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Ok, so I recently asked a very similar question here Finding the Galois group of $f(t)=t^{3}-4t+2$ over $\mathbb{Q}$.

However, $f(t)=t^{3}-5t^{2}+11t-15=(t-3)(t^{2}-2t+5)$. So the roots of $f$ are $3$, $1-2i$, and $1+2i$. Presumably I should consider the extension $\mathbb{Q}(3,1+2i,1-2i)$.

First of all, is this correct?

Second, am I right in saying that $\mathbb{Q}(3,1+2i,1-2i)=\mathbb{Q}(3,1+2i)$?

Third, if this is correct, where do I go from here?

Thank you.

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  • $\begingroup$ $\mathbb{Q}(3,1+2i)=\mathbb{Q}(i)$ $\endgroup$
    – lhf
    Mar 31, 2017 at 11:38
  • $\begingroup$ Ok, yes that seems really obvious now. Thank you. So I have to determine $\text{Gal}(\mathbb{Q}(i)/\mathbb{Q})$? $\endgroup$
    – PercyF2519
    Mar 31, 2017 at 11:40

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You're right in saying that $\mathbb{Q}(3,1+2i,1-2i)=\mathbb{Q}(3,1+2i)$.

Now note that $\mathbb{Q}(3,1+2i)=\mathbb{Q}(i)$.

So the splitting field of $f$ is $\mathbb{Q}(i)$ and you have to find $\text{Gal}(\mathbb{Q}(i)/\mathbb{Q})$, which has order $2$...

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  • $\begingroup$ Since it is of order $2$: $|\text{Gal}(\mathbb{Q}(i)/\mathbb{Q})|=2$, so $\text{Gal}(\mathbb{Q}(i)/\mathbb{Q})\cong\mathbb{Z}/2\mathbb{Z}$ yes? $\endgroup$
    – PercyF2519
    Mar 31, 2017 at 11:48
  • $\begingroup$ @PercyF2519, yes, generated by complex conjugation. $\endgroup$
    – lhf
    Mar 31, 2017 at 11:49
  • $\begingroup$ Woohoo! Thank you or your help. $\endgroup$
    – PercyF2519
    Mar 31, 2017 at 11:50
  • $\begingroup$ Out of interest, I tried another way of doing this - by calculating the discriminant of $f$ which turns out to be $-1024$. What does this tell us? I've not done an example with negative discriminant (I thought it had to be positive). $\endgroup$
    – PercyF2519
    Mar 31, 2017 at 11:52

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