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Let $f(x)$ be an irreducible cubic polynomial in $\mathbb{Q}[x]$ with roots $x_{1}, x_{2}$, and $x_{3}$ (necessarily all distinct since $f(x)$ is irreducible over $\mathbb{Q}$). And let $D$ be the discriminant of $f(x)$.

I don't completely understand why the splitting field of $f(x)$ over $\mathbb{Q}$ is $\mathbb{Q}(x_{1}, \sqrt{D})$.

The proof is basically the following.

We can assume that $f(x)$ is monic since multiplying a polynomial by a constant doesn't change its roots or its discriminant.

Because $x_{1},x_{2}$, and $x_{3}$ are the roots of $f(x)$, $$f(x) = (x-x_{1})(x-x_{2})(x-x_{3}) = (x-x_{1})g(x),$$ where $g(x_{1}) \ne 0$.

Then $\mathbb{Q}(x_{1}, x_{2}, x_{3})= \mathbb{Q}(x_{1}, \sqrt{\text{disc}\, g}) $.

And because $D = g(x_{1})^2 \cdot \text{disc} \, g$, $\mathbb{Q}(x_{1}, \sqrt{\text{disc}\, g}) = \mathbb{Q}(x_{1}, \sqrt{D})$.

I think the reason $\mathbb{Q}(x_{1}, x_{2}, x_{3})= \mathbb{Q}(x_{1}, \sqrt{\text{disc}\, g})$ is because the splitting field of $g(x)$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{\text{disc} \, g})$.

But why does $D = g(x_{1})^2 \cdot \text{disc} \, g$? What property is being used here?

And why does $\mathbb{Q} \left(x_{1}, \sqrt{\frac{D}{g(x_{1})^{2}}} \right) = \mathbb{Q}(x_{1}, \sqrt{D})?$

user232456
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1 Answers1

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Let $\Delta$ be the discriminant of $g$. If you add $\sqrt{\Delta}$ to $\mathbb{Q}(x_1)$, you get the roots of $f$. Conversely, if you have the roots $x_2$ and $x_3$ in an extension of $\mathbb{Q}(x_1)$, this extension contains $\Delta$. So the splitting field for $f$ is indeed $\mathbb{Q}(x_1,\sqrt{\Delta})$.

The discriminant of $f$ is, by definition, $$ D=(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=(g(x_1))^2\Delta $$ because $\Delta=(x_2-x_3)^2$. Note that $g(x_1)\in\mathbb{Q}(x_1)$ by definition of $g$. So $$ \sqrt{D}=\pm g(x_1)\sqrt{\Delta}\in\mathbb{Q}(x_1,\sqrt{\Delta}) $$

egreg
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