Is the proof of these results correct?

[coquelicot]

Hello,
Below are two results with their proof. Of course, there may be several ways to prove these results, but I just need some checking. Can someone check carefully if the math is OK ? (but very carefully, because if there is a failure, I will be murdered :-) ) ? thx.

Claim 1: Let $L/K$ be a field extension. Then $P \in L[X]$ is algebraic over $K[X]$, if and only if the coefficients of $P$ are algebraic over K. Furthermore, the conjugates of $P$ over $K[X]$ obtain by replacing each coefficients $P_i$ of $P$ by a certain conjugate of $P_i$ over $K$.

Proof: Assume that $P = P_kX^k + \cdots + P_1 X + P_0$.
If $P$ is algebraic over $K[X]$, there exists an algebraic relation of the form $a_nP^n + a_{n-1}P^{n-1} + \cdots+ a_1P + a_0 =0$ where $a_i\in K[X]$ and $a_i$ are not all identically equal to 0. Whence it is easy to see that the the free coefficient $P_0$ of $P$ fulfils an algebraic relation of the form $\alpha_n P_0^n + \cdots + \alpha_0$ where the $\alpha_i$ are the free coefficients of $a_i$. So $P_0$ is algebraic over $K$ and a fortiori over $K[X]$.
It follows immediately that the polynomial $(P - P_0)/X = P_kX^{k-1} + \cdots+ P_1$ is algebraic over $K[X]$. The same argument as above then shows that $P_1$ is algebraic over $K$, and so on, $P_2, \cdots, P_k$ are algebraic over $K$.

Conversely, if the coefficients $P_i$ of $P$ are algebraic over $K$. Let
$h(X,Y) = (Y-P)(Y-P')\cdots(Y-P^{(m)})$, where $P^{(i)}$ are the polynomials obtained from $P$ by replacing the coefficients of $P$ by their conjugates over $K$, in all the possible ways. By the fundamental theorem of symmetric functions, $h\in K[X, Y]$ and $h(X, P(X)) = 0$. This proves that $P$ is algebraic over $K[X]$. Also, this makes clear the last assertion of Claim 1.

Claim 2: If $\sigma_1(X1,\ldots, X_n), \ldots, \sigma_n(X_1, \ldots, X_n)$ are the n elementary polynomials over an integral domain $A$, and if P is a polynomial in n variables such that $P(\sigma_1, \ldots, \sigma_n) = 0$, then $P = 0$.

proof: Let $P(S_1,\ldots, S_n)$ be a polynomial of minimal degree in $S_1$ such that $P(\sigma_1, \ldots, \sigma_n) = 0$.
Then substituting 0 for $X_2,\ldots, X_n$ in the relation above leads to $P(X_1, 0, \ldots, 0) = 0$, since
$\sigma_1 = X_1+X_2+\cdots+X_n$,
$\sigma_2 = X_1X_2+X_1X3 + \cdots$ etc.
Hence $S_1$ divides $P$, or $P = S_1 Q$, and there holds $Q(\sigma_1, \ldots, \sigma_n) = 0$. By the minimality of the degree in $S_1$, there must holds $P = Q = 0$.

 

[fresh_42]

What is an algebraic polynomial?

 

[coquelicot]

What is an algebraic polynomial?

I don't say that P is an algebraic polynomial, but that P, as an element of the ring L[X], is algebraic over K[X] (that is, over the field K(X)), which means that P is solution of a polynomial h(Y) with coefficient in A[X].

 

[fresh_42]

I would have written it a bit more reader friendly, e.g. without introducing the $\alpha_i$, without the "easy to see", $h(X,Y)$ more explicitly as the product over the automorphism group, etc. but it looks correct. I assume the integral domain in claim 2 is hidden in the fact that $\sigma_1= S_1(\sigma_1,\ldots ,\sigma_n)$ can't be a zero divisor? At least I haven't seen another location.

 

[coquelicot]

I would have written it a bit more reader friendly, e.g. without introducing the $\alpha_i$, without the "easy to see", $h(X,Y)$ more explicitly as the product over the automorphism group, etc. but it looks correct. I assume the integral domain in claim 2 is hidden in the fact that $\sigma_1= S_1(\sigma_1,\ldots ,\sigma_n)$ can't be a zero divisor? At least I haven't seen another location.

Thank you so many. Indeed, I've mistakenly written "integral domain" in Claim 2. My intension was any commutative ring.

 

[fresh_42]

Thank you so many. Indeed, I've mistakenly written "integral domain" in Claim 2. My intension was any commutative ring.

