Proving $\mathbb R[x]/\langle 1+x^2\rangle$ $\cong$ $\mathbb C$ without using 1st isomorphism theorem

Question

I've seen many the proofs of this by making use of First isomorphism theorem, by considering the map,$$\phi:\mathbb R[x]\rightarrow\mathbb C$$ defined by $\phi(a+bx)=a+bi$.

My questions are

• Why these proofs consider ($a+bx$) as the only form for the memebers of $\mathbb R[x]$?
• Why they are not using ($a+bx+cx^2$) or ($a+bx+cx^2+dx^3$) or any other polynomials of degree greater than 1?

Is there any other proof of this without making use of First isomorphism theorem?

Thank you!!

Answer 1

The equation $\phi(a+bx) = a+bi$ is given along with the assumption that $\phi$ is a homomorphism. This allows you to calculate, for instance, $\phi(a + bx + cx^2) = \phi(a + bx) + \phi(cx^2) = \phi(a+ bx) + \phi(cx)\phi(x) = a + bi + ci^2 = a - c + bi$. In fact, you only really need to specify $\phi(x) = i$ together with the fact that $\phi$ is a ring homomorphism to know the value of $\phi$ for every element of $\mathbb R[x]$.

You don't have to use the first isomorphism theorem, since you can just write down the isomorphism between $\mathbb R[x]/(x^2+1)$ and $\mathbb C$ and then verify that it is in fact an isomorphism.

Answer 2

Because in the quotient $x^2=-1$, so every equivalent class has a representing belong to set of polynomial of degree at most $1$