User:TheSpleen/Sandbox

Theorem
Let $\left({R,+,\circ}\right)$ be a finite ring with unity and no proper zero divisors.

Then $R$ is a field.

Proof
Let $1_R$ denote the unity of $R$.

Consider the two maps from $R$ to itself, for each nonzero $a \in R$:


 * $x \mapsto a \circ x$
 * $x \mapsto x \circ a$

By Ring Element is Zero Divisor iff not Cancellable, all nonzero elements in $R$ are cancellable. Thus:


 * $a \circ x = a \circ y \implies x=y$
 * $x \circ a = y \circ a \implies x=y$

Therefore, both maps are by definition injective. By Same Cardinality Bijective Injective Surjective, the maps are then also surjective.

Since the maps are surjective, it follows that $a \circ x = 1_R$ for some $x$, and $y \circ a = 1_R$ for some $y$. Recalling that the maps were defined for each nonzero $a \in R$, this means that every nonzero element of $R$ has both a left inverse and a right inverse.

By Left Inverse and Right Inverse is Inverse, each nonzero element of $R$ has an inverse, so $R$ is by definition a division ring.

It now follows by Wedderburn's Theorem that $R$ is a field.

Theorem
Let $P \left(x\right)$ be a polynomial over $\Q$ in $x$, where $x \in \C$ or $x \in \R$.

Let $a \in \Q$ and $b \in \Q_+$, such that $\sqrt {b}$ is irrational.

Then $a + \sqrt{b}$ is a root of $P$ iff $a - \sqrt{b}$ is a root of $P$.

Proof
Denote by $D \left(x\right)$ the quadratic polynomial:
 * $x^2 - 2ax + a^2-b$

Since $D\left(x\right)$ and $P\left(x\right)$ have rational coefficients, by Polynomial Long Division there exist unique polynomials $Q\left(x\right), R\left(x\right)$ over $\Q$, such that $\deg \left({R\left(x\right)}\right) < 2$ or $R \left(x\right)$ is null, and:
 * $P\left(x\right)=Q\left(x\right)D\left(x\right)+R\left(x\right)$

Since $\deg \left({R\left(x\right)}\right) < 2$ or $R\left(x\right)$ is null, we can express $R\left(x\right)$ as $cx+d$, for some rational $c,d$. Thus:
 * $P\left(x\right)=Q\left(x\right)D\left(x\right)+cx+d$

Observe that:
 * $x^2 - 2ax + a^2-b = \left( {x - \left( {a + \sqrt{b}} \right) } \right) \left( {x - \left( {a - \sqrt{b}} \right) } \right)$

so by Polynomial Factor Theorem, $\left( {a + \sqrt{b}} \right)$ and $\left( {a - \sqrt{b}} \right)$ are the two roots of $D\left(x\right)$, so $D\left({a + \sqrt{b}}\right) = D\left({a - \sqrt{b}}\right) = 0$

First assume that $\left( {a + \sqrt{b}} \right)$ is a root in $P \left(x\right)$.

Then $P \left({a + \sqrt{b}}\right)=0$, and we get: