Finding $f\in\mathbb{Q}[x]$ such that $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$ and $\deg(f)\leq 3$. What's wrong with my approach?

Question

I'd like to find a polynomial $f(x) \in \mathbb{Q}[x]$ satisfying $$f(\sqrt{2}+\sqrt{3})=\sqrt{2}$$ and $\deg(f) \leq 3$.

What I've been trying is the following:

Since $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$,

then $f(\sqrt{2}+\sqrt{3})-\sqrt{2}=0$.

so think of $g(x)=f(x+\sqrt{3})-x$ as a polynomial over $\mathbb{Q}(\sqrt{3})$.

And I've already known that $x^2 -2$ is irreducible over $\mathbb{Q}(\sqrt{3})$

$g(x)$ has a $\sqrt{2}$ as a root of itself, $x^2 -2$ divides $g(x)$ in $\mathbb{Q}(\sqrt{3})[x]$.

$g(x)$ must be of the form $(x^2 -2)(ex+f)$ where $e, f \in \mathbb{Q}(\sqrt{3})$

and also of the form $a(x+\sqrt{3})^3 +b(x+\sqrt{3})^2 +c(x+\sqrt{3}) +d -x$, where $f(x)=ax^3 +bx^2 +cx+d \in \mathbb{Q}[x]$.

after comparing the coefficients of two polynomials, I found that $f(x)=\frac{1}{4}x^3-\frac{9}{4}x$.

But the actual polynomial is $\frac{1}{2}x^3-\frac{9}{2}x$.

There must be a flaw in the above reasoning.

Where did I do a mistake? Could you point it out?

Thank you.

Answer 1

The error appears to be somewhere in the unwritten details of "after comparing the coefficients of two polynomials, I found that...".

Here's a lowbrow approach:

Hint Denote $\beta := \sqrt{2} + \sqrt{3}$. We're looking to solve $\sum_{i = 0}^3 a_i \beta^i = \sqrt{2}$ in rational numbers $a_i$. Computing gives that $$\beta^2 = 5 + 2 \sqrt{6}, \qquad \beta^3 = 11 \sqrt{2} + 9 \sqrt{3} .$$ Now, $1, \sqrt{2}, \sqrt{3}, \sqrt{6}$ are linearly independent over $\Bbb Q$, but the only power in $\beta^0, \ldots, \beta^3$ in which a nonzero rational multiple of $\sqrt{6}$ occurs is $\beta^2$, so $a_2 = 0$.

Among $\beta^0, \beta^1, \beta^3$ only $\beta^0$ has a nonzero rational summand, so $a_0 = 0$. Thus, we are solving $$a_1 (\sqrt{2} + \sqrt{3}) + a_3 (11 \sqrt{2} + 9 \sqrt{3}) = \sqrt{2} .$$ Linear independence now implies that we can compare coefficients of $\sqrt{2}$ and $\sqrt{3}$.

Answer 2

Let $e = g + \sqrt{3}h$, $f = j + \sqrt{3}k$. $$\begin{eqnarray}(x^2 -2)(ex+f) &=& a(x+\sqrt{3})^3 +b(x+\sqrt{3})^2 +c(x+\sqrt{3}) +d -x \\ (g + \sqrt{3}h)x^3 + (j + \sqrt{3}k)x^2 - 2(g + \sqrt{3}h)x - 2(j + \sqrt{3}k) &=& a(x^3 + 3\sqrt{3}x^2 + 9x + 3\sqrt{3}) +b(x^2 + 2\sqrt{3}x + 3) + cx + \sqrt{3}c +d - x \\ (g + \sqrt{3}h)x^3 + (j + \sqrt{3}k)x^2 - 2(g + \sqrt{3}h)x - 2(j + \sqrt{3}k) &=& ax^3 + (b + 3\sqrt{3}a)x^2 + (9a + c - 1 + 2\sqrt{3}b)x + (3b + d + 3\sqrt{3}a + \sqrt{3}c) \\ \end{eqnarray}$$

So $$\begin{eqnarray}g &=& a \\ h &=& 0 \\ j &=& b \\ k &=& 3a \\ -2g &=& 9a+c-1 \\ -2h &=& 2b \\ -2j &=& 3b + d \\ -2k &=& 3a + c \end{eqnarray}$$

Quickly reduces to $$\begin{eqnarray}h = b = j = d &=& 0 \\ g &=& a \\ k &=& 3a \\ c &=& -9a \\ 2a &=& 1 \\ \end{eqnarray}$$

So there's no flaw in your reasoning: the flaw is in the part you didn't include in the question.

Answer 3

Hint for an alternative approach:

your polynomial $f$ satisfies $$f(\pm \sqrt{2}\pm \sqrt{3})=\pm \sqrt{2}$$ (the signs in front of $\sqrt{2}$ are the same), so it is a Lagrange interpolation polynomial.