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Is there are nice way to prove the primitive element theorem without using field extensions?

The primitive element theorem says that if $x$ and $y$ are algebraic over $F$ and $y$ is separable over $F$, then there exists a $z \in F(x,y)$ such that $F(x,y) = F(z)$. In the case where $F$ is infinite, $z$ can be expressed in the form $x + {\lambda}y$ with $\lambda \in F$. In fact, almost any $\lambda$ will do. There are only a finite number of exceptions. These exceptions are $\frac{\alpha_i - x}{y - \beta_j}$ where $\alpha_i$ and $\beta_j$ range over the (other) roots of the minimal polynomials of $x$ and $y$ respectively.

But in order to talk about $\alpha_i$ and $\beta_j$ we need to build a field extension where the minimal polynomials of $x$ and $y$ split. This is the step I'm hoping to avoid.

Perhaps we can build a polynomial in $F[x]$ whose roots are $\frac{\alpha_{i_1} - \alpha_{i_2}}{\beta_{j_1} - \beta_{j_2}}$ and simply avoid picking $\lambda$s which are roots of this polynomial, and then use whatever properties this polynomials has to prove that this works.

Or maybe there is another completely different way of proving the primitive element theorem while avoiding building field extensions.

I found a nice proof that $x$ is separable over $F$ if and only if every $y \in F(x)$ is separable over $F$ that avoids building field extensions. It uses derivations instead. Now I'm hoping to do the same with the primitive element theorem.

Edit: I'll try to give some motivation. Adding roots of polynomials to fields in constructive mathematics is more difficult than in classical mathematics (because irreducibility is undecidable). It only works for countable fields, and then you have no guarantee that the original field will be a decidable subset of the new field. Yes, it can be done, but it seems like a pain. There is another way too using double negation translations, but it also seems like a pain. Instead I'd rather avoid the whole issue of building splitting fields, if I can, and it seems tantalizingly close to possible. After all, the only reason splitting fields are used here is to build a finite set of elements in the base field to avoid.

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You can build this polynomial using resultants, but I'm not sure no splitting field is used during the proof... –  Homology Jun 27 '10 at 12:48
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Resultants are so much more confusing and computationally intensive than any sort of proof using splitting fields. –  Harry Gindi Jun 27 '10 at 15:58
    
@Russell: I think that the proof of the primitive element theorem in Zariski and Samuel's "Commutative Algebra" (Page 84 in my 1965 edition) gives the kind of construction you are looking for. –  SJR Aug 8 '13 at 13:52

2 Answers 2

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OK, I have a proof which meets your conditions. I relied on this write up of the standard proof as a reference.

Lemma 1: Let $K/F$ be an extension of fields, and let $f(x)$ and $g(x)$ be polynomials in $F[x]$. Let $d_F(x)$ be the GCD of $f$ and $g$ in $F[x]$ and let $d_K(x)$ be the GCD of $f$ and $g$ in $K[x]$. Then $d_F(x)$ and $d_K(x)$ coincide up to a scalar factor.

Proof: Since $d_F(x)|f(x)$ and $d_F(x)|g(x)$ in $K[x]$, we have $d_F(x)|d_K(x)$. Now, there are polynomials $p(x)$ and $q(x)$ in $F[x]$ such that $f(x) p(x) + g(x) q(x) = d_F(x)$. So $d_K(x) | d_F(X)$. Since $d_F(x)$ and $d_K(x)$ divide each other, they only differ by a scalar factor.

Lemma 2: Let $f(x)$ and $g(x)$ be polynomials with $g(0) \neq 0$. Then, for all but finitely many $t$, the polynomials $f(tx)$ and $g(x)$ have no common factor.

