# Algebraic element - integral domain

Let $K$ a field and $L$ a subfield of $K$. Let the set $\overline{L}:= \{k \in K: k$ is algebraic over $L$ $\}$ is another subfield of $K$. Show that $\overline{\overline{L}}=\overline{L}$. [Hint : If $c_0+c_1\alpha+ \dots + c_n\alpha^n=0$ where $\alpha \in K$ and $c_0,c_1, \dots, c_n \in \overline{L}$, then consider the field $L[c_0, c_1, \dots,c_n]$.]

I think we can use the following

Theorem : Let $A$ be an integral domain. There exists a field $F$ containing $A$ and such that if $G$ is another field containing $A$, then there exists a monomorphism $\phi : F \to G$ (i.e. $G$ contain an image isomorphic to $F$).

This exercise is hard to solve (for me). Is anyone is able to give me a good hint to resolve this question?

• Hint: any element $a$ in K which is algebraic over $\overline{L}$ satisfies a polynomial in coefficients from $\overline{L}$. These finite coefficients are themselves algebriac over L. So the extension inside K over L that contains the element $a$ is finite hence algebraic so it is already inside $\overline{L}$. – DBS Feb 3 '16 at 2:48

I assume that you know how to show that $\bar L$ is a field. (Tell me if this is unclear for you).

In order to show $\overline{\bar{L}}=\bar{L}$, you have to prove two parts :

• $\bar{L} \subseteq \overline{\bar{L}}$. This part is easy since for any subfield $M \subseteq K$, one has $M \subseteq \overline M$.

• $\overline{\bar{L}} \subseteq \bar{L}$. This is trickier. Let $x \in \overline{\bar{L}}$. I want to prove that there is some non-zero polynomial $P \in L[X]$ which has $x$ as root, i.e.

$$\exists (b_0,\dots, b_{r}) \in L^r \setminus \{\overrightarrow 0\} \quad b_rx^r+ \cdots b_1x+b_0=0 \tag 0$$

— By hypothesis, there exists some non-zero polynomial $Q(X)=a_nX^n+\cdots+a_0 \in \overline L[X]$ that vanishes at $X=x$.

— The idea is to use some linear algebra. More precisely, tha aim is to use the following result :

If $V$ is a finite dimensional vector space over a field $L$, then any collection $\{v_1, \dots, v_d, v_{d+1}\}$ of $d+1$ elements (with $d=\dim_L(V)$) is linearly dependent over $L$.

This means that you can find a vector $(b_1,\dots,b_{d+1})≠\overrightarrow{0}$ in $L$ such that $$b_1 \cdot v_1+\cdots + b_{d+1} \cdot v_{d+1}=0 \tag 1$$

— How can you link this result with your exercise ? Here are some hints : your vector space $V$ over the field $L$ will be $L[a_0, \dots , a_n][x]$. You have to show that it is finite dimensional over $L$, say with dimension $d$. First, you can show that $L' = L[a_0, \dots , a_n]$ is finite dimensional over $L$ :

Try to prove it by induction on $n$. When $n=0$, as $a_0$ is algebraic over $L$, it has a minimal polynomial $m$ of degree $r$. Then you prove that $[L[a_0] : L] = r < \infty$.

Similarly, you can show that $L'[x]$ is finite dimensional over $L'$ (think that $x$ is algebraic over $L'$). Therefore $V=L'[x]$ is finite dimensional over $L$.

— Then you have to "create", using your element $x \in \bar L$, a collection of $d+1$ elements in $V$. By the previous result , these $d+1$ elements will satisfies a relation as in $(1)$. Notice that $(0)$ is similar to $(1)$...

Think to $\{1,x,x^2,\dots, x^d\}$...

• dependent, not dependant – Pedro Tamaroff Feb 3 '16 at 21:46
• What is the difference between $L[a_0, \dots , a_n][x]$ and $L[a_0, \dots , a_n]$? Does $L[a_0, \dots , a_n][x] = L[a_0, \dots , a_n,x]$? – Taj Mohamed Bandalandabad Feb 4 '16 at 4:45
• @Watson We proved in class that $L[\alpha]$ is a finite vector space on $L$, and we also proved that $L[\alpha, \beta] = L[\alpha][\beta]$. Since two days, I try to resolve this problem. Are you able to complete the proof? – Taj Mohamed Bandalandabad Feb 4 '16 at 5:11
• @John : it is true that $L[a_0,…,a_n][x]=L[a_0,…,a_n,x]$. However, you only have $L[a_0,…,a_n] \subseteq L[a_0,…,a_n][x]$ at the beginning of the proof. But if you take $Q(X)$ as the minimal polynomial of $x$ over $\bar L$, then (since we prove that $x \in \bar L$) one has $Q(X) = X-x$, so that $L' = L'[x]$. But you don't know it before proving that $x \in \bar L$. Notice that my proof works even if $Q(X)$ isn't the minimal polynomial of $x$ over $\bar L$ ; in such a case you could have for instance $K= \Bbb R, L = \Bbb Q, x=\sqrt 2, Q(X)=X^2-2, L' = L[1,-2]=L ≠ L'[x]$. – Watson Feb 4 '16 at 10:24
• @John : for your second question, let $M=L[\alpha]$. Then you can prove that $M[\beta]$ is finite dimensional over $M$. Then the dimension of $L[\alpha, \beta] = M[\beta]$ over $L$ (often denoted by $[L[\alpha, \beta] : L]$) is the product $[L[\alpha, \beta] : L[\alpha] ] \cdot [L[\alpha] : L]$ which is finite. – Watson Feb 4 '16 at 10:29