# Math4302 Modern Algebra (Lecture 28)

## Rings

### Field of quotients

Let $R$ be an integral domain ($R$ has unity and commutative with no zero divisors).

Consider the pair $S=\{(a,b)|a,b\in R, b\neq 0\}$.

And define the equivalence relation on $S$ as follows:

$(a,b)\sim (c,d)$ if and only if $ad=bc$.

We denote $[(a,b)]$ as set of all elements in $S$ equivalent to $(a,b)$.

Let $F$ be the set of all equivalent classes. We define addition and multiplication on $F$ as follows:

$$
[(a,b)]+[(c,d)]=[(ad+bc,bd)]
$$

$$
[(a,b)]\cdot[(c,d)]=[(ac,bd)]
$$

<details>
<summary>The multiplication and addition is well defined </summary>

Addition:

If $(a,b)\sim (a',b')$, and $(c,d)\sim (c',d')$, then we want to show that $(ad+bc,bd)\sim (a'd+c'd,b'd)$.

Since $(a,b)\sim (a',b')$, then $ab'=a'b$; $(c,d)\sim (c',d')$, then $cd'=dc'$,

So $ab'dd'=a'bdd'$, and $cd'bb'=dc'bb'$.

 $adb'd'+bcb'd'=a'd'bd+b'c'bd$, therefore $(ad+bc,bd)\sim (a'd+c'd,b'd)$.

---

Multiplication:

If $(a,b)\sim (a',b')$, and $(c,d)\sim (c',d')$, then we want to show that $(ac,bd)\sim (a'c',b'd')$.

Since $(a,b)\sim (a',b')$, then $ab'=a'b$; $(c,d)\sim (c',d')$, then $cd'=dc'$, so $(ac,bd)\sim (a'c',b'd')$

</details>

#### Claim (F,+,*) is a field

- additive identity: $(0,1)\in F$
- additive inverse: $(a,b)\in F$, then $(-a,b)\in F$ and $(-a,b)+(a,b)=(0,1)\in F$
- additive associativity: bit long.

- multiplicative identity: $(1,1)\in F$
- multiplicative inverse: $[(a,b)]$ is non zero if and only if $a\neq 0$, then $a^{-1}=[(b,a)]\in F$.
- multiplicative associativity: bit long

- distributivity: skip, too long.

Such field is called a quotient field of $R$.

And $F$ contains $R$ by $\phi:R\to F$, $\phi(a)=[(a,1)]$.

This is a ring homomorphism.

- $\phi(a+b)=[(a+b,1)]=[(a,1)][(b,1)]\phi(a)+\phi(b)$
- $\phi(ab)=[(ab,1)]=[(a,1)][(b,1)]\phi(a)\phi(b)$

and $\phi$ is injective.

If $\phi(a)=\phi(b)$, then $a=b$.

<details>
<summary>Example</summary>

Let $D\subset \mathbb R$ and

$$
\mathbb Z \subset D\coloneqq \{a+b\sqrt{2}:a,b\in \mathbb Z\}
$$

Then $D$ is a subring of $\mathbb R$, and integral domain, with usual addition and multiplication.

$$
(a+b\sqrt{2})(c+d\sqrt{2})=(ac+2bd)+(ad+bc)\sqrt{2}
$$

$$
-(a+b\sqrt{2})=(-a)+(-b)\sqrt{2})
$$

...

$D$ is a integral domain since $\mathbb R$ has no zero divisors, therefore $D$ has no zero divisors.

Consider the field of quotients of $D$. $[(a+b\sqrt{2},c+d\sqrt{2})]$. This is isomorphic to $\mathbb Q(\sqrt2)=\{r+s\sqrt{2}:r,s\in \mathbb Q\}$

$$
m+n\sqrt{2}=\frac{m}{n}+\frac{m'}{n'}\sqrt{2}\mapsto [(mn'+nm'\sqrt{2},nn')]
$$

And use rationalization on the forward direction.

</details>

#### Polynomial rings

Let $R$ be a ring, a polynomial with coefficients in $R$ is a sum

$$
a_0+a_1x+\cdots+a_nx^n
$$

where $a_i\in R$. $x$ is indeterminate, $a_0,a_1,\cdots,a_n$ are called coefficients. $a_0$ is the constant term.

If $f$ is a non-zero polynomial, then the degree of $f$ is defined as the largest $n$ such that $a_n\neq 0$.

<details>
<summary>Example</summary>

Let $f=1+2x+0x^2-1x^3+0x^4$, then $deg f=3$

</details>

If $R$ has a unity $1$, then we write $x^m$ instead of $1x^m$.

Let $R[x]$ denote the set of all polynomials with coefficients in $R$.

We define multiplication and addition on $R[x]$.

$f:a_0+a_1x+\cdots+a_nx^n$

$g:b_0+b_1x+\cdots+b_mx^m$

Define,

$$
f+g=a_0+b_0+a_1x+b_1x+\cdots+a_nx^n+b_mx^m
$$

$$
fg=(a_0b_0)+(a_1b_0)x+\cdots+(a_nb_m)x^m
$$

In general, the coefficient of $x^m=\sum_{i=0}^{m}a_ix^{m-i}$.

> [!CAUTION]
>
> The field $R$ may not be commutative, follow the order of computation matters.

We will show that this is a ring and explore additional properties.
