Let $ax^2+bx+c$ be in $\mathbb{Q}[x]$, where $a\ne 0$. I want to find a primitive element for a splitting field of $f$ over $\mathbb{Q}$.

I'm not sure if I get the idea completely. To find the splitting field, I need to do the following:

(1) Find the zeroes of $f$, which there are two, $\alpha_1, \alpha_2$.

(2) Adjoin the two roots to make the extension $\mathbb{Q}(\alpha_1, \alpha_2)$ and show that $\mathbb{Q}(\alpha_1, \alpha_2)=\mathbb{Q}(\alpha_1+ \alpha_2)$, correct?

(3) Thus $\alpha_1+\alpha_2$ is a primitive element of the field?

Would appreciate if someone could please clarify this for me.


You can factor your polynomial as

$$ax^2 + bx + c = a(x-\alpha_1)(x-\alpha_2) = a(x^2 - (\alpha_1 + \alpha_2)x + \alpha_1 \alpha_2)$$

From the last expression, you can see that the coefficient of $x$ is $b = -a(\alpha_1 + \alpha_2)$, i.e., the sum of the roots is $-b/a \in \mathbb{Q}$.

And so when you adjoin just one of the roots, say, $\alpha_1$, you are able to generate the other one as well: $-b/a = \alpha_1 + \alpha_2 \implies -b/a - \alpha_1 = \alpha_2$. Since just one of the roots suffices to get you the other, you may conclude that $\mathbb{Q}(\alpha_1) = \mathbb{Q}(\alpha_1, \alpha_2)$.

Note that the above shows $\mathbb{Q}(\alpha_1 + \alpha_2) = \mathbb{Q}$, so this will not be the same as the splitting field $\mathbb{Q}(\alpha_1)$ if $\alpha_1 \not\in \mathbb{Q}$. You can adjoin either root to get the other, but adjoining their sum gets you nothing new!

  • $\begingroup$ If $f$ has two roots $\alpha_1, \alpha_2$, then shouldn't we be able to express it as $(x-\alpha_1)(x-\alpha_2)$? I'm wondering why you're expressing it as $a(x-\alpha_1)(x-\alpha_2)$. Also, what if $f$ has the roots $\sqrt{2}$ and $\sqrt{3}$? Then $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$. $\endgroup$ – sequence Mar 30 '17 at 18:19
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    $\begingroup$ Now I see what why $\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$. The difference is that a polynomial with these two roots has coefficients that are irrational. However, for the first of my question above, it appears that we can factor a 2nd degree polynomial as $(x-\alpha_1)(x-\alpha_2)$ only if it is monic. But if it is not, then there is a different procedure of factoring it... $\endgroup$ – sequence Mar 30 '17 at 20:38
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    $\begingroup$ If it is not monic, then factor out the $a$ first. $\endgroup$ – Benjamin Dickman Mar 30 '17 at 21:03
  • $\begingroup$ What if f has no roots? $\endgroup$ – Maggie Mak Mar 31 '17 at 2:13
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    $\begingroup$ @MaggieMak By the fundamental theorem of algebra, the polynomial has a root in $\mathbb{C}$. For example, if $f$ is defined by $x \mapsto x^2 + 1$ then its roots are $\pm \sqrt{-1} = \pm i$. And, indeed, we would have $\mathbb{Q}(i) = \mathbb{Q}(i, -i)$ as the desired splitting field, whereas adding the roots together would give $-i + i = 0$, so that the extension $\mathbb{Q}(-i + i) = \mathbb{Q}(0) = \mathbb{Q}$ does not contain the roots of $f$. $\endgroup$ – Benjamin Dickman Mar 31 '17 at 2:32

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