ABasic algebra

A.1 Commutative rings

Let $R$ be a commutative ring. Recall that $u\in R$ is a unit if there exists $v\in R$ , such that $u v = 1$ . The subset of units in $R$ is an abelian group with the multiplication from $R$ . It is denoted $R^*$ . We will mostly be dealing with integral domains i.e., commutative rings with no zero divisors.

We will assume that the definition and construction of $R[x]$ , the commutative ring of polynomials with coefficients in $R$ , is known. Also, recall that a non-zero polynomial $f\in R[x]$ can have at most $\deg(f)$ roots if $R$ is an integral domain.

A.1.1 Quotients

One exceedingly important construction is the quotient ring. If $I$ is an ideal in $R$ , then

$x \equiv y \pmod I \iff x - y \in I$ is an equivalence relation. The equivalence class containing $x\in R$ is simply $x + I$ (why?). We denote this $[x]$ . The set of equivalence classes is denoted $R/I$ and becomes a commutative ring via

$\begin{aligned} [a] + [b] &= [a + b]\\ [a] [b] &= [a b]. \end{aligned}$ Even though this is so simple to write up, it can sometimes be difficult to grasp in real life. It is a very powerful definition.

A.1.2 Homomorphisms and ideals

Let $S$ be another commutative ring and $f: R\rightarrow S$ a ring homomorphism. If $I\subset R$ is an ideal, then $f(I)$ is not necessarily an ideal in $S$ . We let $I S$ denote the ideal generated by $f(I)$ in this context.

Show that $f(I)$ is an ideal in $S$ if $f$ is surjective. Give an example of a ring homomorphism $f: R\rightarrow S$ and an ideal $I\subset R$ , such that $f(I)$ is not an ideal in $S$ .

For a surjective ring homomorphism $f: R\rightarrow S$ with kernel $K$ , there is a 1-1 inclusion preserving correspondence between ideals in $S$ and ideals in $R$ containing $K$ . This correspondence is given by

$J\mapsto f(J) = \{f(x) \mid x\in J\}, \tag{A.1}$ where $J\supseteq K$ is an ideal containing $K$ .

Show that the inverse map to (A.1) is

$I\mapsto f^{-1}(I) = \{x\in R\mid f(x)\in I\},$ where $I$ is an ideal in $S$ (as $f$ is surjective, $f(f^{-1}(I)) = I$ holds, but what about $f^{-1}(f(J)) = J$ ?).

A.1.3 Fields

A commutative ring $k$ is called a field if $k^* = k\setminus \{0\}$ . An integral domain $R$ is contained in its field $K$ of fractions, which is given by

$K = \left\{\frac{f}{g}\, \middle| \, f\in R, g\in R\setminus\{0\}\right\},$ where

$\frac{f}{g}\sim \frac{r}{s} \iff f s = g r.$ The above is an equivalence relation and behaves well with the usual addition and multiplication of fractions. We view $R$ as a subring of $K$ by mapping $r$ to $r/1$ .

The fraction field of the polynomial ring $k[x]$ is denoted $k(x)$ . These are the classical rational functions i.e., quotients of polynomials.

A.1.4 Algebraically closed fields

A field $k$ is called algebraically closed if every non-constant polynomial $f\in k[x]$ has a root i.e., there exists $\alpha\in k$ with $f(\alpha) = 0$ .

Prove that an algebraically closed field must be infinite.

The most famous example of an algebraically closed field is the complex numbers $\mathbb{C}$ . It is, however, not easy to prove that $\mathbb{C}$ is algebraically closed. Do you recall a proof?

Every field $k$ is contained in a unique (up to isomorphism fixing $k$ ) algebraically closed field $\overline{k}$ , such that the field extension $k\subset \overline{k}$ is algebraic (what do I mean by algebraic here?).

$\overline{\mathbb{R}} = \mathbb{C}.$

How would you describe $\overline{\mathbb{F}}_p$ , where $\mathbb{F}_p$ is the field with $p$ (prime) elements?

A.1.5 Prime ideals

An ideal $P\subset R$ is called a prime ideal if

$x y\in P\implies x\in P\quad\text{or}\quad y\in P$ for every $x, y\in R$ . This happens if and only if $R/P$ is an integral domain.

A.1.6 Maximal ideals

A maximal ideal is an ideal $M$ in $R$ , which is maximal among proper ideals i.e., if $I$ is a proper ideal with $M\subseteq I$ , then $M=I$ . An ideal $M$ is a maximal ideal if and only if $R/M$ is a field.

Prove that $M$ is a maximal ideal if and only if $M + R x = R$ for every $x\not\in M$ .

One can prove that every commutative ring contains a maximal ideal (Zorn's lemma).

A.1.7 The correspondence respects max and prime

The correspondence in (A.1) respects maximal and prime ideals:

$M\, \text{is a maximal (prime) ideal in}\, R\, \text{if and only if}\, f(M)\, \text{is a maximal (prime) ideal in}\, S$

A.2 Local rings

A commutative ring $R$ is called a local ring if it contains precisely one maximal ideal.

Show that a commutative ring $R$ is local if and only if $R\setminus R^*$ is an ideal (which then is the unique maximal ideal in $R$ ).

Let $p$ be a prime number.

Show that

$\mathbb{Z}_{(p)} = \left\{\frac{a}{b}\, \middle| \, a, b\in \mathbb{Z}, p\nmid b\right\}$ is a subring of $\mathbb{Q}$ containg $\mathbb{Z}$ as a subring:

$\mathbb{Z}\subset \mathbb{Z}_{(p)} \subset \mathbb{Q}.$ Prove that $\mathbb{Z}_{(p)}$ is a local ring with unique maximal ideal $(p) \mathbb{Z}_{(p)}$ . Is $\mathbb{Z}_{(p)}$ a field?