What are the polynomials $f$ in $\mathbb{R}[x]$ such that $f(2+i)=0$?

Question

In other words what is the kernel of the homomorphism $ev_{2+i}:\mathbb{R}[x]\rightarrow \mathbb{C}$ given by $ev_{2+i}(f) = f(2 + i)$

Answer 1

The set of these polynomials is an ideal, call it $I$, and the ring in question is PID, so there must be a generator for $I$. The polynomial

$$p(x) =(x-(2+i))(x-(2-i)) = x^2-4x+5$$

is a generator, since there isn't any degree $1$ polynomial in $I$ (it would be a real multiple of $(x-(2+i))$ and these always have nonreal numbers in them).

Hence $I=p\mathbb{R}[X]$.

Answer 2

The polynomial $(x-2)^2 + 1 = x^2 - 2x + 5$ is clearly in the kernel of this map.

The key is that it generates the kernel, i.e., that every real polynomial vanishing on $2+i$ is just a multiple of this one. We will use that $\mathbb{R}[x]$ is a principal ideal domain. By considering degrees, it then suffices to show that this polynomial is of minimal degree vanishing on $2+i$.

But a linear polynomial with real coefficients can't vanish on $2+i$ (if $a\cdot(2+i) + b = 0$ we conclude $a = 0$ since $a$ and $b$ are real).

Answer 3

Since $\mathbb R [X]$ is a PID, we know that this ideal is generated by a single element. Furthermore, since the image is a field, the ideal should be maximal $\Leftrightarrow$ the generating polynomial should be irreducible. Irreducibles in $\mathbb R[X]$ can be of degree 1 or 2. Degree 1 is not an — this would mean that $2+i \in \mathbb R$, — so we are looking for a degree-2 polynomial. In fact, any polynomial $p$ of degree 2 will work as long as $p(2+i)=0$, since it will generate the same ideal.

Now, to find such $p$, just square things a bit:

$$x = 2 + i$$ $$x - 2 = i$$ $$(x - 2)^2 = -1$$ $$x^2-4x+5=0$$ So, take $p(x)=x^2-4x+5$, and the whole kernel is $p \cdot \mathbb R [X] = \{ p\cdot q \,|\, q \in \mathbb R[X] \}$

Answer 4

If a polynomial in $\mathbf R[x]$ vanish at $2+i$, it also vanishes at $2-i$, hence it is divisible in $\mathbf C[x]$ by $x-2-i$ and by $x-2+i$. Therefore, as these polynomials are coprime, it is divisible by their product $$(x-2-i)(x-2+i)=(x-2)^2-i^2=x^2-4x+5.$$ Conversely this polynomial vanishes at $x=2+i$. As it is in $\mathbf R[x]$, this proves $x^2-4x+5$ is a generator of the ideal $Z(2+i)$.