MHB Is t Algebraic Over F(S) in Field Extension E/F?

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Hey! :o

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? (Wondering)
 
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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? (Wondering) 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 $h(s_1, \ldots , s_n)\neq 0$ ? (Wondering) 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 $F(S)$? (Wondering)
 
I thought about it again...

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? (Wondering)
 
mathmari said:
Hey! :o

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 $F$?"

mathmari said:
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.

mathmari said:
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 $F$.
 
caffeinemachine said:
Do you mean that "Let $t\in E$, and $S\cup\{t\}$ be algebraically dependent over $F$?"

Yes, you are right! (Blush)

Is my idea of #3 correct? (Wondering)
caffeinemachine said:
What you infer is right but this is much weaker than what algeraic independence means.

What do you mean? (Wondering)
 
mathmari said:
Is my idea of #3 correct? (Wondering)

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".

mathmari said:
What do you mean? (Wondering)

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 $f=0$.
 
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