I'm trying to find all homomorphisms from $\mathbb{R[x]}/(X^2+1)$ to $\mathbb{C}$. I'm using first isomorphisms theorem, as said here Homomorphisms from quotient polynomial rings to some $\mathbb{Z_n}$ and I know how to show that there exists a homomorphisms $ \phi$ and how to show that $ \langle X^2 + 1 \rangle \subset ker{\phi}$.

But I still can't show two things: that it is onto and that $ker{\phi} \subset \langle X^2 + 1 \rangle$ - I think I omitted this part, my intuition tells me that showing that the polynomial is in kernel may be insufficient to say that the polynomial is equal to the kernel.

  • $\begingroup$ Are you given that the hom restricts to the identity map on $\,\Bbb R\,?\ \ $ $\endgroup$ – Bill Dubuque Jan 13 '14 at 23:17
  • $\begingroup$ To be honest, I'm only told the same that states wikipedia: en.wikipedia.org/wiki/Ring_homomorphism $\endgroup$ – surykatka Jan 13 '14 at 23:42

Homomorphisms $f:\Bbb R[X]/(X^2+1)\to \Bbb C$ correspond to homomorphisms $\phi:\Bbb R[x]\to\Bbb C$ which has $(X^2+1)\,\subseteq\,\ker\phi$ (as $f$ has to take 'all forms of' zero to zero).

Now a $\phi:\Bbb R[X]\to\Bbb C$ must map $1$ to $1$ (because, I guess, unitarity of rings is assumed) and, a priori, it can map $X$ to anywhere in $\Bbb C$ but that already determines the whole homomorphism $\phi$.

Then, $(X^2+1)\subseteq\ker\phi\ \iff\ X^2+1\in\ker\phi\ \iff\ \phi(X)^2=-1$, that means that either $\phi(X)=i$ or $\phi(X)=-i$. So we get exactly two such homomorphisms.

  • $\begingroup$ It's said too fast for me. Do you even use here this first isomorphisms theorem and if so, where? I understand that there would be two homomorphisms for each root of this polynomial, but I'm lost in formalizing. $\endgroup$ – surykatka Jan 13 '14 at 23:06
  • $\begingroup$ My first sentence is basically the first isomorphism theorem. We don't (directly) care the roots of these polynomials, we handle the polynomials themselves as element in the ring $\Bbb R[X]$. And we want that the element $X^2+1$ go to $0$ at $\phi$. And here comes the root thing: that means exactly that the element $\phi(X)\in\Bbb C$ is a root of the given polynomial $X^2+1$. $\endgroup$ – Berci Jan 13 '14 at 23:10
  • $\begingroup$ So I don't see where is it said that $ker{\phi} \subset \langle X^2 + 1 \rangle$. I think it is crucial to show both that $ker{\phi} \subset \langle X^2 + 1 \rangle$ and that $\langle X^2 + 1 \rangle \subset ker{\phi}$ to say that $\langle X^2 + 1 \rangle = ker{\phi}$. If not, please deliberate on the use of the theorem, because I really don't get this part. $\endgroup$ – surykatka Jan 13 '14 at 23:15
  • $\begingroup$ What is your aim? Well, we talk about exactly two maps (one extends $X\mapsto i$ to homomorphism, the other extends $X\mapsto -i$). Check the claim for both. $\endgroup$ – Berci Jan 13 '14 at 23:23
  • $\begingroup$ I think I finally understood this. Nevertheless, the aim of asking about kernels was driven by my tutor, who solved similar problem during classes and insisted on showing that "$ker \phi = (a~polynomial)$", like there existed no more elements of the kernel. $\endgroup$ – surykatka Jan 13 '14 at 23:38

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