Kyungtak Hong

Algebraically closed fields are infinite

Wed, 24 Aug 2022 11:00:00 +0000

The proof is analogous to the Euclid’s proof of the infinitude of primes.

Theorem Algebraically closed fields are infinite.

proof. Let $F=\{a_{0}, \ldots, a_{n}\}$ be a finite field. Then the polynomial $f(x) = 1+\prod_{i=0}^{n} (x-a_{i})$ does not have a root in $F$. Thus $F$ is not algebraically closed. $\blacksquare$

Essential extensions of integral domains

Sun, 21 Aug 2022 11:00:00 +0000

The notion of “maximal” essential extension of a module is in general defined only when it is contained in another module. However, such containment is not necessary for the case of integral domains since their quotient fields are in fact the final objects in the category of essential extensions.

Lemma Let $D$ be an integral domain and $F$ be its quotient field. Then $F$ is an injective $D$-module.

proof. Using Baer’s criterion, it is enough to show that any $D$-module homomorphism $f : I \to F$ from an ideal $I$ of $D$ has an extension $\bar{f} : D \to F$.

The case $I=0$ is clear, so assume that $I \neq 0$ and let $r \in I$ be a nonzero element. Since $F$ is a divisible $D$-module, there exists $x \in F$ such that $f(r) = rx$. Now define $\bar{f} : D \to F$ to be $\bar{f}(a) = ax$. For any $a \in I$, the equation

\[\begin{equation}\label{eq:1} r\bar{f}(a) \;=\; rax \;=\; af(r) \;=\; f(ar) \;=\; rf(a) \end{equation}\]

holds. Since $F$ is a torsion-free $D$-module, ($\ref{eq:1}$) implies that $\bar{f}(a) = f(a)$ and therefore $\bar{f}$ is an extension of $f$. $\blacksquare$

Remark The statement in general holds for the case when $F$ is a divisible torsion-free $D$-module.

Theorem Every essential extension of an integral domain can be embedded into its quotient field.

proof. Let $D$ be an integral domain, $F$ be its quotient field and $E$ be an essential extension of $D$.

Since $F$ is injective, the inclusion $D \to F$ has an extension $\phi : E \to F$. The equation

\[\begin{equation}\label{eq:2} \mathrm{Ker}(\phi) \cap D \;=\; \mathrm{Ker}(D \to F) \;=\; 0 \end{equation}\]

together with $E$ being an essential extension of $D$ implies that $\mathrm{Ker}(\phi) = 0$, and therefore $\phi$ is injective. $\blacksquare$

The equivalence of norms in finite dimensional vector spaces

Thu, 04 Aug 2022 11:00:00 +0000

Theorem Let $V$ be a finite dimensional $\mathbb{R}$-vector space. Then any two norms on $V$ induce the same topology.

proof. Without loss of generality, let $V = \mathbb{R}^{d}$. It is enough to show that any norm $\lVert\cdot\rVert$ on $V$ is equivalent to the $2$-norm $\lVert\cdot\rVert_{2}$.

Let $\{ \mathbf{e}_{1}, \ldots, \mathbf{e}_{d} \}$ be the standard basis of $V$ and $M = \max_{1 \leq i \leq d} \lVert\mathbf{e}_{i}\rVert$. The inequality

\[\begin{equation}\label{eq:1} \lVert\mathbf{x}\rVert \;\leq\; \sum_{i=1}^{d} \lVert x_{i}\mathbf{e}_{i}\rVert \;\leq\; M \sum_{i=1}^{d} \lvert x_{i}\rvert \;\leq\; Md \lVert\mathbf{x}\rVert_{2} \end{equation}\]

where $\mathbf{x} = \sum_{i=1}^{d} x_{i}\mathbf{e}_{i} \in V$ shows that the identity map $(V,\lVert\cdot\rVert_{2}) \to (V,\lVert\cdot\rVert)$ is continuous.

For the inverse direction, let $S=\{\mathbf{x} \in V \;|\; \lVert x \rVert = 1\}$ be the unit sphere in $(V,\lVert\cdot\rVert_{2})$ and consider the function $f : \mathbf{x} \mapsto \lVert \mathbf{x} \rVert$ on $S$. Since $f$ is continuous by ($\ref{eq:1}$) and $S$ is a compact set disjoint from $\mathbf{0} \in V$, $f$ obtains the minimal value $c > 0$. Thus, for any $\mathbf{x} \neq \mathbf{0}$,

\[\begin{equation} \frac{\lVert\mathbf{x}\rVert}{\lVert\mathbf{x}\rVert_{2}} = f\left(\frac{\mathbf{x}}{\lVert\mathbf{x}\rVert_{2}}\right) \geq c \end{equation}\]

holds and therefore the identity map $(V,\lVert\cdot\rVert) \to (V,\lVert\cdot\rVert_{2})$ is also continuous. $\blacksquare$