In this case with all the fields around you certainly have both. But there is another question I have:
At the end, when you have $P(X_1,0\ldots,0)=0$, how do you get $S_1 \mid P$ and not $(S_1-X_1) \mid P\,?$

In this case you have
$$P(\sigma_1,\ldots,\sigma_n)=0=(S_1-X_1)(\sigma_1,\ldots,\sigma_n)\cdot Q(\sigma_1,\ldots,\sigma_n)=(\sigma_1-X_1)\cdot Q(\sigma_1,\ldots,\sigma_n)=(X_2+\ldots+X_n)\cdot Q(\sigma_1,\ldots,\sigma_n)$$ Does the degree in $S_1$ have changed despite of the corrected factor?

 

[coquelicot]

In this case with all the fields around you certainly have both. But there is another question I have: At the end, when you have $P(X_1,0\ldots,0)=0$, how do you get $S_1 \mid P$ and not $(S_1-X_1) \mid P\,?$

Wow! you've saved my life: I've realized that the proof is completely wrong. Let me try to propose another proof: The result is obviously true for n = 1. Assume inductively it is true at rank n-1. And let $P$ be a non zero polynomial of minimal degree in $S_n$ satisfying $P(\sigma_1, \ldots, \sigma_n) = 0$. Substitute 0 for $X_n$ in this relation. Under this substitution, $\sigma_n$ vanishes, and $\sigma_i$ passes to $\sigma'_i$ for all $i < n$, where $\sigma'_i$ is the i-th elementary symmetric polynomial of $n-1$ variables.
So we have $P(\sigma'_1,\ldots, \sigma'_{n-1}, 0) = 0$. In other words, with
$Q(S_1, \ldots, S_{n-1}) = P(S_1,\ldots, S_{n-1}, 0)$, there holds
$Q(\sigma'_1, \ldots, \sigma'_{n-1}) = 0$.
By the induction hypothesis, $Q = 0$, or $P(S_1, \ldots, S_{n-1}, 0) = 0$.
That means that $P$ is multiple of $S_n$. But then, etc.

 

[fresh_42]

I've realized that the proof is completely wrong.

I'm not convinced that it is completely wrong. You still have $Q(\sigma_1,\ldots,\sigma_n) =0$ since $X_2+\ldots+X_n$ is clearly no zero divisor. So the question is, whether $P=(S_1-X_1)\cdot Q$ which I think it is, i.e. I haven't run it through, but it should be the case, and if the degree of $Q$ as a polynomial in $S_1$ is (one) less than the degree of $P$ in $S_1$.

Without having read it in detail (it's late here and I'll read it tomorrow), I think your induction argument is basically the same as the argument by degree, which means I think they are either both wrong or both true. I tend to the latter: both true.

 

[coquelicot]

Wait! I think you interpret what I've written in a way I've not intended to. Let me explain myself: by hypothesis, $P$ is a polynomial in n variables $S_1, \ldots, S_n$. The $\sigma_i$ are the elementary polynomials in n variables $X_1, \ldots, X_n$. So $h:=P(\sigma_1, \ldots, \sigma_n)$ is a polynomial in the variables $X_i$ (the $S_i$ are absent in it). Claim 2 says that if $h=0$, then $P=0$. In the first proof above, I've incorrectly asserted that if $P(X_1, 0,\ldots, 0) =0$, or what is the same (since$X_1$ is a variable), $P(S_1,0,\ldots,0) = 0$, then $S_1$ divides $P$ (I don't know why). I've tried to fix this in the second proof.

 

[fresh_42]

Let me write what I thought you've meant in your first attempt, i.e. how I read it.

First let us forget the additional variables $S_i$. Instead let us introduce the map $\varphi \, : \, A[X_1,\ldots,X_n] \longrightarrow A[\sigma_1,\ldots,\sigma_n]$ which is a ring homomorphism.
Claim $2$ now says, that $\varphi$ is injective and a polynomial $P \in \operatorname{ker}\varphi$ is assumed.

Now your idea was to consider another ring homomorphism, namely $\psi\, : \,A[X_1,\ldots,X_n] \longrightarrow A[X_1]$ by substitution of $X_i=0\;(i>1)$. Both homomorphisms commute, so
$$
0=\psi(0)=\psi(\varphi (P)) = \varphi (\psi(P))=\varphi(P(X_1,0,\ldots,0))=P(\sigma_1,0\ldots,0)\stackrel{(*)}{=}P(X_1,0\ldots,0)
$$
which means $X_1 \mid \psi(P)$ or $\psi(P)=X_1 \cdot \chi(X_1)$ for a polynomial $\chi(X_1) \in A[X_1]$ and $\operatorname{deg}\chi(X_1) < \operatorname{deg}\psi(P)$. However, we still have $0=\varphi(\psi(P))=\varphi(X_1\cdot \chi(X_1))= \sigma_1 \cdot \chi(\sigma_1)$ and thus $\chi(\sigma_1)=0$ (integral domain?!). So with your choice of a polynomial of minimal degree in $X_1$ we're done.

