Algebraic conjugates
Andrew Mclaughlin 
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
$\endgroup$3 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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 $\alpha$.)
$\endgroup$ 0 $\begingroup$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 (theremight 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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