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$\phi:\mathbb{R}[x] \to \mathbb{R}$ defined by $f(x)\to f(1+\sqrt {2}$)

I noticed that in the solution, we say that $1+\sqrt{2}$ is a root for the polynomials we consider. Can someone explain why that's the case and show me the steps for this problem briefly?

user26857
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Lana
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  • $\ker\phi={f\in\mathbb R[x]:\phi(f)=0}={f\in\mathbb R[x]:f(1+\sqrt2)=0}$ – user26857 Jan 21 '17 at 10:32
  • I know that definition, but f could be anything.. so when you solve it, how do you know what the root is before you even know the polynomial? – Lana Jan 22 '17 at 15:41

1 Answers1

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Maybe you could try to solve some "easier" problems before the one you were given, for instance:

1) What is the kernel of $$\phi_1 : \Bbb R[X] \to \Bbb R, \qquad f \mapsto f(1)$$ i.e. what is the set of all the polynomials $f \in \Bbb R[X]$ such that $f(1)=0$ ? Try first to find an example of such a polynomial. For instance...

$f(X)=X-1$.

Can you find another one?

$f(X)=2X-2$, or $f(X)=(X^{2017}+\pi)(X-1)$.

Now you should see that $(X-1) \subset \ker(\phi_1)$. Conversely, if $f(1)=0$, can you say that $f$ is a multiple of $X-1$ in $\Bbb R[X]$? I let you think about it.

2) What is the kernel of $$\phi_2 : \Bbb R[X] \to \Bbb R, \qquad f \mapsto f(\sqrt 2)$$ As before, first try to give me an example of polynomial $f \in \Bbb R[X]$ such that $f(\sqrt 2)=0$, except $f=0$ of course ;-)

$f(X)=X^2-2$.

You can probably see that $(X^2-2) \subseteq \ker(\phi_2)$. Conversely, if $f(\sqrt 2)=0$, then try to perform an euclidean divison by $X^2-2$: $$f(X)=(X^2-2)q(X)+r(X), \qquad \deg(r)<\deg(X^2-2)=2$$

Now evaluate at $x=\sqrt 2$ in order to get $$f(\sqrt 2)=0 = 0\cdot q(\sqrt2)+r(\sqrt2)=a+b\sqrt 2,$$ where $r(X)=a+bX$ has degree $1$. It must follow that $a=b=0$ since $\sqrt 2$ is irrational (it implies that $1,\sqrt 2$ are linearly independent over $\Bbb Q$). You just got $r(X)=0$, which yields $f(X)=(X^2-2)q(X)$, and that's it! You proved that $f(X) \in (X^2-2)$, and therefore $\ker(\phi_2) = (X^2-2)$.

What about $1+\sqrt 2$ now? There are two ways:

  • Assume that you computed the kernel of $f \mapsto f(a)$, and you proved that it is equal to $(P(X))$ (true since $\Bbb R[X]$ is Euclidean, hence a PID). Then the kernel of $f \mapsto f(1+a)$ is simply $(P(X-1))$. Here, in part 2) above, we found what is the kernel of $f \mapsto f(\sqrt 2)$, so you immediately get the kernel of $f \mapsto f(1 +\sqrt 2)$.

  • Another way is to "imitate" what we did in 1) and 2). First find a non-zero polynomial $f$ such that $f(1+\sqrt 2)$, and with minimal degree. It was $X-1$ for $a=1$ in 1) and it was $X^2-2$ for $a=\sqrt 2$ in 2). I let you find it for $a=1+\sqrt 2$. Then apply some Euclidean division as we did, and you'll find the kernel of $\phi$.

Here is the answer:

$$\ker(\phi) = \{g(X)(X^2-2X-1) \mid g \in \Bbb R[X]\} = ((X-1)^2-2) = (X^2-2X-1),$$ so a generator is $X^2-2X-1$ (but $3X^2-6X-3$ is also a generator).

Watson
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