1
$\begingroup$

Let $k$ be a field and $f(x)\in k[x]$. Let $g(x) = f(\alpha x + \beta)$ for some $\alpha, \beta \in k, \alpha\neq 0$. Prove that $f(x)$ and $g(x)$ have the same discriminants and Galois groups.

I have evaluated the case for when the discriminant is 0, but I'm confused as where to go with the non-zero case.....

  • 0
    One thing to notice is that the map $k[x] \rightarrow k[x]$ that takes $x \mapsto \alpha x + \beta$ is an isomorphism because $\alpha \neq 0$ and $k$ is a field, and $g(y)$ is the image of $f(x)$ under this isomorphism.2012-11-19
  • 0
    Also, if $\gamma$ is a root of $f$, then $\frac{\gamma - \beta}{\alpha}$ is a root of $g$.2012-11-19
  • 0
    So, there is a bijection between the roots of $f(x)$ and the roots of $g(x)$?2012-11-19
  • 0
    They have the same splitting field Mike.2012-11-19
  • 2
    Are you sure the question asks you to prove they have the same discriminant? I think my observation implies that $Disc(g) = \frac{1}{\alpha^2}Disc(f)$, so may be I am doing something incorrect.2012-11-19
  • 0
    I noticed this as I did the calculation myself, and yes, it asks to show they have the same discriminants in the question. Is there an explicit example we could try to disprove this?2012-11-19
  • 0
    Correcting the exponent in an earlier comment of mine: as @Rankeya and Mike M. observe, there is a power of $a$ appearing in the discriminant. As in the answer below, the power is ${1\over 2}n(n-1)$...2012-11-19
  • 0
    I made a comment to the answer below, but I am still wondering if this makes a difference. If $a\neq 1$, I think the discriminants are not equal as stated in my problem.2012-11-19
  • 0
    I apologize for getting the power of $\alpha$ wrong.2012-11-19
  • 0
    I mentioned it to my prof and the statement that the discriminants are the same is false. Thank you everyone for the input2012-11-20

1 Answers 1

1

Up to a sign the discriminant is the product of the differences of the roots.

If $r_i$ are the roots of $f$ the the roots of $g$ are $(r_i-b)/a$ so the differences are $(r_i-r_j)/a$.

Thus the discriminant of $g$ is $1/a^m$ times the discriminant of $f$, where $m=n(n-1)$, and $n$ is the degree.

  • 0
    But isn't the discriminant $f$ defined as $\prod_{i where $\gamma_k$'s are the roots of $f$, so then the factor of $1/\alpha^m$ would reduce to $1/\alpha^{n(n-1)}$ as you define it?2012-11-19
  • 0
    @MikeM. you're right2012-11-19