Analog to the primitive element theorem for transcendental extensions?

The Primitive Element Theorem states that if $E/F$ is a finite separable field extension, then there exists an element $a$ such that $E=F(a)$.

There's a similar result I found, that I don't quite fully understand. For instance, let $F$ be a field and $F[a_1,\dots,a_n]$ be a finite separable extension. Suppose also that $F_u=F(u_1,\dots,u_n)$ is a purely transcendental extension of $F$, with $u_1,\dots,u_n$ algebraically independent over $F$. Why is it true that $F_u[a_1,\dots,a_n]=F_u[u_1a_1+\cdots+u_na_n]$?

I get that $u_1a_1+\cdots+u_na_n\in F_u[a_1,\dots,a_n]$, and so $F_u[u_1a_1+\cdots+u_na_n]\subseteq F_u[a_1,\dots,a_n]$. However, why is the converse true? I tried reproducing the argument for the primitive element theorem without success. Thank you.

Since the $u_i$ are algebraically independent over $F$, and $F_u[a_1,\dots,a_n]$ is finitely generated over $F_u$, I was trying to use the corollary Bill Cook pointed me to, to conclude that by setting $y_1=\sum_{j=1}^n u_ja_j$, then $F_u[a_1,\dots,a_n]$ is integral over $F_u[y_1]$. From this can I conclude the equality? I wary of how to proceed, as I do not know where separability comes into use.

• I wonder if you're supposed to assume that $u_1, \ldots, u_n$ are algebraically independent (i.e. the transcendence degree of $F_u / F$ is $n$)? – Zhen Lin Feb 3 '12 at 18:13
• Dear @Zhen Lin, that hypothesis slipped my mind. I have now added it, thank for pointing it out. – Waldott Feb 3 '12 at 20:59
• In other words, $u_1,\ldots,u_n$ is a transcendence basis for $F_u$ over $F$. – Arturo Magidin Feb 3 '12 at 21:05
• You might want to take a look at Serge Lang's Algebra: Chapter VIII Theorem 2.2 and surrounding text -- it's very closely related to your question. In my "revised 3rd edition" it's on page 358. – Bill Cook Feb 3 '12 at 21:24

I think this statement is proven exactly as the primitive element theorem is proven. I'll give you the argument in the case $n=2$, courtesy of J. S. Milne's notes on field theory, from which the general case follows.

Let $E=F[a,b]$ with $b$ separable over $F_u$. Let $f$ and $g$ be the minimum polynomials of $a$ and $b$ respectively over $F$. Let $a=a_1, a_2, \ldots ,a_s$ be the roots of $f$ in some field $K$ containing $E$, and let $b=b_1, b_2, \ldots ,b_t$ be the roots of $g$. For $j \neq 1$, we have $b_j \neq b_1$ so $$a_i +Xb_j=a_1 +Xb_1$$ has exactly one solution: $X= \displaystyle \frac{a_j -a_i}{b_1-b_j}$. This means that $a_i + \frac{u_2}{u_1} b_j \neq a + \frac{u_2}{u_1} b$ in $K_u$ unless $i=1=j$.

Now let $\gamma = a+\frac{u_2}{u_1}b$. Then the polynomials $g(X)$ and $f(\gamma-\frac{u_2}{u_1}X)$ have coefficients in $F_u[\gamma]$, and have $b$ as a root: $$g(b)=0, f(\gamma-\frac{u_2}{u_1}b)=f(a)=0.$$

Moreover, $b$ is their only common root, because $\gamma - \frac{u_2}{u_1}b_j \neq a_i$ unless $i=1=j$. Therefore $$\gcd(g(X),f(\gamma-\frac{u_2}{u_1}X))=X-b.$$

Though we've computed the gcd in some field that splits $fg$, it is a simple fact that the gcd of two polynomials has coefficients in the same field as the coefficients of the polynomials. Hence $b \in F_u[\gamma]$, and this implies that $a=\gamma - \frac{u_2}{u_1}b$ lies in $F_u[\gamma]$. Hence, $F_u[a,b]=F_u[\gamma]=F_u[a+\frac{u_2}{u_1}b]=F_u[au_1+bu_2]$.

Let $\alpha$ be a primitive element for $F[a_1,...,a_n]/F$ and let $d=[F(\alpha):F]$. By looking at the proof of the PET, we can take $\alpha$ to be an $F$-linear combination of the $a_i$, $\alpha=\sum b_i a_i$.

It suffices to prove that $[F_u[a_1 u_1+\cdots + a_n u_n]:F_u]=d$. So suppose $f(a_1 u_1+\cdots + a_n u_n)=0$ where $f(T)\in F_u[T]$ is nonzero. Let $k=deg_T(g)$. Viewing $f(a_1 u_1+\cdots + a_n u_n)$ as an element $g(u_1,...,u_n)\in F[a_1,...,a_n](u_1,...,u_n)$, the transcendence of the $u_i$'s implies that $g=0$. In particular, $g(b_1,...,b_n)=0$. But $g(b_1,...,b_n)$ is a degree $k$ polynomial in $\alpha=\sum b_i a_i$, hence $k\geq d$.