Proof: Let $f(x) = f_m x^m + \cdots + f_1 x + f_0$ and $g(x) = g_n x^n + \cdots + g_1 x + g_0$. If $f(tx)$ and $g(x)$ HAVE a nontrivial common factor, then there are polynomials $p(x)$ and $q(x)$, of degrees $\leq n-1$ and $\leq m-1$, such that $$f(tx) p(x)=g(tx) q(x).$$ This is $m+n$ linear equations on the $m+n$ coefficients of $p$ and $q$. Writing this out in coefficients, the matrix $$\begin{pmatrix} f_m t^m & \cdots & f_1 t & f_0 & 0 & 0 & \cdots & 0 \\ 0 & f_m t^m & \cdots & f_1 t & f_0 & 0 & \cdots & 0 \\ \ddots \\ 0 & 0 & \cdots & 0 & f_m t^m & \cdots & f_1 t & f_0 \\ g_n & \cdots & g_1 & g_0 & 0 & 0 & \cdots & 0 \\ 0 & g_n & \cdots & g_1 & g_0 & 0 & \cdots & 0 \\ \ddots \\ 0 & 0 & \cdots & 0 & g_n & \cdots & g_1 & g_0 \end{pmatrix}$$ has nontrivial kernel. The determinant of this matrix is a polynomial in $t$, with leading term $(f_m)^n (g_0)^n t^{mn} + \cdots$. (Recall that $g_0 \neq 0$.) So, for all but finitely many $t$, this matrix has nonzero determinant and $f(tx)$ and $g(x)$ are relatively prime. QED.

Now, we prove the primitive element theorem (for infinite fields). Let $\alpha$ and $\beta$, in $K$, be algebraic and separable over $F$, with minimal polynomials $f$ and $g$. We will show that, for all but finitely many $t$ in $F$, we have $F(\alpha - t \beta) = F(\alpha, \beta)$.

Let $f(x) = (x -\alpha) f'(x -\alpha)$ and $g(x) = (x - \beta) g'(x - \beta)$. Since $\beta$ is separable, we know that $g'(0) \neq 0$. Note that $f'$ and $g'$ have coefficients in $K$. By Lemma 2, for all but infinitely many $t$ in $F$, the polynomials $f'(ty)$ and $g'(y)$ have no common factor. Choose a $t$ for which no common factor exists. Set $F' = F(\alpha - t \beta)$; our goal is to show that $F'=F(\alpha, \beta)$.

Set $h_t(x) = f(tx + \alpha-t \beta)$. Note that $h_t(x)$ has coefficients in $F'$. We consider the GCD of $h_t(x)$ and $g(x)$.

Working in $K$, we can write $h_t(x) = t (x - \beta) f'(t (x- \beta))$ and $g(x) = (x-\beta) g'(x-\beta)$. By the choice of $t$, the polynomials $f'(t (x- \beta))$ and $g'(x-\beta)$ have no common factor, so the GCD of $h_t(x)$ and $g(x)$, in the ring $K[x]$, is $x-\beta$.

By Lemma 1, this shows that $x - \beta$ is in the ring $F'[x]$. In particular, $\beta$ is in $F'$. Clearly, $\alpha$ is then also in $F'$, as $\alpha= (\alpha - t \beta) + t \beta$. We have never written down an element of any field larger than $K$. QED.

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That said, let me reiterate my objection to your plan. In a course on Galois theory, you will want to show that every finite separable extension is contained in a finite Galois extension. Without this result, Galois theory is almost useless. And I don't see how you will prove this without doing something equivalent to constructing splitting fields. –  David Speyer Jun 28 '10 at 13:18
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@David: here is a way to express the same proof of Lemma 2 without the big matrix: since $g$ is not a monomial (as $g(0) \ne 0$), $g(x)$ and $f(Tx)$ in $K[T,x]$ are relatively prime. Passing to $K(T)[x]$ allows us to write $h_1 g(x) + h_2 f(Tx) = 1$ in $K(T)[x]$ for some $h_1, h_2 \in K(T)[x]$. This all now makes sense in $K[T]_ h[x]$ for some nonzero $h \in K[T]$ (a "common denominator" of $h_1$ and $h_2$). Then we can specialize $T$ to any $t \in K$ that is not a zero of $h$, so $g(x)$ and $f(tx)$ are relatively prime in $K[x]$ away from that finite set of $t$. –  Boyarsky Jun 28 '10 at 13:38
    
