4

Maybe somebody knows how to proove the following algebraic theorem:

$C \subset U$ is a field extension and $N \subset U$ so, that all $x \in N$ are algebraic over $C$ and $C[N]=\left\lbrace f(x_1,\dotsc,x_n); n \in \mathbb{N}, f \in C[X_1,\dotsc,X_n], x_1,\dotsc,x_n \in N \right\rbrace$. Then:

(A) $C[N]$ is a field extension of $C$.

(B) All $x \in C[N]$ are algebraic over $C$.

Jyrki Lahtonen
  • 133,153
math_space
  • 604
  • For part (A) it should be clear that $C[N]$ is closed under multiplication, addition, and subtraction. Also because $C[N] \subseteq U$ and $U$ is a field every nonzero element in $C[N]$ has an inverse in $U$, you just need to show that that inverse is contained in $C[N]$. – Jim Mar 20 '14 at 17:04
  • I also think that part (A) is quite clear. But I don't know how to show, that the inverse is contained in $C[N]$. I think the difficult part is (B). My idea is: If we can show that $C[N]$ is a finite field extension of $C$ we could follow that all $x \in C[N]$ are algebraic over $C$. How could I do this? – math_space Mar 20 '14 at 17:12
  • Unless $N$ is finite $C[N]$ need not be a finite extension. – Jim Mar 20 '14 at 21:47
  • What is true though is that for every element $a \in C[N]$ there is a finite field extension $C \subseteq A$ with $a \in A$. That gives you that each element of $C[N]$ is algebraic over $C$. – Jim Mar 20 '14 at 21:55
  • Once you know each element of $C[N]$ is algebraic over $C$ then you can show that $C[N]$ is closed under inverses. – Jim Mar 20 '14 at 21:56
  • I see - $C[N]$ need not to be a finite extension. I only have to show that every $a \in C[N]$ is algebraic over $C$. But I cannot see that there exists such a finite field extension $C \subset A$ with $a \in A$. And why should this imply that $a$ is algebraic over $C$? – math_space Mar 20 '14 at 23:00
  • If $a \in C[N]$ then $a = f(x_1, \ldots, x_n)$ for some $f, n$, and $x_1, \ldots, x_n \in N$. Then $a \in A = C(x_1, \ldots, x_n)$ which is algebraic because all the $x_i$ are. – Jim Mar 20 '14 at 23:04
  • You definie $A$ as $A:=C[x_1,...,x_n]=\left\lbrace f(x_1,...,x_n): f \in C[X_1,...,X_n]\right\rbrace$? And this $A$ would be an finite field extension. – math_space Mar 20 '14 at 23:11
  • Is this correct? – math_space Mar 20 '14 at 23:20
  • Actually I define it as $C(x_1, \ldots, x_n)$, which means I take all the elements in $C[x_1, \ldots, x_n]$ as well as their inverses. I believe it's actually the case that these are equal but if I just say that I'm including the inverses then I don't have to worry about whether $A$ is actually a field or not, it will be automatically. – Jim Mar 21 '14 at 00:37

1 Answers1

1

First prove (B): every element of $C[N]$ is a polynomial expression $f(x_1,\dots,x_n)$ in elements $x_1, \dots, x_n$ that are algebraic over $C$ and hence is itself algebraic over $C$.

As already noted in the comments, for (A), it is clear that $C[N]$ is closed under multiplication, addition, subtraction; what is left is to show that it is closed under taking inverses.

Start with $0 \neq a \in C[N]$ and look at the subring $C[a]$ of $C[N]$. By (B), $a$ is algebraic over $C$, which makes $C[a]$ into an algebraic field extension of $C$. So $a^{-1} \in C[a] \subseteq C[N]$.

(Note that the fact that $C[x]$ is an algebraic field extension doesn't directly - or with induction - prove both (A) and (B), as $N$ could be infinite; the point of this exercise seems to be that you have to restrict attention to a finite subset of $N$.)

If you don't (yet) know that $C[a]$ is actually a field, I'll show why $a^{-1} \in C[a]$, which is enough here.

Say $g(x) = c_n x^n + c_{n-1} x^{n-1} + \dots + c_0 \in C[x]$ is the minimum polynomial of $a$ over $C$. Then $$a(c_n a^{n-1} + c_{n-1} a^{n-1} + \dots + c_1) = -c_0.$$ Note that $c_0 \neq 0$ because $g$ is the minimum polynomial and hence $$a^{-1} = -c_0^{-1}(c_n a^{n-1} + c_{n-1} a^{n-1} + \dots + c_1) \in C[a].$$

Magdiragdag
  • 15,049
  • 1
    Thank you, I think now I understand part (A). For part (B) I have one more question: If we take a $x \in C[N]$ there exists a polynomial $g \in C[X_1,...,X_n]$ in several variables such that $x=g(x_1,...,x_n)$ for some $n \in \mathbb{N}$ and some $x_1,...,x_n \in N$. We also know that those $x_1,...,x_n \in N$ are algebraic over $C$. But how can I exactly argue now that $C[N]$ itself is algebraic over $C$? You said that I have to restrict attention to a finite subset of $N$. – math_space Mar 21 '14 at 10:05
  • The finite subset I meant is ${x_1,\dots,x_n}$. Because the $x_1,\dots,x_n$ are algebraic over $C$, $C[x_1,\dots,x_n]$ is algebraic over $C$. (This is a standard, but not an immediately obvious, argument. See, for instance, http://math.stackexchange.com/questions/155122/how-to-prove-that-the-sum-and-product-of-two-algebraic-numbers-is-algebraic) – Magdiragdag Mar 21 '14 at 10:16