Define a function $\phi: \mathbb{R}[x] \to \mathbb{C}$ by $\phi(f(x)) = f(i), \forall f(x) \in \mathbb{R}[x]$.

  1. Is $\phi$ a ring homomorphism?

  2. Find a concrete $f(x) \in \mathbb{R}[x]$ such that $deg(f(x)) = 3$ and $deg(\phi(f(x))) = 0.$

  3. Find a concrete $g(x) \in \mathbb{R}[x]$ such that $(\phi(g(x))) = 13 -45i$.

My attempt: I know this is a very elementary level question, but I need to make sure I am doing it right. So, here it goes:

  1. I am slightly confused here. Can we just say that $\phi$ is a ring homomorphism because $\phi(f(x)+g(x))=\phi((f+g)(x)) = (f+g)(i) = f(i)+g(i) = \phi(f(x)) + \phi(g(x))$ and $\phi(f(x)g(x))=\phi((fg)(x)) = (fg)(i) = f(i)g(i) = \phi(f(x)\phi(g(x))?$

  2. Let, $f(x) = x^3 + x $. Then, $\phi(f(x)) = i^3 + i = -i + i = 0.$ Is this correct?

  3. Let, $g(x) = 13 - 45x$. Then, $\phi(g(x)) = 13 - 45i$

  • 2
    $\begingroup$ yes$~~~~~~~~~~~$ $\endgroup$ – user87543 Mar 13 '15 at 18:33
  • $\begingroup$ Seems correct, however for 1 check if your def of ring homomorphism requires $f(1)=1$. It's satisfied but you may need to say it for completeness. $\endgroup$ – Gregory Grant Mar 13 '15 at 18:34
  • $\begingroup$ I guess I could check, but the definition we've learned doesn't require us to check $f(1) = 1$ $\endgroup$ – Jellyfish Mar 13 '15 at 18:37
  • $\begingroup$ @GregoryGrant Since the codomain is a field (domain would suffice) and the map is not constantly $0$, the fact that $\phi(1)=1$ is automatic. $\endgroup$ – egreg Mar 13 '15 at 18:41
  • $\begingroup$ @egreg it may be automatic but that still needs to be noted in the solution if indeed it's required by the particular definition used. $\endgroup$ – Gregory Grant Mar 13 '15 at 19:56
  1. Yes. Alternatively, you can use the universal property of polynomial rings to note that "evaluating at a point" is always a ring homomorphism

  2. What is the degree of a complex number, anyway?

  3. Yes

  • $\begingroup$ @OP beware that $(1)$ is true only for commutative rings (or $R$-algebras). Indeed evaluating $\,rx = xr\,$ at $\,s\in S\,$ yields $\,rs = sr\,$ so $\,r\,$ commutes with all elements of $\,S.\,$ This necessary condition turns out to be sufficient for evaluation maps to be ring homs. $\endgroup$ – Bill Dubuque Mar 13 '15 at 20:32

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