1

I'm studying Rotman's Advanced Modern Algebra, and have trouble understanding example 3.118 on page 187. It states: "Consider the evaluation map $\varphi:\mathbb{R}[x]\to\mathbb{C}$ given by $\varphi:f(x)\to f(i)$." Then surjectivity is proved, then it is proved that $(x^2+1)\subseteq\ker\varphi$. These I understand. But then it states: "For the reverse inclusion, take $g(x)\in\ker\varphi$. Now $i$ is a root of $g(x)$, and so $\gcd(g,x^2+1)\not=1$ in $\mathbb{C}[x]$; therefore, $\text{gcd}(g,x^2+1)\not=1$ in $\mathbb{R}[x]$."

I get that $\gcd(g,x^2+1)\not=1$ in $\mathbb{C}[x]$ because they both are divisible by $x-i$ because $i$ is a root of both of them. But why does this carry over to $\mathbb{R}[x]$ where $i$ does not exist?

Here is a picture of the whole example, with the part I have trouble with yellowed.

user1381133
  • 41
  • 2
    If $i$ is a root of a polynomial and the polynomial is real, then what is another root of this polynomial? – Eman Yalpsid Nov 22 '24 at 12:03
  • 1
    The gdc does not depend on field extension – julio_es_sui_glace Nov 22 '24 at 12:06
  • 1
    To calculate $\gcd(g(x),x^2+1)$ you can use the Euclidean Algorithm, since both polynomials have real coefficients, the result will be a polynomial with real coefficients. – jjagmath Nov 22 '24 at 12:07
  • This is an answer to the question, not just a comment. Please notice that comments should only be used to clarify, not answer the question. See How do comments work for more information. Therefore, I suggest to post your answer as an answer. This brings extra visibility to the answer und puts the question off the unanswered list. – Martin Brandenburg Nov 22 '24 at 13:20
  • 1
    Answering Eman Yalpsid's comment: If $i$ is a root and the polynomial is real, then according to the complex conjugate root theorem, $-i$ is also a root. This means $g(x)=(x-i)(x+i)f(x)$ for some $f(x)\in\mathbb{C}[x]$. Because $(x-i)(x+i)=x^2+1$, now $x^2+1\mid g(x)$. So $\gcd(g,x^2+1)\not=1$ in $\mathbb{R}[x]$. – user1381133 Nov 22 '24 at 14:44
  • I also found this https://math.stackexchange.com/questions/84551/gcd-in-polynomial-rings-with-coefficients-in-a-field-extension. – user1381133 Nov 22 '24 at 14:53

0 Answers0