# Minimum polynomial

• Feb 9th 2008, 06:39 AM
Minimum polynomial
Let $\displaystyle K = \mathbb {Q} [ \sqrt {m} ]$, where m is a square-free integer. Let $\displaystyle a=r+s \sqrt {m}$ be an algebraic integer in K.

a) Show that the minimum polynomial of a is $\displaystyle x^2-2rx+r^2-ms^2$

Now, I have verified that $\displaystyle (a)^2 - 2r(a)+r^2-ms^2 = 0$

And I have to prove it is irreducible over K, should I use the norm, but the norm is not defined here yet, can I use N(ax+b) = a^2 + b^2?

I factored the polynomial into $\displaystyle [x-(r- \sqrt {m}s)][x+(r- \sqrt {m}s)] - 2rx$, will this help?

b) Show that $\displaystyle 2r, r^2-ms^2$ are integers.
• Feb 9th 2008, 02:24 PM
ThePerfectHacker
Quote:

Let $\displaystyle K = \mathbb {Q} [ \sqrt {m} ]$, where m is a square-free integer. Let $\displaystyle a+r+s \sqrt {m}$ be an algebraic integer in K.

a) Show that the minimum polynomial of a is $\displaystyle x^2-2rx+rs^2$

I assume you mean $\displaystyle a = r+s\sqrt{m}$.
But then, $\displaystyle (r+s\sqrt{m})^2 - 2r (r+s\sqrt{m}) + rs^2 \not = 0$ if $\displaystyle r\not = 0$. Thus, it cannot be a minimal polynomial.
• Feb 9th 2008, 02:38 PM
When I plug in a into the poly, I have:

$\displaystyle (a)^2 - 2r(a) + r^2 - ms^2 = (r+s \sqrt {m})^2-2r(r+s \sqrt {m}) + r^2 - ms^2$
$\displaystyle =r^2+2sr \sqrt {m} + s^2m - 2r^2 - 2s \sqrt {m}r + r^2 - ms^2 = 0$

So it does equal to zero.
• Feb 9th 2008, 02:44 PM
ThePerfectHacker
Quote:

$\displaystyle (a)^2 - 2r(a) + r^2 - ms^2 = (r+s \sqrt {m})^2-2r(r+s \sqrt {m}) + r^2 - ms^2$
$\displaystyle =r^2+2sr \sqrt {m} + s^2m - 2r^2 - 2s \sqrt {m}r + r^2 - ms^2 = 0$
Okay, so then you should change $\displaystyle x^2 - 2rx + rs^2$ to $\displaystyle x^2 - 2rx + (r^2 - ms^2)$ in your first post. Now, you are saying, "minimal polynomial" remember you need to specify over what field it is a minimal polynomial over, I assume it should be $\displaystyle \mathbb{Q}$. In that case you have to prove that this polynomial has no rational zeros. Can you show that?
Since m is a square-free integer, $\displaystyle \sqrt {m}$ is not a real number, so it is not a rational number. However, the only way to eliminate the m in the polynomial is to let $\displaystyle x=a+b \sqrt {m}$, otherwise the m would remain. So there cannot be any rational zeros.