Nomenclature

**elninio** · December 11th 2009, 01:49 PM

How exactly does this read:
(prove that) R[x]/<x^2+2> (is isomorphic to C)

And what exactly does the "/" imply algebraicly?

Thanks.

**sfspitfire23** · December 11th 2009, 10:18 PM

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**sfspitfire23** · December 11th 2009, 10:19 PM

Here, you are creating the complex numbers! This is very elegant. The $\mathbb{R}$ shows that you are in the reals, so $x^2$ can have any coefficient in the reals. BUT, $x^2=-2$ There is nothing when squared in the reals equal to 2, so you have to go imaginary. So, in the multiplication table, whenever you see an $x^2$ replace it with a -2. The $/$ means you are "modding out by"

**Shanks** · December 12th 2009, 01:56 AM

sfspitfire23 give us such a elegant explanation.
So, follow his thread, and you will find the isomorphism.
Good luck!

**proscientia** · December 12th 2009, 03:19 AM

Originally Posted by elninio

How exactly does this read:
(prove that) R[x]/<x^2+2> (is isomorphic to C)

It reads:

(prove that) the quotient ring of $\mathbb R[x]$ [the ring of polynomials with real coefficients] by the ideal generated by $x^2+2$ (is isomorphic to $\mathbb C)$

To prove this, given any $f(x)\in\mathbb R[x],$ use the division algorithm to write $f(x)=q(x)(x^2+2)+ax+b$ where $q(x)\in\mathbb R[x]$ and $a,b\in\mathbb R$ are uniquely determined by $f(x).$ Define a map $\varphi:\mathbb R[x]\to\mathbb C$ by $\varphi\left(f(x)\right)=b+i\sqrt2a.$ All that remains is to show that $\varphi$ is a ring epimomophism (surjective homomorphism) with kernel $\left\langle x^2+2\right\rangle.$

Originally Posted by elninio

And what exactly does the "/" imply algebraicly?