MATH 415 Lecture 18

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Embedded Quotient Fields

Continuing from previous lecture, after proving that $F$ is a field, we want to show that integral domain $D$ is isomorphic to the field $F$.

Construct $i:D\to F$ such that $i(a)=\left[(a,1)\right]$. We claim $i$ is an isomorphism between $D$ and $i(D)$. Indeed, $i$ is

1. homomorphism
   • $i(a)+i(b)=[(a+b,1)]=i(a+b)$
   • $i(a)\cdot i(b)=[(a\cdot b,1)]=i(a\cdot b)$
2. one-to-one (if $i(a)=i(b)$, then $(a,1)\sim (b,1)$ implies $a=b$)
3. onto $i(D)$ (tautology by definition of image; trivial)

Thus we justify the following theorem:

Theorem 21.5

Any integral domain $D$ can be enlarged to (or embedded in) a field $F$ such that every element of $F$ can be expressed as a quotient of two elemens of $D$.

Such $F$ is called a field of quotients of $D$.

Note: This theorem explains existence. There is another theorem in the book that explains uniqueness.

Rings of Polynomials

We begin with a ring $R$.

Let $x$ be an indeterminate or unknown variable.

We denote the ring of polynomials in one indeterminate with coefficients in $R$ as $R[x]$

In general, a polynomial is an infinite formal sum of form:

$p(x)=\sum _{i=0}^{\mathrm{\infty }}{a}_i{x}^i=a_0+a_{1}x+\dots +a_n{x}^n+\dots$

Where $a_i\in R$ and $a_i=0$ for all but a finite number of values of $i$.

If $a_n\ne 0$, then we call $n$ the degree of $p(x)$.

• A polynomial of the form $p(x)=a_0$ (degree 0) is called a constant polynomial
• $p(x)=a_0+a_{1}x$ (degree 1) is called linear
• $p(x)=a_0+a_{1}x+a_2{x}^2$ (degree 2) is called quadratic
• etc.

We define addition and multiplication as follows:

Let $f,g\in R[x]$ with

$\begin{array}{rl}f(x)& =\sum _{i=0}^{\mathrm{\infty }}{a}_i{x}^i\\ g(x)& =\sum _{i=0}^{\mathrm{\infty }}{b}_i{x}^i\end{array}$

Then

$\begin{array}{rlr}(f+g)(x)& =\sum _{i=0}^{\mathrm{\infty }}(a_i+b_i)x^i(f\cdot g)(x)& =\sum _{i=0}^{\mathrm{\infty }}(\sum _{j=0}^i{a}_j\cdot b_{i-j})x^i\end{array}$

Note that $ab\ne ba$ in general.

But if $R$ is commutative, then $R[x]$ is also commutative.

Similarly, $R[x]$ has unity if and only if $R$ has unity, and unity in $R[x]$ will be constant polynomial $p(x)=1$.

$R$ is isomorphic to subring $R_1\subset R[x]$, where $R_1$ is the set of constant polynomials.

Example

${\mathbb{\mathbb{Z}}}_2$ is a (finite) field, but ${\mathbb{\mathbb{Z}}}_2[x]$ is an infinite field.

$(x+1{)}^2=x^2+2x+1$, but $2x\equiv 0(\mathrm{mod}2)$, so $(x+1{)}^2=x^2+1$ in ${\mathbb{\mathbb{Z}}}_2[x]$.

Multiple Indetermintes

We write $R[x,y]$ to represent all polynomials in two indeterminates with coefficients in $R$ and are expressed as follows:

$p(x,y)=\sum _{i=0}^{\mathrm{\infty }}\sum _{j=0}^{\mathrm{\infty }}{a}_{ij}{x}^i{y}^j$

In this case, $xy=yx$.

Evaluation Homomorphisms

Let $E$ and $F$ be fields with $F\le E$ ($F$ is a subfield of $E$).

Theorem 22.4

If $F\le E$, $\alpha \in E$, then the map ${\varphi }_{\alpha }:F[x]\to E$ defined by

${\varphi }_{\alpha }(a_0+a_{1}x+\dots +a_n{x}^n)=a_0+a_1\alpha +\dots +a_n{\alpha }^n$

for $a_0+a_{1}x+\dots +a_n{x}^n\in F[x]$ is a homomorphism of $F[x]$ into $E$. Furthermore, ${\varphi }_{\alpha }(x)=\alpha$, and ${\varphi }_{\alpha }$ maps $F$ isomorphically by the identity map; that is, ${\varphi }_{\alpha }(a)=a$ for $a\in F$.

The homomorphism ${\varphi }_{\alpha }$ is an evaulation at $\alpha$.

To take a page from Haskell, an evaluation homomorphism has the following type signature:

eval :: (E a, E b) => (a -> b) -> a -> b

E {$E }  ; phialpha [ below of = E] {$\phi_\alpha(x) = \alpha \in \phi_\alpha(F[x])$}  ; Fx [ left of = phialpha ] {$p \in F[x]$}  ; F [ below of = Fx ] {$a \in F$}  ; phia [ below of = phialpha] {$\phi_\alpha(a) = a \in F$}  ;

\draw [->] (Fx) (phialpha) {$\phi_\alpha$}  ; \draw [--] (Fx) (F)  ; \draw [->] (F) (phia) {identity map}  ; \draw [--] (E) (phialpha)  ; \draw [--] (phialpha) (phia)  ;