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Given an irreducible monic polynomial $f(X) \in \mathbb{Q}[X]$ of degree $d$, we can construct extension of $\mathbb{Q}$ as a quotient field $S = \mathbb{Q}[t] / \langle f(t) \rangle $. Considered as polynomial in $S[X]$, $f[X]$ has some roots (at least one).

I wonder, whether there is a systematic way to list roots of $f[X]$ which belong to $S$ represented as elements of $\mathbb{Q}[t] / \langle f(t) \rangle$ (i.e. polynomials over $t$ with coefficients in $\mathbb{Q}$ of degree less than $d$)? Obviously, one of the roots is just $t$ since

$$ f(t) \equiv 0 \mod f(t) $$

so, can I find the other roots (if exist) in $S$ without doing a complete factorization?


Example 1: for $f(X) = X^4 + 1$ and extension field $S = \mathbb{Q}[t] / \langle t^4 + 1 \rangle $, the roots are $\pm t$, $\pm t^3$

Example 2: for $f(X) = X^2 + 2X + 3$ and extension field $S = \mathbb{Q}[t] / \langle t^2 + 2t + 3 \rangle $, the roots are $t$, $- 2 - t$

Example 3 (thanks @dan_fulea): for $f(X) = X^3 - 2$ and extension field $S = \mathbb{Q}[t] / \langle t^3 - 2 \rangle $, there are no other roots than $t$

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  • $\begingroup$ The other conjugates may not live in $\Bbb Q[t]/(f)$. (Here i am using the standard notation $(f)$ for the ideal generated by $f$.) Which is exactly the question (in this situation, when only some or no roots live in this particular extension)? $\endgroup$ – dan_fulea Jun 8 '18 at 20:45
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    $\begingroup$ Sorry, I don't understand your comment. Splitting field, by definition, is that the poly $f$ splits into linear factors, that is has all roots living in the splitting field. So all the roots should live in $\mathbb{Q}[t]/(f(t))$ (given that $f$ is irreducible over $\mathbb{Q}$). Or I miss something? $\endgroup$ – Stanislav Poslavsky Jun 8 '18 at 20:52
  • $\begingroup$ For example $t \in \mathbb{Q}[t]/(f)$ is always a root of $f(X)$ $\endgroup$ – Stanislav Poslavsky Jun 8 '18 at 20:53
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    $\begingroup$ Why is the splitting field exactly $\Bbb Q[t]/(f)$? OK, let us use an example, $f=X^3-2$. Then $K=\Bbb Q[X]/(X^3-2)$ contains exactly one root of $f$. We can embeed this $K$ into $\Bbb Q[\sqrt [3]2]$ by mapping the class of $X$ to $a:=\sqrt [3]2$. To get the full splitting we need to adjoin the third root of unity, $\Bbb Q(a):\Bbb Q$ is not Galois. $\endgroup$ – dan_fulea Jun 8 '18 at 20:59
  • $\begingroup$ you are right and I was wrong! I've corrected the question. $\endgroup$ – Stanislav Poslavsky Jun 8 '18 at 21:12

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