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Let $\alpha = a + b \sqrt{d} \in \mathbb{Q} \left(\sqrt{d} \right) = \{a+b \sqrt{d}:a,b \in \mathbb{Q} \}.$

The minimal polynomial $m(x)$ of an algebraic number $\alpha \in \mathbb{C}$ is the monic polynomial of smallest degree, with coefficients in $\mathbb{Q}$ such that $m(\alpha) = 0.$ How do we find the minimal polynomial of $\alpha = a + b \sqrt{d}$ over $\mathbb{Q}$?

user26857
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St Vincent
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  • $x-(a+b\sqrt{d})$. – André Nicolas May 28 '15 at 03:54
  • @AndréNicolas The phrasing is confusing, but the OP does state that the minimal polynomial is supposed to have coefficients in $\mathbb{Q}$. – Viktor Vaughn May 28 '15 at 03:55
  • Yes, but OP also said "in the field" $\mathbb{Q}(\sqrt{d})$ and I interpreted "in" as meaning "over". Maybe mine was a feeble joke. – André Nicolas May 28 '15 at 03:57
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    St Vincent: I have edited your post to hopefully reflect what you meant to ask, in order to avoid confusion on part of those answering. Please roll back if this is NOT what you are asking, in which case André Nicolas' answer is what you are looking for. – Alex Wertheim May 28 '15 at 03:58

4 Answers4

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$$\alpha-a=b\sqrt{d}$$ $$(\alpha-a)^2=b^2d$$ $$\alpha^2-2\alpha a+a^2=b^2d$$ $$f(x)=x^2-2x a+a^2-b^2d$$

Zelos Malum
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  • How do we know $x^2-2xa+a^2-b^2 d$ is irreducible? – user5826 Jul 10 '20 at 01:49
  • because I say so :P If you do the minimal operation it is irreducible. And we know also from here, that due to squareroot that the minimal polynomial must be of degree 2, so any reducibility would make it lower. Of course if d is the square of some rational number, then it is indeed reducible because you've done an unnecessary squaring. – Zelos Malum Jul 10 '20 at 04:30
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Hint: If $\alpha \in K$ is a root of an irreducible polynomial $f \in F[x]$, then $\sigma(\alpha)$ is a root of $f$ for all field automorphisms $\sigma \in \text{Aut}(K/F)$. Can you think of an automorphism $\sigma$ of $\mathbb{Q}(\sqrt{d})$ that fixes $\mathbb{Q}$? What are the possibilities for $\sigma(\sqrt{d})$?

Viktor Vaughn
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Another way to determine the minimal polynomial, which is nice in the quadratic case, is to compute the trace and norm of the element. $${\rm Tr}(a-b\sqrt{d})=(a-b\sqrt{d})+(a+b\sqrt{d})=2a$$ $${\rm N}(a-b\sqrt{d})=(a-b\sqrt{d})(a+b\sqrt{d})=a^2-b^2d$$ Then the coefficient of $x^{n-1}$, in our case $x$, is $-{\rm Tr}(a-b\sqrt{d})$ and the constant coefficient if $(-1)^{n}{\rm N}(a-b\sqrt{d})$.

(For reference, see Robert Ash's book on Algebraic Number Theory.)

- There are many other references in fact, but this is one that I have found helpful.

Erik Holmes
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Let $x= a+ b\sqrt{d}$. Then $x- a= b\sqrt{d}$ so $(x- a)^2= b^2 d$. Those are now all integers and since the number has a square root, this is the minimal polynomial.

(Of course, that is the same as $x^2- 2ax+ a^2- b^2d$.)

Viktor Vaughn
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HallsofIvy
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