ab)&=&\lbrack (ab)x,x\rbrack \\ &=&\lbrack abx,x\rbrack \\ &=&\lbrack abxx,xx\rbrack ~~~~~~~~~\text{ note that we can multiply by }x\neq 0!\\ &=&\lbrack ax,x\rbrack \circ\lbrack bx,x\rbrack \\ &=&f(a)\circ f(b). \end{array}$$ Together, we have shown the isomorphism $$(S, \ast, \circ)\simeq (R, +, \cdot)$$.

### Step 4: Demonstrate that $$(F,\ast,\circ)$$ fulfills property (2)

We shall show that $$(F,\ast,\circ)$$ is minimal with the property (1). By construction of $$S\subseteq F$$, $$F$$ contains all elements of the form $$\lbrack ax,x\rbrack$$ for all $$x\in R \setminus \{0\}$$. Because $$F$$ is a field, it must also contain the inverse elements $$\lbrack x,ax\rbrack$$ for all $$a\in R \setminus \{0\}$$. Because $$(F,\ast,\circ)$$ is closed under the operation $$\circ$$, we also have that

$$\lbrack ax,x\rbrack \circ\lbrack x,bx\rbrack =\lbrack a \cancel{xx},b\cancel{xx}\rbrack =\lbrack a,b\rbrack ,~~~~~~(a\in R,~b\in R \setminus \{0\}).$$ This demonstrates that we cannot reduce $$F$$ further more, e.g. by taking away any single equivalence class $$\lbrack a,b\rbrack$$. Therefore $$F$$ is minimal with the property (1).

Github: ### References

#### Bibliography

1. Kramer Jürg, von Pippich, Anna-Maria: "Von den natürlichen Zahlen zu den Quaternionen", Springer-Spektrum, 2013