# What are $\mathbb Q$-conjugates?

I‘m reading this paper from Keith Conrad and in Example 1.1 he states:

I don‘t the term "$$\mathbb Q$$-conjugates" - what does that mean? I don‘t get why $$\sqrt 2 \mapsto \sqrt 3$$ is not possible in this case.

• $\mathbb Q(\sqrt2)$ is not isomorphic to $\mathbb Q(\sqrt3)$, because the former has a solution to $x^2=2$ and the latter does not Commented Feb 13 at 21:09
• Numbers are $\mathbf Q$-conjugates when they have the same minimal polynomial over $\mathbf Q$. An automorphism over a field $K$ sends numbers to other numbers that are roots of the same polynomial in $K[x]$, so they must go to numbers with the same minimal polynomial in $K[x]$. I'm curious: where are you taking a course that is discussing Galois extensions and Galois groups without yet hearing the term "$K$-conjugate" in a field extension $L/K$?
– KCd
Commented Feb 13 at 21:25

If $$K/F$$ is a field extension, which is to say $$K$$ is a field that contains $$F$$ as a subfield, and $$\alpha$$ is an element of $$K$$ but not $$F$$, then the $$F$$-conjugates of $$\alpha$$ are the elements that satisfy all polynomial expressions in $$F$$ that $$\alpha$$ also satisfies. Equivalently, you consider the Galois group $$G$$ of the extension, and say that $$\beta$$ is a conjugate of $$\alpha$$ if there's some $$\sigma \in G$$ such that $$\sigma(\alpha) = \beta$$.
In this specific case, we have $$K = \mathbb{Q}(\sqrt 2, \sqrt 3)$$. The $$\mathbb{Q}$$-conjugates of $$\sqrt 2$$ are all the $$\mathbb{Q}$$-linear combinations of $$\sqrt 2, \sqrt 3,$$ and $$1$$ that satisfy the same rational polynomials that $$\sqrt 2$$ does. The automorphism $$\sigma$$ defined by
$$\sigma(a + b\sqrt 2 + c\sqrt 3) = a - b\sqrt 2 + c\sqrt 3$$
For example, we have $$(\sqrt 2)^2 - 2 = 0$$, and $$(-\sqrt 2)^2 - 2 = 0$$. In fact, any polynomial with rational coefficients that has $$\sqrt 2$$ as a root also has $$-\sqrt 2$$ as a root - you can prove this as an exercise. By contrast, $$(\sqrt 3)^2 - 2 = 1 \neq 0$$, so $$\sqrt 3$$ is not a $$\mathbb{Q}$$-conjugate of $$\sqrt 2$$. Equivalently, there is no $$\tau \in \mathrm{Gal}(\mathbb{Q}(\sqrt 2, \sqrt 3)/\mathbb{Q})$$ such that $$\tau(\sqrt 2) = \sqrt 3)$$.