MHB It is an algebraically dependent set over F

[mathmari]

Hey!

Let $E/F$ be a field extension, $S\subseteq E$ and let $t\in E$ be algebraic over $F(S)$.

I want to show that there are distinct $s_1, \ldots , s_r\in S$, different from $t$, such that $\{s_1, \ldots , s_r, t\}$ is an algebraically dependent set over $F$.
I have done the following:
We have that $t\in E$ is algebraic over $F(S)$, i.e., there is a non-zero $f\in F(S)[x]$ such that $f(t)=0$, right? (Wondering)

To show that $\{s_1, \ldots , s_r, t\}$ is an algebraically dependent set over $F$, we have to show that there is a non-zero polynomial $g\in F[x_1, \ldots , x_r, y]$ such that $g(s_1, \ldots , s_r, t)=0$, right? How can we use the polynomial $f$ for that? (Wondering)

 

[caffeinemachine]

Hey!

Let $E/F$ be a field extension, $S\subseteq E$ and let $t\in E$ be algebraic over $F(S)$.

I want to show that there are distinct $s_1, \ldots , s_r\in S$, different from $t$, such that $\{s_1, \ldots , s_r, t\}$ is an algebraically dependent set over $F$.
I have done the following:
We have that $t\in E$ is algebraic over $F(S)$, i.e., there is a non-zero $f\in F(S)[x]$ such that $f(t)=0$, right? (Wondering)

To show that $\{s_1, \ldots , s_r, t\}$ is an algebraically dependent set over $F$, we have to show that there is a non-zero polynomial $g\in F[x_1, \ldots , x_r, y]$ such that $g(s_1, \ldots , s_r, t)=0$, right? How can we use the polynomial $f$ for that? (Wondering)

The coefficients of $f$ are elements of $F(S)$. Thus each coefficient of $f$ looks like a ratio of two polynomials over $F$ evaluated on some members of $S$. Thus there are $s_1, \ldots, s_r\in S$ such that each coefficient of $f$ is of the form $p(s_1, \ldots, s_r)/q(s_1, \ldots, s_r)$, where $p$ and $q$ are polynomials over $F$. Now if you "rationalize" $f$ you will get your required polynomial showing $s_1, \dots, s_r, t$ are algebraically dependent over $F$.

 

[mathmari]

The coefficients of $f$ are elements of $F(S)$. Thus each coefficient of $f$ looks like a ratio of two polynomials over $F$ evaluated on some members of $S$. Thus there are $s_1, \ldots, s_r\in S$ such that each coefficient of $f$ is of the form $p(s_1, \ldots, s_r)/q(s_1, \ldots, s_r)$, where $p$ and $q$ are polynomials over $F$. Now if you "rationalize" $f$ you will get your required polynomial showing $s_1, \dots, s_r, t$ are algebraically dependent over $F$.

We have that $t\in E$ is algebraic over $F(S)$, so for a positive integer $r$ there are elements $s_1, \ldots , s_r\in S$ and a non-zero polynomial $f\in F(s_1, \ldots , s_r)[x]$ such that $f(t)=0$.

Since $f\in F(s_1, \ldots , s_r)[x]$ the polynomial is of the form $$f=\frac{p_0(s_1, \ldots , s_0)}{q_0(s_1, \ldots , s_r)}+\frac{p_1(s_1, \ldots , s_r)}{q_1(s_1, \ldots , s_r)}x+\ldots +\frac{p_n(s_1, \ldots , s_r)}{q_n(s_1, \ldots , s_r)}x^n$$ with $p,q$ polynomials in $F$ and with $\frac{p_n(s_1, \ldots , s_r)}{q_n(s_1, \ldots , s_r)}\neq 0$.

Let $h$ be the lcm of the $q_i$'s. It holds that $h\neq 0$.

When we multiply $f$ by $h$, the coefficients of $f$ are now polynomials. So,
$$\tilde{f}(x)=f\cdot h=h_0(s_1, \ldots , s_r)+h_1(s_1, \ldots , s_r)x+\ldots +h_n(s_1, \ldots , s_r)x^n$$
Let $g(s_1, \ldots , s_r, x):=\tilde{f}(x)$. Since $\tilde{f}$ is non-zero, it follows that $g$ is non-zero.

So, we have a non-zero polynomial $g\in F[x_1, \ldots , x_r, y]$ with $g(s_1, \ldots , s_r, t)=\tilde{f}(t)=0$. This is the definition that the set $\{s_1, \ldots , s_r, t\}$ is algebracailly dependent.

Is everything correct? (Wondering)

 

[caffeinemachine]

We have that $t\in E$ is algebraic over $F(S)$, so for a positive integer $r$ there are elements $s_1, \ldots , s_r\in S$ and a non-zero polynomial $f\in F(s_1, \ldots , s_r)[x]$ such that $f(t)=0$.

Since $f\in F(s_1, \ldots , s_r)[x]$ the polynomial is of the form $$f=\frac{p_0(s_1, \ldots , s_0)}{q_0(s_1, \ldots , s_r)}+\frac{p_1(s_1, \ldots , s_r)}{q_1(s_1, \ldots , s_r)}x+\ldots +\frac{p_n(s_1, \ldots , s_r)}{q_n(s_1, \ldots , s_r)}x^n$$ with $p,q$ polynomials in $F$ and with $\frac{p_n(s_1, \ldots , s_r)}{q_n(s_1, \ldots , s_r)}\neq 0$.

Let $h$ be the lcm of the $q_i$'s. It holds that $h\neq 0$.

When we multiply $f$ by $h$, the coefficients of $f$ are now polynomials. So,
$$\tilde{f}(x)=f\cdot h=h_0(s_1, \ldots , s_r)+h_1(s_1, \ldots , s_r)x+\ldots +h_n(s_1, \ldots , s_r)x^n$$
Let $g(s_1, \ldots , s_r, x):=\tilde{f}(x)$. Since $\tilde{f}$ is non-zero, it follows that $g$ is non-zero.

So, we have a non-zero polynomial $g\in F[x_1, \ldots , x_r, y]$ with $g(s_1, \ldots , s_r, t)=\tilde{f}(t)=0$. This is the definition that the set $\{s_1, \ldots , s_r, t\}$ is algebracailly dependent.

Is everything correct? (Wondering)

Yes.

 

[mathmari]

Yes.

Thank you so much! (Smile)