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Suppose $L/K$ is an algebraic field extension. Take $\alpha_1 \in L$. Then $\alpha_1$ has minimal polynomial $f(x)$ over $K$. Let $\alpha_2, ... \alpha_k$ be the other roots of $f$ in $L$. The $\alpha_i$ are known as algebraic conjugates.

Suppose I am given an $\alpha \in L$ and wish to find the minimal polynomial of $\alpha$ over $K$. It'd be helpful if I knew something about how the algebraic conjugates of $\alpha$ behaved.

Let's take a concrete example. Suppose $L = \mathbb Q(\sqrt{3})$ and $K = \mathbb Q$, and I want to know the minimal polynomial of $2 + \sqrt{3}$. I could work it out by writing $ x = 2 + \sqrt{3}$, squaring, rearranging and squaring again. But this would quickly become cumbersome if we chose a more complicated extension and complicated $\alpha$. I know its a good idea to look at the product $ \prod (x - \alpha_i) $, where the $\alpha_i$ are algebraic conjugates of $\alpha$. What are the algebraic conjugates of $ 2 + \sqrt{3} $? I 'know' they are $ \pm 2 \pm \sqrt{3}$, but why is this?

I suppose my question really comes down to:

i) Why are the conjugates of $2 + \sqrt{3}$ those mentioned above? Why do the symmetric sums and products involved in Vieta's formulas yield rationals in this case?

ii) Are there any other 'types' of $\alpha$ (and/or $L$) for which the form of its conjugates are known?

Thanks

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3 Answers 3

Another, trivially different, way of looking at it: Let $\alpha = 2 + \sqrt3$. Then $\mathbf Q(\alpha) = \mathbf Q(\sqrt3)$. Quadratic extensions of $\mathbf Q$ are all Galois, and you know that the only non-trivial automorphism of $\mathbf Q(\sqrt3)$ is the one that performs $a + b\sqrt3 \mapsto a - b\sqrt3$.

I think the easiest way to see that symmetric polynomials in the conjugates are rational uses separability. More specifically, say I have an algebraic number with conjugates $\beta_1, \ldots, \beta_n$. If $f$ is a symmetric polynomial and $\sigma$ is some embedding of $\mathbf Q(f(\beta_1, \ldots, \beta_n))$ into an algebraic closure, then $\sigma$ permutes the conjugates and hence fixes $f(\beta_1, \ldots, \beta_n)$. By separability the degree of this field extension is equal to the number of such embeddings, and we've shown that there is exactly one. Hence this expression is rational.

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I like this answer - an honest answer to this question has to mention Galois theory somewhere, at least implicitly, in my opinion - but from the language used, I'm slightly concerned the thread starter might not have done much Galois theory. –  Billy Jan 8 '12 at 18:03
    
@Billy I had this concern as well. If the OP could comment on his background then that might be best. Certainly Clive's explanation for the first point is better in that case (and I didn't want to repeat it), but I find it hard to imagine a simpler reason for the second bit. I could sketch how separability works over $\mathbf Q$, if the OP is interested. –  Dylan Moreland Jan 8 '12 at 18:06

Actually the minimal polynomial of $2+\sqrt{3}$ is $(x-2)^2-3$. Hence the algebraic conjugates are $\{ 2 + \sqrt{3}, 2-\sqrt{3} \}$.

How did I know this? Well in this case, you can see this by setting $y=x-2 = \sqrt{3}$. Then $x$ is a root of $P$ if and only if $y$ is a root of $Q$, where we define $Q(y)=P(y+2)$. Clearly these have the same degree and one is monic iff the other is monic, so $Q$ is the minimal polynomial of $y$ if and only if $P$ is the minimal polynomial of $x$. Clearly $y$ has minimal polynomial $y^2-3$, so $x$ must have minimal polynomial $(x-2)^2-3$.

In general there aren't really any short tricks for finding the minimal polynomial of an element of an extension of $\mathbb{Q}$. If it were so simple then we'd know whether or not $e+\pi$ is rational, for example!

But I suppose a trick that could save you some time is knowing that if $x=a+b\alpha$ then the minimal polynomial of $x$ has the same degree as the minimal polynomial of $\alpha$, and its conjugates are $a+b\beta$ for all $\beta$ conjugate to $\alpha$. (This is thanks to the fact that $\mathbb{Q}$-homomorphisms of $\mathbb{Q}(\alpha)$ correspond to permutations of algebraic conjugates of $\alpha$.)

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Why are the conjugates of $2+\sqrt{3}$ those mentioned above?

The elements of $\mathbb Q(\sqrt{3})$ are of the form $a + b\sqrt{3}$ where $a,b \in \mathbb Q$. Let $c + d\sqrt{3}$ be a conjugate (there might be more than one) of $a + b\sqrt{3}$, and consider the polynomial $$\begin{align*} (x-(a+b\sqrt{3})(x-(c+d\sqrt{3})) &= x^2 -x((a+c) + (b+d)\sqrt{3}) + (a+b\sqrt{3})(c+d\sqrt{3})\\ &= x^2 -x((a+c) + (b+d)\sqrt{3}) + ((ac+3bd)+(ad+bc)\sqrt{3}) \end{align*}$$ This is a polynomial in $\mathbb Q[x]$ if $d = -b$ and $ad+bc = 0$, that is, $c = a$. In other words, $$x^2 -2ax + (a^2-3b^2) = (x-(a+b\sqrt{3})(x-(a-b\sqrt{3}))$$ is a quadratic polynomial in $\mathbb Q[x]$ that has $a+b\sqrt{3}$ as a root, and so $x^2 -2ax + (a^2-3b^2)$ is the minimal polynomial of $a+b\sqrt{3}$ over $\mathbb Q$. The other root of this polynomial is the (unique) conjugate $a - b\sqrt{3}$ of $a+b\sqrt{3}$.

In particular, $2-\sqrt{3}$ is the conjugate of $2+\sqrt{3}$, and $\pm 2 \pm \sqrt{3}$ are not conjugates as you claim. The four elements constitute two sets of conjugate elements: $\{2+\sqrt{3}, 2-\sqrt{3}\}$ and $\{-2+\sqrt{3}, -2-\sqrt{3}\}$

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