Write an equation with the lowest degree , one of whose roots are $\sqrt{3}-i\cdot \sqrt{2}$, with integral coefficients

Question

I know that with the conjugate root theorem, the conjugate is also a root. However, the integral coefficients part trips me up, as my polynomial has a irrational coefficient.

Answer 1

$$ x = \sqrt 3 + i \sqrt 2$$ $$ \iff x - (\sqrt 3 + i \sqrt 2) = 0$$ $$ \Rightarrow (x - \sqrt 3 + i \sqrt 2)\times (x - \sqrt 3 - i \sqrt 2) = 0$$ $$ \iff (x - \sqrt 3 + i \sqrt 2)\times (x - \sqrt 3 - i \sqrt 2) = 0$$ $$ \iff (x - \sqrt 3)^2 - (i \sqrt 2)^2 = 0$$ $$ \iff (x - \sqrt 3)^2 - (i \sqrt 2)^2 = 0$$ $$ \iff x^2 - 2 \sqrt 3 x + 5 = 0 $$ That's probably what you had got. Some of its coefficients are still irrational, we should continue, using similar approach. $$ \Rightarrow (x^2 - 2 \sqrt 3 x + 5)\times (x^2 + 2 \sqrt 3 x + 5) = 0 $$ $$ \Rightarrow (x^2 + 5) ^2 - (2 \sqrt 3 x) ^2 = 0 $$

Answer 2

So using your idea, $$ P(x) = \left( x - \left(\sqrt{3} + \mathrm{i}\sqrt{2} \right) \right) \left( x - \left(\sqrt{3} - \mathrm{i}\sqrt{2} \right) \right) = x^2 - 2\sqrt{3} x + 5 \text{.} $$ What does this tell us? If you already have $\sqrt{3}$ as a legal coefficient, then you have got the identified root using a degree two (or degree one, since we haven't verified that we actually need a quadratic) polynomial. However, $\sqrt{3}$ is not a legal coefficient since it is not an integer. How do we clear that up? We want the roots of $$ x^2 - 2 \sqrt{3} x + 5 = 0 $$ so $$ x^2 + 5 = 2 \sqrt{3} x \text{,} $$ giving $$ x^4 + 10 x^2 + 25 = (x^2 + 5)^2 = (2 \sqrt{3} x)^2 = 12 x^2 \text{.} $$ Then our polynomial is $$ x^4 -2 x^2 + 25 \text{,} $$ it has the specified root, and it has integer coefficients.

How do we verify that a lower degree polynomial would not suffice? Let $z = \sqrt{3} - \mathrm{i}\sqrt{2}$. Construct a table of powers. We only need to go up to $z^3$ since we already know a degree four solution to our problem. \begin{align*} z^0 &= 1 \\ z^1 &= \sqrt{3} - \mathrm{i}\sqrt{2} \\ z^2 &= 1 - \mathrm{i} 2 \sqrt{2}\sqrt{3} \\ z^3 &= -3 \sqrt{3} - \mathrm{i} 7 \sqrt{2} \end{align*} If we can find integer combinations (not all zero) of the powers less than $k$ (for $k=0,1,\dots$) that sum to zero, then we have found a lower degree polynomial that has the required root.

• Clearly, the only integer multiple of $z^0$ giving zero is $0 \cdot z^0$, which is not "not all zero".
• Of $z^0$ and $z^1$, the only integer multiple of $z^1$ with no imaginary part is $0$, so there is no polynomial if degree one.
• Of $z^0$, $z^1$, and $z^2$, if the coefficient of $z^1$ is nonzero, then the real part of the sum has a nonzero multiple $\sqrt{3}$, which is not an integer. This would prevent the sum being zero. Therefore, the coefficient of $z^1$ is zero, leaving no way to eliminate the imaginary part of $z^2$ unless its coefficient is also zero. So there is no polynomial of degree two having the specified root.
• Of $z^0, \dots, z^3$, we must have $3$-times as many $z^1$ as $z^3$ to cancel the $\sqrt{3}$ in the real part. But this produces an imaginary part that is an integer multiple of $10 \sqrt{2}$, which cannot be cancelled by any integer multiple of $z^2$ or $z^0$. So there is no polynomial of degree three having the specified root.

Consequently, the lowest degree polynomial having $z$ as a root has degree four and we have already found such a polynomial.