From 69ccf7443392851eead00961fa6da3ec4f89b2ab Mon Sep 17 00:00:00 2001 From: Indigo5684 <159226326+Indigo5684@users.noreply.github.com> Date: Tue, 30 Sep 2025 13:19:29 -0500 Subject: [PATCH] TODOs --- docs/math/abstract-algebra/DF-13-fields.md | 4 +++- docs/physics/electrostatics/6 -polarization-magnetization.md | 3 +++ docs/physics/electrostatics/7-time-dependent-em-fields.md | 3 +++ 3 files changed, 9 insertions(+), 1 deletion(-) create mode 100644 docs/physics/electrostatics/6 -polarization-magnetization.md create mode 100644 docs/physics/electrostatics/7-time-dependent-em-fields.md diff --git a/docs/math/abstract-algebra/DF-13-fields.md b/docs/math/abstract-algebra/DF-13-fields.md index f9d8443..a9150ff 100644 --- a/docs/math/abstract-algebra/DF-13-fields.md +++ b/docs/math/abstract-algebra/DF-13-fields.md @@ -16,4 +16,6 @@ **Definition**. If $K$ is generated by $F(\alpha)$, then $K$ is a *simple extension* of $F$. -**Theorem**. Let $F$ be a field, $p(x) \in F[x]$ be irreducible. Then, if $\alpha$ is a root of $p(x)$ and $K$ is an extension of $F$ containing $\alpha$, then $F(\alpha) \cong \frac{F[x]}{(p(x))}$. \ No newline at end of file +**Theorem**. Let $F$ be a field, $p(x) \in F[x]$ be irreducible. Then, if $\alpha$ is a root of $p(x)$ and $K$ is an extension of $F$ containing $\alpha$, then $F(\alpha) \cong \frac{F[x]}{(p(x))}$. + +TODO diff --git a/docs/physics/electrostatics/6 -polarization-magnetization.md b/docs/physics/electrostatics/6 -polarization-magnetization.md new file mode 100644 index 0000000..3e8ceb8 --- /dev/null +++ b/docs/physics/electrostatics/6 -polarization-magnetization.md @@ -0,0 +1,3 @@ +# Chapter 6 - Polarization and Magnetization + +TODO diff --git a/docs/physics/electrostatics/7-time-dependent-em-fields.md b/docs/physics/electrostatics/7-time-dependent-em-fields.md new file mode 100644 index 0000000..63455cc --- /dev/null +++ b/docs/physics/electrostatics/7-time-dependent-em-fields.md @@ -0,0 +1,3 @@ +# Chapter 7 - Time Dependent Electric and Magnetic Fields + +TODO