# If $f_{K(\alpha)}^{\beta}\big| f_{K}^{\beta}$, then $\deg\left(f_{K(\alpha)}^{\beta}\right)| \deg\left(f_{K}^{\beta}\right)$?

Let $$K\subset K(\alpha)\subset K(\alpha,\beta)$$ be field extensions. Then the question is whether the minimal polynomials of $$\beta$$ over $$K$$ and $$\beta$$ over $$K(\alpha)$$ have degrees where one divides the other; in other words, is it always true that

$$f_{K(\alpha)}^{\beta}\big| f_{K}^{\beta}\implies \deg\left(f_{K(\alpha)}^{\beta}\right)| \deg\left(f_{K}^{\beta}\right)?$$

I'm having trouble finding a counter-example (if it exists). Thanks for any suggestions or hints!

I have found something weird. I have proved the following: $$\gcd\bigg([K(\alpha):K],[K(\beta):K]\bigg)=1 \implies [K(\alpha,\beta):K]=[K(\alpha):K][K(\beta):K].$$ Now suppose that $$\gcd\bigg([K(\alpha):K],[K(\beta):K]\bigg)=1$$ which means that: $$K(\alpha)\cap K(\beta)=K$$, right?

W.l.o.g. assume that $$[K(\alpha):K]<[K(\beta):K]$$.

By the tower rule and the definition of minimal polynomials, we have $$[K(\alpha,\beta):K]=[K(\alpha,\beta):K(\alpha)]\cdot[K(\alpha):K]=\deg\left(f_{K(\alpha)}^{\beta}\right)\cdot\deg\left(f_{K}^{\alpha}\right).$$ By the theorem above, we get $$[K(\alpha,\beta):K]=[K(\alpha):K]\cdot[K(\beta):K]=\deg\left(f_{K}^{\alpha}\right)\cdot\deg\left(f_{K}^{\beta}\right).$$ Now we have $$\deg\left(f_{K(\alpha)}^{\beta}\right)=\deg\left(f_{K}^{\beta}\right)$$ as expected. So at least there is a certain condition for which the titled problem is true. How about the inequality where one of these degrees divides the other degree? Could someone help me with that?

• If I understood correctly, won't the following be a counterexample? Let $K=\Bbb{Q}$, $\alpha=\root3\of2$, $\beta=\omega\root3\of2$ where $\omega=e^{2\pi i/3}$ is the primitive third root of unity. The minimal polynomial of both $\alpha$ and $\beta$ over $K$ is $x^3-2$. But the minimial polynomial of $\beta$ over $K(\alpha)$ is $$(x^3-2)/(x-\alpha)=x^2+\alpha x+\alpha^2.$$ So the degrees are $3$ and $2$. – Jyrki Lahtonen Mar 7 at 5:32
• On the other hand, if $K(\alpha)/K$ is Galois, then the factors of $f_K^\beta(x)$ over $K(\alpha)$ are permuted by the Galois group, and you do get divisibility of degrees. This latter part at least has been covered on our site a few times. – Jyrki Lahtonen Mar 7 at 5:34
• @JyrkiLahtonen Indeed this example is sufficient, because we now have $$[K(\alpha)(\beta):K(\alpha)]=\deg\left(f^{\beta}_{K(\alpha)}\right)=\deg\left(x^2+\alpha x+\alpha^2\right)=2$$ and $$[K(\beta):K]=\deg\left(f^{\beta}_{K}\right)=\deg\left(x^3-2\right)=3$$ and $2\nmid3$. Thanks for suggesting! – Algebear Mar 7 at 17:14