Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
Are you given that the hom restricts to the identity map on $\,\Bbb R\,?\ \ $ – Bill Dubuque Jan 13 '14 at 23:17
To be honest, I'm only told the same that states wikipedia: – surykatka Jan 13 '14 at 23:42

1 Answer 1

up vote 3 down vote accepted

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.

share|cite|improve this answer
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. – surykatka Jan 13 '14 at 23:06
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 $X^2+1$. – Berci Jan 13 '14 at 23:10
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. – surykatka Jan 13 '14 at 23:15
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. – Berci Jan 13 '14 at 23:23
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. – surykatka Jan 13 '14 at 23:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.