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

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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.

  • $\begingroup$ $\mathbb Q(\sqrt2)$ is not isomorphic to $\mathbb Q(\sqrt3)$, because the former has a solution to $x^2=2$ and the latter does not $\endgroup$ Commented Feb 13 at 21:09
  • 3
    $\begingroup$ 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 $L/K$? $\endgroup$
    – KCd
    Commented Feb 13 at 21:25

1 Answer 1


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

$$\sigma(a + b\sqrt 2 + c\sqrt 3) = a - b\sqrt 2 + c\sqrt 3$$

demonstrates this conjugacy.

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 $\tau(\sqrt 2) = \sqrt 3)$.


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