Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.....

share|improve this question
    
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. –  Rankeya Nov 19 '12 at 7:30
    
Also, if $\gamma$ is a root of $f$, then $\frac{\gamma - \beta}{\alpha}$ is a root of $g$. –  Rankeya Nov 19 '12 at 7:41
    
So, there is a bijection between the roots of $f(x)$ and the roots of $g(x)$? –  Mike M. Nov 19 '12 at 7:54
    
They have the same splitting field Mike. –  JSchlather Nov 19 '12 at 8:23
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. –  Rankeya Nov 19 '12 at 8:30
show 5 more comments

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
    
But isn't the discriminant $f$ defined as $\prod_{i<j}(\gamma_i - \gamma_j))^2$ 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? –  Mike M. Nov 19 '12 at 15:06
    
@MikeM. you're right –  i. m. soloveichik Nov 19 '12 at 15:14
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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