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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.....

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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
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
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.

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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

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