This is how I read your proof. Have I made a mistake? Maybe $\varphi \circ \psi = \psi \circ \varphi$ and $(*)$ needs a closer look and should be proven by induction or a recursive solution of the equation system from $n$ down.
The induction is just another way to write it in my opinion.

 

[coquelicot]

Let me write what I thought you've meant in your first attempt, i.e. how I read it.

Yes, you have correctly read my first proof. But the problem is: why do you think (and did I think) that $P(X_1, 0, \ldots, 0)=0$ implies that $P$ is multiple of $X_1$. If for example $P(X_1, \ldots, X_n) = X_1X_2 + X_2$ then $P(X_1, 0,\ldots, 0) = 0$ but $P$ is not multiple of $X_1$. This is why I tried to find another way.

 

[mathwonk]

this isn't quite the same but suppose L is algebraic over K. Then since the element X already belongs to K[X], hence is algebraic over it, and the algebraic elements of any extension form a field, so also is any algebraic combination of elements of L and the symbol X algebraic over K[X]. I.e. L(X) is algebraic over K(X).

Applying this reasoning to the subfield of elements of L which are algebraic over K should do it. What do you think? the other direction looks a little harder, since if L contains a transcendental element over K, we have to prove that L(X) cannot be algebraic over K(X). I don't think that's quite as obvious. but maybe it implies a transcendence basis of L(X) over K has at least two elements, hence the extension cannot be rendered algebraic by adding in only one element to K? sorry I didn't read your proof, but it is more fun to make one than to read one.

 

[coquelicot]

this isn't quite the same but suppose L is algebraic over K. Then since the element X already belongs to .

Yes, the fact that $L[X]/K[X]$ is algebraic as soon as $L/K$ is, is evident, and proves one side of the first assertion. This does not prove the "Furthermore" assertion though. You can of course use Galois theory to prove it, but my aim is to give a very elementary proof.

 

[mathwonk]

i was not using galois theory but the theory of transcendence degree of field extensions. I.e. just as with the case of linearly independent elements over a field, the maximal cardinality of a set of algebraically independent elements over a field, i.e. the cardinality of a transcendence basis, is independent of their choice. so if the transcendence degree of L over K is at least one, then th tranmscendence degree of L(X) over k is at least 2. so it cannot be algebraic over K(X). but I admire that you wish a proof without using this theory. It is rather basic to the whole theory of transcendental extensions of course.

what is your definition of conjugate of P? is it an element of L(X) which is an image of P under a K(X) - automorphism of L(X)? or is it allowed to be an element of the splitting field of the polynomial satisfied by P over K(X)? and must the K - conjugates of the coefficients of P be elements of L? if so, then is the point to show that every automorphism of L(X) fixing K(X) must map L isomorphically onto itself? Is that clear?, since (if L is algebraic over K), an automorphism of L(X) that fixes K(X) must map L to a subfield of L(X) that is algebraic over K, hence that cannot contain any non constant polynomial in X.

 

[coquelicot]

what is your definition of conjugate of P? is it an element of L(X) which is an image of P under a K(X) - automorphism of L(X)? or is it allowed to be an element of the splitting field of the polynomial satisfied by P over K(X)? and must the K - conjugates of the coefficients of P be elements of L?

There is no assumption that $L$ be normal over $K$. The conjugates of $P$ are the conjugates of $P$ as an element of the field $L(X)$ over the field $K(X)$, that is, any root of the minimal polynomial of $P$ over $K(X)$, in the algebraic closure of $K(X)$. Equivalently, it is the image of P under a K(X)-monomorphism of $L(X)$ inside the algebraic closure of $L(X)$.

 

[mathwonk]

then what about just enlarging L to contain the algebraic closure of K?

 

[coquelicot]

then what about just enlarging L to contain the algebraic closure of K?

Still only one half of the first assertion of Claim 1.

Can you check also the proof of Claim 2 I posted in reply #7 (Monday 5:29 PM) ? (the proof in the Question is incorrect).

 

[fresh_42]

Still only one half of the first assertion of Claim 1.

Can you check also the proof of Claim 2 I posted in reply #7 (Monday 5:29 PM) ? (the proof in the Question is incorrect).

I think there's still the same problem as before. At the point where $S_n \mid P$ or formerly $S_1\mid P$, we may conclude that $P=S_i \cdot Q$ with $\operatorname{deg}_{S_i} Q < \operatorname{deg}_{S_i} P$ and thus $0 = \sigma_i \cdot Q(\sigma_1,\ldots ,\sigma_n)$. What I don't see is, what happens in the case $Q=0$ because your induction hypothesis used $P \neq 0\,?$ If you adjust the induction in this respect, then the question is: what if $Q$ simply doesn't depend on $S_n\,?$

Maybe I overlooked something and the induction can be done without the assumption $\operatorname{deg}_{S_n} P > 0$. But in a quick attempt to do the induction step $1 \mapsto 2$ I haven't seen the solution. At least the etc. deserves a closer look. My suggestion also went wrong on the inductive reproduction of the minimum situation, not on the division part.