@David: I believe that the theorem that every finite separable extension is contained in a finite Galois extension is not going to hold constructively in general. It will hold for those fields where it is decidable whether not polynomials from F[X] are irreducible or not, and this is the case for most applications of Galois theory. I could restrict myself to fields of this type, but I don't like adding unnecessary hypothesis to my theorems. If it makes you feel better, I'm not teaching a course, but making a proof in Coq (but your objection still applies, so I don't dismiss it out of hand). –  Russell O'Connor Jun 28 '10 at 13:51
    
@Boyarsky: thanks for your contribution. I believe I can apply your argument (or at least something very close). –  Russell O'Connor Jun 28 '10 at 14:14
    
@Boyarsky: Why is it clear that $g(x)$ and $f(Tx)$ are relatively prime? –  Russell O'Connor Jun 29 '10 at 9:47

I assume you mean that you want to avoid constructing splitting fields; you can't avoid talking about field extensions when proving a theorem about field extensions! Here's a sort of a proof, but I think it is probably cheating.

Lemma: Let $F$ be an infinite field. Let $V$ be a $K$ vector space, containing two linearly independent vectors $u$ and $v$. Let $W_1$, $W_2$, ..., $W_r$ be a finite list of subspaces of $V$, none of which contain the space spanned by $u$ and $v$. Then there are scalars $a$ and $b$ in $F$ such that $au+bv$ is not in any $W_i$.

Proof: Let $R$ be the plane spanned by $u$ and $v$. Since $W_i \not \supseteq R$, we know that $W_i \cap R$ has dimension at most $1$. So there is at most one ratio of $a/b$ for which $au+bv$ lies in $W_i$. Since $F$ is infinite, there are infinitely values of the ratio for which $au+bv$ is not in any $W_i$.

Now, take $V$ to be the field $F(\alpha, \beta)$, and take $W_i$ to be the list of proper subfields of $V$ containing $F$.

The problem is that this proof relies on the fact that a finite seperable extension has finitely many sub-extensions. I'm not sure it is possible to prove this without introducing splitting fields at some point.

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Yes, I want to avoid constructing splitting field. More generally I'm hoping to avoid the construction of adding arbitrary roots of polynomials to given fields. Assume that $x$ and $y$ live in some given (finite) field extension $E/F$ that we know nothing about. So we can talk about $K(z)$ and such for intermediate fields $K$ and elements $z \in E$, but not much else. What you propose would work, but the theorem about finite separable extensions seems to require Galois theory, which is what I ultimately what I want to prove using the primitive element theorem. –  Russell O'Connor Jun 27 '10 at 15:24
    
Russell, two points. (1) Having written the notes you linked to, I read them again and fixed a gap (concerning the distinction between uniquely extending derivations on a field step by step through a particular tower of fields and knowing there is only one extension to the top field). (2) If your goal is to develop Galois theory using the primitive element theorem, what is your reason for wanting to avoid splitting fields for the proof of the prim. elt. theorem? Presumably splitting fields will be in your proof of the Galois correspondence, as you indicate, so why not use them earlier too? –  KConrad Jun 27 '10 at 17:43
    
And a third point: that a finite separable extension has finitely many intermediate fields does not need Galois theory. Sure, it is easy to understand why that theorem works from Galois theory, but a proof without Galois theory is definitely possible. I think such an approach is in Artin's Notre Dame Galois theory notes (before he gets to Galois theory itself). –  KConrad Jun 27 '10 at 17:45
    
@KConrad: are you sure Artin's proof of finiteness of the set of intermediate extensions doesn't use the primitive element theorem? (Not that there's anything wrong with that; to be honest, I don't really understand the motivation behind the OP, since the usual arguments seem perfectly natural and elementary already.) –  BCnrd Jun 27 '10 at 18:06
    
@KConrad: re: (2). Perhaps I haven't studied the proof of the fundamental theorem of Galois theory closely enough, but I don't see anywhere where objects outside the given Galois extensions are used. –  Russell O'Connor Jun 27 '10 at 18:36

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