Complex Numbers as Quotient Ring of Real Polynomial

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\C$ be the set of complex numbers.

Let $P \left[{x}\right]$ be the set of polynomials over real numbers, where the coefficients of the polynomials are real.

Let $\left\langle{x^2 + 1}\right\rangle = \left\{ {Q \left({x}\right) \left({x^2 + 1}\right): Q \left({x}\right) \in P \left[{x}\right]}\right\}$ be the ideal generated by $x^2 + 1$ in $P \left[{x}\right]$.

Let $D = P \left[{x}\right] / \left\langle{x^2 + 1}\right\rangle$ be the quotient of $P \left[{x}\right]$ modulo $\left\langle{x^2 + 1}\right\rangle$.


Then:

$\left({\C, +, \times}\right) \cong \left({D, +, \times}\right)$


Proof

By Division Algorithm of Polynomial, any set in $D$ has an element in the form $a + b x$.

Define $\phi: D \to \C$ as a mapping:

$\phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right) = a + b i$

We have that:

$\forall z = a + b i \in \C : \exists \left[\!\left[{a + b x}\right]\!\right] \in D$

such that:

$\phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right) = a + b i = z$

So $\phi$ is a surjection.


To prove that it is a injection, we let:

$\phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right) = \phi \left({\left[\!\left[{c + d x}\right]\!\right]}\right)$

So:

\(\displaystyle \phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right)\) \(=\) \(\displaystyle \phi \left({\left[\!\left[{c + d x}\right]\!\right]}\right)\)
\(\displaystyle \iff \ \ \) \(\displaystyle a + b i\) \(=\) \(\displaystyle c + d i\) Definition of $\phi$
\(\displaystyle \iff \ \ \) \(\displaystyle a = c\) \(\land\) \(\displaystyle b = d\) Equating both real part and imaginary part
\(\displaystyle \iff \ \ \) \(\displaystyle a + b x\) \(=\) \(\displaystyle c + d x\)
\(\displaystyle \iff \ \ \) \(\displaystyle \left[\!\left[{a + b x}\right]\!\right]\) \(=\) \(\displaystyle \left[\!\left[{c + d x}\right]\!\right]\)

So $\phi$ is an injection and thus a bijection.


It remains to show that $\phi$ is a homomorphism for the operation $+$ and $\times$.

\(\displaystyle \phi \left({\left[\!\left[{a + b x}\right]\!\right] + \left[\!\left[{c + d x}\right]\!\right]}\right)\) \(=\) \(\displaystyle \phi \left({\left[\!\left[{\left({a + c}\right) + \left({b + d}\right) x}\right]\!\right]}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({a + c}\right) + \left({b + d}\right) i\) Definition of $\phi$
\(\displaystyle \) \(=\) \(\displaystyle \left({a + b i}\right) + \left({c + d i}\right)\) Definition of Complex Addition
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right) + \phi \left({\left[\!\left[{c + d x}\right]\!\right]}\right)\)


\(\displaystyle \phi \left({\left[\!\left[{a + b x}\right]\!\right] \times \left[\!\left[{c + d x}\right]\!\right]}\right)\) \(=\) \(\displaystyle \phi \left({\left[\!\left[{\left({a + b x}\right) \times \left({c + d x}\right)}\right]\!\right]}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({\left[\!\left[{a \times c + \left({a \times d + b \times c}\right) x + b \times d \, x^2}\right]\!\right]}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({\left[\!\left[{a \times c + \left({a \times d + b \times c}\right) x + b \times d \, x^2 - b \times d \left({x^2 + 1}\right)}\right]\!\right]}\right)\) Definition of $D$ as a quotient ring modulo $\left\langle{x^2 + 1}\right\rangle$
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({\left[\!\left[{\left({a \times c - b \times d}\right) + \left({a \times d + b \times c}\right) x}\right]\!\right]}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({a \times c - b \times d}\right) + \left({a \times d + b \times c}\right) i\) Definition of $\phi$
\(\displaystyle \) \(=\) \(\displaystyle \left({a + b i}\right) \times \left({c + d i}\right)\) Definition of Complex Multiplication
\(\displaystyle \) \(=\) \(\displaystyle \phi \left({\left[\!\left[{a + b x}\right]\!\right]}\right) \times \phi \left({\left[\!\left[{c + d x}\right]\!\right]}\right)\) Definition of $\phi$


Thus $\phi$ has been demonstrated to be a bijective ring homomorphism and thus by definition a ring isomorphism.

$\blacksquare$