0

Is the quotient ring $\Bbb C[x]/(x^2+1)$ an integral domain?

My solution goes like this:

If possible let us assume that $\Bbb C[x]/(x^2+1)$ an integral domain. This means $(x^2+1)$ is a prime ideal in $\Bbb C[x].$ Now, $x^2+1=(x+i)(x-i)\in (x^2+1)=I$ due to which, $x+i\in I$ or $x-i\in I.$

If $x+i\in I$ then $x+i=(x^2+1)p(x)$ for some $p(x)\in \Bbb C[x].$ Now, $p(x)\neq 0\implies \deg (p(x))+2=1,$ a contradiction. So, $x+i\notin I.$ Similarly, $x-i\notin I.$

This means, $(x^2+1)$ is not a prime ideal, again a contradiction. So, $\Bbb C[x]/(x^2+1)$ is not an integral domain.

However, when I approached in the following way, I found out that $\Bbb C[x]/(x^2+1)$ is supposed to be an integral domain:

Let $f:\Bbb C[x]\to \Bbb C$ such that $f(p(x))=p(i),\forall p(x)\in \Bbb C[x].$

Let $p(x),q(x)\in\Bbb C[x]$ then, $$f(p(x)+q(x))=f((p+q)(x))=(p+q)(i)=p(i)+q(i)=f(p(x))+f(q(x))$$ and $$f(p(x)q(x))=f(pq(x))=pq(i)=p(i)q(i)=f(p(x))f(q(x)).$$ Furthermore if $a+bi\in \Bbb C$ then $f(a+bx)=a+bi,$ where $a+bx\in\Bbb C[x].$ So, $f$ is an onto homomorphism.

Now, we see that if $p(x)\in (x^2+1)$ then $f(p(x))=p(i)=0\implies f(x)\in \ker(f)$. Next, if $p(x)\in\ker(f)$ then $f(p(x))=p(i)=0\implies x-i|p(x)\implies x+i|p(x)$ so, $x^2+1|p(x)\implies p(x)\in (x^2+1).$ Thus, $\ker f=(x^2+1).$

Hence, $\Bbb C[x]/(x^2+1)$ is isomorphic to $\Bbb C.$ But, $\Bbb C$ is an integral domain and so is, $\Bbb C[x]/(x^2+1).$

Both of the solution seems fine to me, but the conclusions I am arriving at, are different and this is absurd. I am in a big dilemma regarding which of the approach is the valid one. Both of the solutions can't be correct at the same time. So, I want to know the reason why the other approach is incorrect. Any help regarding this apparent issue will be greatly appreciated.

Thomas Finley
  • 3,625
  • An element $p(x)$ will be $\in\ker(f)$ if and only if $p(x)\in(i- x)$. This can be shown by solving the equation $f(p(x))=0$ for $p(x)$. – J.G.131 Jun 04 '24 at 15:32
  • No @JAG131. $\ker f=(x-i)$. Your candidate ${\alpha+\alpha x\mid\alpha\in\Bbb C}$ is not even an ideal. Oh, now you agree with my $(x-i)$. Fine! – Anne Bauval Jun 04 '24 at 15:33
  • @AnneBauval Yes, that's correct. (Sorry, edited that in a little late.) – J.G.131 Jun 04 '24 at 15:36
  • 1
    $x-i\mid {\color{violet} {x-i}}$ and $x+i\nmid {\color{violet} {x-i}}$. – Asigan Jun 04 '24 at 15:45
  • The 2nd proof is wrong: by the linked dupe, the kernel ideal is generated by any (monic) polynomial of minimal degree, i.e. the minimal polynomial of $i$ over $\Bbb C,,$ which is $,x-i,,$ not $,x^2+1.,$ You seem to be computing the minimal polynomial over $\Bbb R,,$ not over $\Bbb C.\ \ $ – Bill Dubuque Jun 04 '24 at 16:07
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Jun 04 '24 at 16:10

1 Answers1

2

Your 1st approach is correct (but could be simplified: you don't need to reason by contradiction. Just notice that $x−i,x+i∉I$ and their product $∈I$).

The 2nd one is not correct: $x−i\mid p(x)$ does not imply $x+i\mid p(x)$, because $p(x)$ is not assumed to have real coefficients. Actually, $\ker f=(x-i)$.

Anne Bauval
  • 49,005
  • Please don't answer off-topic (SV) questions. Making the answer CW doesn't alleviate the problems they pose (e.g. they clutter search results greatly complexifying searching for answers). – Bill Dubuque Jun 04 '24 at 15:59