# Element algebraic over field iff algebraic over field extension

Let $F$ be a field, and $K$ a finite extension of $F$. I want to show that any element algebraic over $K$ is algebraic over $F$, and conversely.

Well, if $a$ is algebraic over $F$, we can write $c_0+c_1a+\ldots+c_na^n=0$ for some $c_0,\ldots,c_n\in F$. But $c_0,\ldots,c_n\in K$, so $a$ is algebraic over $K$.

Now suppose $a$ is algebraic over $K$, so we can write $c_0+c_1a+\ldots+c_na^n=0$ for some $c_0,\ldots,c_n\in K$. What to do then?

• @GitGud $\sqrt{2}$ is algebraic over $\mathbb{Q}$... – Kunal Jan 18 '14 at 2:14
• My bad, sorry and thanks. – Git Gud Jan 18 '14 at 2:17

Hint: $$F(a) \subset K(a)$$
And $[K(a):F]=[K(a):K] [K:F] < \infty$. From here you should be able to get that
$$[F(a):F]< \infty \,.$$
• The last one using $[K(a):F]=[K(a):F(a)][F(a):F]$, right? – Kunal Jan 18 '14 at 2:15
• @Kunal Or even simpler, any subspace of a finite dimensional $F$-vector space is finite... Which in this case, is the same as saying that a subextension of a finite extension must be finite ... – N. S. Jan 18 '14 at 2:18