# Thread: t is algebraic over F(S)

1. Hey!!

We have field extension $E/F$. We have that $S\subseteq E$ is algebraically independent over $F$. Let $t\in E$, then $S\cup \{t\}$ is algebraically independent over $F$.
I want to show that $t$ is algebraic over $F(S)$.

Since $S\subseteq E$ is algebraically independent over $F$ we have that there exist $s_1, \ldots s_n\in S$ (different from each other) and non-zero $f\in F[x_1, \ldots , x_n]$ such that $f(s_1, \ldots , s_n)\neq 0$.

Since $S\cup \{t\}$ is algebraically independent over $F$ we have that there exist $s_1, \ldots s_n, t\in S\cup \{t\}$ (different from each other) and non-zero $g\in F[x_1, \ldots , x_n, y]$ such that $g(s_1, \ldots , s_n, t)=0$.

To show that $t$ is algebraic over $F(S)$, we have to show that there exist $s_1, \ldots , s_n\in S$ and non-zero $h\in F[x_1, \ldots , x_n]$ such that $t=h(s_1, \ldots , s_n)$.
But how exactly can we show the existence of such a polynomial?

I tried now the following:

Since $S\cup \{t\}$ is algebraically independent over $F$ we have that there exist $s_1, \ldots s_n, t\in S\cup \{t\}$ (different from each other) and non-zero $g\in F[x_1, \ldots , x_n, y]$ such that $g(s_1, \ldots , s_n, t)=0$.

We collect all the terms with the same powers of $y$ and then we can write the polynomial as follows:
$$g(x_1, \ldots , x_n, y)=b_0+b_1y+\ldots b_m y^m$$ where $b_i\in F[x_1, \ldots , x_n]$.

Since $g$ is a non-zero polynomial we have that for exampe $b_m$ is non-zero. It cannot be that $b_0\neq 0$ and $b_i=0, i=1, \ldots , m$ because then we would have the polynomial $g(x_1, \ldots , x_n)$ such that $g(s_1, \ldots , s_n)=0$, and that contradicts to the fact that $S$ is algebraically independent over $F$, right?

Form the definition that $S\subseteq E$ is algebraically independent over $F$ do we have that for every non-zeropolynomial $h\in F[x_1, \ldots , x_n]$ there exist $s_1, \ldots s_n\in S$ (different from each other) such that $h(s_1, \ldots , s_n)\neq 0$ ?

If yes, then it would hold for the non-zero polynomial $b_m\in F[x_1, \ldots , x_n]$, so $b_m(s_1, \ldots , s_n)\neq 0$.

Then we would have the following $$g(s_1, \ldots , s_n, t)=0 \Rightarrow b_0(s_1, \ldots , s_n)+b_1(s_1, \ldots , s_n)t+\ldots b_m(s_1, \ldots , s_n) t^m=0$$

Therefore, we have a non-zero polynomial, where $t$ is a root.

Do we conclude from that that $t$ is algebraic over $F(S)$?

Since $S\cup \{t\}$ is algebraically independent over $F$ we have that there exist $s_1, \ldots s_n\in S$ (different from each other) and a non-zero polynomial $g\in F[x_1, \ldots , x_n, y]$ such that $g(s_1, \ldots , s_n, t)=0$.

We collect all the terms with the same powers of $y$ and then we can write the polynomial as follows:
$$g(x_1, \ldots , x_n, y)=b_0+b_1y+\ldots b_m y^m$$ where $b_i\in F[x_1, \ldots , x_n]$.

It cannot be that $b_i=0, i=1, \ldots , m$ because then we would have the polynomial $g(x_1, \ldots , x_n)$ such that $g(s_1, \ldots , s_n)=0$, and that contradicts to the fact that $S$ is algebraically independent over $F$, right?

So, we have that $b_i\neq 0$, for at least one $i\in \{1, \ldots , m\}$.

Then $t$ is a root of a non-zero polynomial of $F(S)$.

Therefore, $t$ is algebraic over $F(S)$.

Is everything correct?

4. Originally Posted by mathmari
Hey!!

We have field extension $E/F$. We have that $S\subseteq E$ is algebraically independent over $F$. Let $t\in E$, then $S\cup \{t\}$ is algebraically independent over $F$.
Do you mean that "Let $t\in E$, and $S\cup\{t\}$ be algebraically dependent over $F$?"

Originally Posted by mathmari
Since $S\subseteq E$ is algebraically independent over $F$ we have that there exist $s_1, \ldots s_n\in S$ (different from each other) and non-zero $f\in F[x_1, \ldots , x_n]$ such that $f(s_1, \ldots , s_n)\neq 0$.
What you infer is right but this is much weaker than what algeraic independence means.

Originally Posted by mathmari
Since $S\cup \{t\}$ is algebraically independent over $F$ we have that there exist $s_1, \ldots s_n, t\in S\cup \{t\}$ (different from each other) and non-zero $g\in F[x_1, \ldots , x_n, y]$ such that $g(s_1, \ldots , s_n, t)=0$.
Again, it seems you want to say that $S\cup\{t\}$ is algebraically dependent over $F$.

Originally Posted by caffeinemachine
Do you mean that "Let $t\in E$, and $S\cup\{t\}$ be algebraically dependent over $F$?"
Yes, you are right!!

Is my idea of #3 correct?

Originally Posted by caffeinemachine
What you infer is right but this is much weaker than what algeraic independence means.
What do you mean?

6. Originally Posted by mathmari
Is my idea of #3 correct?
You are certainly on the right track. But thee again, you seem to have typos. You write "algebraically independent" when your intention seems to be "algebraically dependent".

Originally Posted by mathmari
What do you mean?
Here is what I mean. Given an extension $E:F$, and a subset $S$ of $E$, we say that $S$ is algebraically independent over $F$ to mean the following: For any finite subset $\{s_1, \ldots, s_n\}$ of $S$, and any polynomial $f(x_1, \ldots, x_n)$ over $F$, if $f(s_1, \ldots, s_n)=0$, then $f=0$.