MATH 3231 Class Notes 01-22-2025

noteabstract-algebra

What is Galois Theory?

Problem:

Write $\frac{1}{2+\sqrt{7}}$ as $a+b\sqrt{7}$ for $a,b\in \mathbb{Q}$.

Solution:

$$\frac{1}{2+\sqrt{7}}=\frac{2-\sqrt{7}}{(2-\sqrt{7})(2+\sqrt{7})}=-\frac{2}{3}+\frac{1}{3}\sqrt{7}.$$

Definition:

$\mathbb{Q}[\sqrt{7}]=\{a+b\sqrt{7}|a,b\in \mathbb{Q}\}$

Observations:

1. $\mathbb{Q}\subset \mathbb{Q}[\sqrt{7}]$ (set $b=0$) so $0,1\in \mathbb{Q}[\sqrt{7}]$
2. $\mathbb{Q}[\sqrt{7}]$ is stable under addition (and subtraction).
3. $\mathbb{Q}[\sqrt{7}]$ is stable under multiplication.
4. $a,b\in \mathbb{Q}$ then $a+b\sqrt{7}=0⟺a=b=0$.
5. $\mathbb{Q}[\sqrt{7}]$ is also stable (closed) under division by nonzero elements.

We will say that $\mathbb{Q}[\sqrt{7}]$ is a field.

Proof of 4:

"$⟹$" is easy.

"$⟸$"

$a,b\in \mathbb{Q}$ $a+b\sqrt{7}=0⟹b\sqrt{7}=-a⟹7b^2=a^2$.

$\exists n\in \mathbb{N}$ with $n>0$ and $A=an,B=bn\in \mathbb{Z}$

$7b^2=a^2⟹7b^2{n}^2=a^2{n}^2⟹7(bn{)}^2=(an{)}^2⟹7B^2=A^2$, $A,B\in \mathbb{Z}$.

Set $d=\gcd (A,B)$ (only well defined if $A,B$ are not both 0, which is equivalent to $a,b$ not both $0$), so $A=d\bar{a},B=d\bar{b}$, where $\bar{a},\bar{b}\in \mathbb{Z}$ and $\gcd (\bar{a},\bar{b})=1$.

$7(d\bar{b}{)}^2=d\bar{a}{)}^2⟹7d^2{\bar{b}}^2=d^2{\bar{a}}^2⟹7{\bar{b}}^2={\bar{a}}^2⟹7\mid {\bar{a}}^2⟹7\mid \bar{a}⟹7{\bar{b}}^2=(7\tilde{a}{)}^2⟹\dots$ contradiction

Observation:

$Q[\sqrt{7}]$ is a Vector Space over $\mathbb{Q}$, it has basis $1,\sqrt{7}$, so it has dimension $2$.

Conjugation:

Set $\phi :\mathbb{Q}[\sqrt{7}]\to \mathbb{Q}[\sqrt{7}]$ $\phi (a+b\sqrt{7})=a-b\sqrt{7}$

Properties of $\phi$:

1. $\phi (0)=\phi (0+0\sqrt{7})=0-0\sqrt{7}=0$.
2. $\phi (1)=\phi (1+0\sqrt{7})=1-0\sqrt{7}=1$. More generally, $\phi (a)=a$ for all $a\in \mathbb{Q}$.
3. $\phi$ preserves addition. $\phi ((a+b\sqrt{7})+(c+d\sqrt{7}))=\phi ((a+c)+(b+d)\sqrt{7})=(a+c)-(b+d)\sqrt{7}$ $=(a-b\sqrt{7})+(c-d\sqrt{7}))=\phi (a+b\sqrt{7})+\phi (c+d\sqrt{7})$.
4. $\phi$ preserves multiplication.
5. $\phi$ is a bijection. ${\phi }^{-1}=\phi$

$\phi$ is an example of an automorphism of the field $\mathbb{Q}[\sqrt{7}]$.

${\mathrm{id}}_{\mathbb{Q}[\sqrt{7}]},\phi$ are both automorphisms.

The set of all automorphisms of $\mathbb{Q}[\sqrt{7}]$ is the Galois Group of $\mathbb{Q}[\sqrt{7}]$.

$$\{{\mathrm{id}}_{Q[\sqrt{7}]},\phi \}\subseteq \mathrm{Gal}({}^{\mathbb{Q}[\sqrt{7}]}{/}_{\mathbb{Q}}).$$

This inclusion is in fact an equality.

Proof:

Say $f:\mathbb{Q}[\sqrt{7}]\to \mathbb{Q}[\sqrt{7}]$ is an automorphism. Preserve $\mathbb{Q}$ ($f(a)=a\forall a\in \mathbb{Q}$), preserves addition and multiplication, and is a bijection.

We want that $f={\mathrm{id}}_{\mathbb{Q}[\sqrt{7}]}$ or $f=\phi$.

Observe $f(\sqrt{7})\in \mathbb{Q}[\sqrt{7}]$ uniquely determines $f$. $f(\sqrt{7})=u+v\sqrt{7}$, $u,v\in \mathbb{Q}$.

$f(a+b\sqrt{7})=f(a)+f(b\sqrt{7})=a+f(b)f(\sqrt{7})$ $=a+(b)(u+v\sqrt{7})=(a+bu)+(bv)\sqrt{7}$ (this is from the fact that $f$ preserves addition and $f$ preserves $\mathbb{Q}$ and multiplication).

$(\sqrt{7}{)}^2=7⟹(f(\sqrt{7}){)}^2=f((\sqrt{7}{)}^2)=f(7)=7$ $⟹(f(\sqrt{7}){)}^2=7$ $⟹f(\sqrt{7})=\pm \sqrt{7}$

Thus $f(\sqrt{7})=\{\begin{array}{ll}\sqrt{7},& f={\mathrm{id}}_{Q[\sqrt{7}]}\\ -\sqrt{7},& f=\phi \end{array}$, and these are the only two possibilities.

Application:

Say $P(x)=x^n+c_{n-1}{x}^{n-1}+\dots +c_0$, with $c_i\in \mathbb{Q}\forall 0\le i\le n-1$. If $P(x)$ has $a+b\sqrt{7}$ as root, then $a-b\sqrt{7}$ is also a root.

Say $\alpha =a+b\sqrt{7}$ is a root of $P(x)$, then ${\sum }_{k=0}^n(c_k(\phi (\alpha ){)}^k)=\phi (P(\alpha ))=\phi (0)=0$, so ${\sum }_{k=0}^n(c_k({\alpha }^{\prime }{)}^k)=0$, ${\alpha }^{\prime }$ is a root.