To find the minimal polynomial $1+\sqrt 3$ over $\mathbb{Q}$

Question

I have a rough idea about this question. First, let $x=1+\sqrt3$, then I found that $x^2-2x-2=0$ Then, by applying Eisenstein test with the prime $2$, it is irreducible, hence, the minimal polynomial over $\mathbb{Q}$ is $x^2-2x-2$. Is my answer right?

Answer 1

Hint: find the minimal polynomial $g(x)$ for $\sqrt{2}$ and then let $f(x)=g(x-1)$. Irreducibility is preserved by automorphisms, so $f$ is still irreducible. And $f$ is just $g$ shifted one unit to the right. So if $g$ had a root at $\sqrt{2}$, then $f$ has a at $1+\sqrt{2}$.

Added: your answer is wrong, and you don't even need to go through the computation to know it. The number $1+\sqrt{2}$ is a unit in $\mathbb{Z}[\sqrt{2}]$ (multiply it by minus its conjugate). Hence its norm is $\pm 1$. But its norm (modulo sign) is the independent term of the minimal polynomial.

Answer 2

(Edit: OP meant $1+\sqrt 3$, not $1+\sqrt 2$)

Yes, your solution seems correct! Since you also know that $1\pm\sqrt{3}\not\in\mathbb{Q}$ is a root of $x^2-2x-2$, and that a quadratic polynomial over a field has at most $2$ solutions you can see immediately that it is irreducible.

Answer 3

If $x=1+\sqrt 2$ then $x^2 = 3+2\sqrt 2$, and so $$ \left[\begin{array}{cc} 1 & 1 \\ 3 & 2 \end{array}\right] \left[\begin{array}{c}1 \\ \sqrt 2 \end{array}\right]= \left[\begin{array}{c}x \\ x^2 \end{array}\right]$$

The matrix on the left is non-sigular, so we can invert: $$ \left[\begin{array}{c}1 \\ \sqrt 2 \end{array}\right]= \left[\begin{array}{cc} -2 & 1 \\ 3 & -1 \end{array}\right] \left[\begin{array}{c}x \\ x^2 \end{array}\right]$$

Hence $1=-2x+x^2$ and $\sqrt 2 = 3x-x^2$.

The minimal polynomial over $\mathbb Q$ would be $x^2-2x-1$.