I am looking to find the Galois group of $x^3-x+t$ over $\mathbb{C}(t)$, the field of rational functions with complex coefficients. I have shown that the automorphisms of the rational function field $F(t)$ for fixed $F$ are precisely the fractional linear transformations that is $t \rightarrow \frac{at +b}{ct+d}$ for $a,b,c,d \in \mathbb{C}$. Is this useful? Also is there anyway to factor $x^3-x+t$ nicely?

I slept on this for a little bit and developed an idea to show this. I used Cardano's method to explicitly solve for the roots of this polynomial and show that there exist no linear factors in $\mathbb{C}(t)$ and $f(x)$ is therefore irreducible. This is because the polynomial is cubic, and if there are no linear factors then there cannot be any quadratic factors. Thus, you have to adjoin some root let's call it $\theta$ to $\mathbb{C}(t)$. The degree of this field over $\mathbb{C}(t)$ is a Galois extension and must have degree 3. The only group with order 3 is $\mathbb{Z}_3$, which implies this is the Galois group.

  • 2
    $\begingroup$ You can easily determine the Galois group of a cubic polynomial simply by computing its discriminant and deciding whether it is a square or not. $\endgroup$ Feb 17, 2015 at 23:33
  • $\begingroup$ See www.math.uconn.edu/~kconrad/blurbs/galoistheory/cubicquarticallchar.pdf $\endgroup$ Feb 17, 2015 at 23:36
  • $\begingroup$ Ah because if the discriminant is square then the Galois group is isomorphic to $A_3$ and $S_3$ otherwise. Thank you! $\endgroup$ Feb 17, 2015 at 23:42
  • 1
    $\begingroup$ What you wrote is not correct. Your argument is, essentially, that as the polynomial is irreducible, adjoining one of its roots gives a normal extwnsion, and that is very false. $\endgroup$ Feb 24, 2015 at 5:26
  • 1
    $\begingroup$ Chiming in with Mariano. Why do you get a Galois extension by adjoining $\theta$? $\endgroup$ Feb 24, 2015 at 7:12

2 Answers 2


I don't see how to use Eisenstein here. But the polynomial is a cubic, so if it factors, then one of the factors is linear, and the polynomial would have a zero $x=\frac{p(t)}{q(t)}\in\Bbb{C}(t)$ with some polynomials $p(t),q(t)\in\Bbb{C}[t]$, $q\neq0$.

To exclude this possibility we use a line of reasoning analogous to the familiar rational root test - taking advantage of the fact that $\Bbb{C}[t]$ is a UFD. So assume that all common factors of $p(t)$ and $q(t)$ have been cancelled. Then $$ x^3-x+t=\frac{p^3-pq^2+tq^3}{q^3}=0. $$ Here the numerator has to be zero, so from $$ p^3-pq^2=-tq^3 $$ we can conclude that $p$ divides the left hand side, hence also the right hand side. But $p$ has no common factors with $q$, so it has to be a factor of $t$.

Similarly from $$ p^3=pq^2-tq^3 $$ we see that $q$ divides the right hand side, hence also $p^3$. But, as above, this implies that $q$ must be a constant.

The non-zero constants are the units of $\Bbb{C}[t]$, so we can conclude that $x=p/q$ is either a constant or, $x=at$ for some $a\in\Bbb{C}$.

I'm sure that you can show that neither of those work. Therefore this cubic has no linear factors over $\Bbb{C}(t)$ and hence it is irreducible.

  • $\begingroup$ thank you for your help. I offer an alternative solution, do you see any problems with it? $\endgroup$ Feb 24, 2015 at 5:21

Hint: Show that the polynomial is irreducible over $\mathbb C[t]$ (which shows irreducibility over $\mathbb C(t)$ by Gauss) and then compute the discriminant as usual.

  • 1
    $\begingroup$ I know showing irreducibility over $\mathbb{Q}$ can be done using Eisenstein's Criterion. Is there a similar trick for $\mathbb{C}$? $\endgroup$ Feb 17, 2015 at 23:43
  • 1
    $\begingroup$ Eisenstein's criterion works over the field of fractions of a UFD, such as $\mathbb Z$ or $\mathbb C[t]$. $\endgroup$ Feb 17, 2015 at 23:47
  • $\begingroup$ After substituting $x+a$ for some integer $a$, I believe then I will be able to show the polynomial to satisfy Eisenstein's critertion. Thank you @MarianoSuárez-Alvarez and @Mesih! $\endgroup$ Feb 17, 2015 at 23:53
  • $\begingroup$ By Gauss, irreducibility in $\mathbf C(T)[X]$ is (essentially) equivalent to irreducibility in $\mathbf C[T,X]$, which (here) follows from the fact that the degree in $T$ is $1$. $\endgroup$
    – ACL
    Jun 26, 2019 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.