Another possible way could be to write $P=\sum p_k(S_2,\ldots,S_n)S_1^k$ and then rearrange the expression by using the fact, that $\sigma_1^k$ is again a polynomial in $\sigma_k$.

 

[coquelicot]

I think there's still the same problem as before. etc. .

First, thank you so many for the checking of my proof. Indeed, I have somewhat abridged it but I still think it is correct. Let me write down all the details.

The result is obviously true for n = 1 (that is, for polynomials of one variable $S_1$). Assume inductively it is true at rank n-1. Suppose, to obtain an absurdity, that $P$ is a non zero polynomial of n variables $S_1,\ldots, S_n$ satisfying $P(\sigma_1, \ldots, \sigma_n) = 0$. By the induction hypothesis, $P$ must depend on $S_n$. We can suppose that $P$ is of degree minimal with respect to $S_n$ such that $P(\sigma_1, \ldots, \sigma_n) = 0$. The substitution $X_n=0$ in $\sigma_i$ sends $\sigma_n$ to $0$ and $\sigma_i$ to $\sigma'_i$ for all $i < n$, where $\sigma'_i(X_1, \ldots, X_{n-1})$ is the i-th elementary symmetric polynomial of $n-1$ variables.
So, substituting $X_n=0$ in $P(\sigma_1, \ldots, \sigma_n)=0$, we have
$P(\sigma'_1,\ldots, \sigma'_{n-1}, 0) = 0$. In other words, with
$Q(S_1, \ldots, S_{n-1}) = P(S_1,\ldots, S_{n-1}, 0)$, there holds
$Q(\sigma'_1, \ldots, \sigma'_{n-1}) = 0$.
By the induction hypothesis, $Q = 0$, or what is the same, $P(S_1, \ldots, S_{n-1}, 0) = 0$.
That means that $P$ is multiple of $S_n$: $P = S_n P'$, with $P' \not = 0$ and $P'(\sigma_1, \ldots, \sigma_n) = 0$ ($\sigma_n$ is not a divisor of 0 in $A[X_1, \ldots, X_n]$, even if $A$ is not an integral domain).
Since $deg_{S_n}P > deg_{S_n}P'$, this is the desired contradiction. Hence Claim 2 is true for polynomials in $n$ variables, completing the induction.

Note: According to this proof, it is not necessary that $A$ be an integral domain, but only a commutative ring or even a commutative Z-algebra.

 

[fresh_42]

Sounds good, but the simple fact, that you haven't really used the specific form of the $\sigma_i$ makes me skeptic and look for a flaw.

By the induction hypothesis, $P$ must depend on S_n

Why? Can't it be $P(S_1,\ldots,S_n)=P(S_1,\ldots,S_{n-1})$ and yet $P(\sigma_1,\ldots,\sigma_n)=0\,?$ This isn't covered by the induction hypothesis, as the $X_n$ like all others are still contained in the remaining $\sigma_i\,.$

($\sigma_n$ is not a divisor of $0$ in $A[X_1, \ldots, X_n]$, even if $A$ is not an integral domain).

It would help to mention the special form of $\sigma_n = X_1\ldots X_n$. For the other $\sigma_i$ it is not as obvious, at least to me. Maybe it is, would be a nice lemma about the Frobenius homomorphism.

I would it write as:

(induction basis) Either $P=0$ and we are done or there is a $P\neq 0$ of minimal degree in $S_n$.
No idea yet, how the case $P\in A[S_1,\ldots,S_{n-1}] \subseteq A[S_1,\ldots,S_{n}]$ should be dealt with, but it belongs here.
$\ldots$ as you wrote but explicitly mention $\sigma_n= X_1\ldots X_n$ isn't a zero divisor $\ldots$
$P=S_nP'$ with $P'(\sigma_1,\ldots\sigma_n)=0$.
Now it follows by choice of $P$ that $P'=0$ and thus $P=0$ contradicting our assumption.

I find the last conclusion easier to follow this way than to think about why $P'$ should be different from the zero polynomial. I first started to think about $P'\neq 0$ and found myself in the middle of the induction.

 

[coquelicot]

Why? Can't it be $P(S_1,\ldots,S_n)=P(S_1,\ldots,S_{n-1})$ and yet $P(\sigma_1,\ldots,\sigma_n)=0\,?$ This isn't covered by the induction hypothesis, as the $X_n$ like all others are still contained in the remaining $\sigma_i\,.$

Arghh! You are right. I now understand that the induction hypothesis does not imply immediately that $P = 0$. But this can be seen in the following way: if $P$ does not depend on $S_n$, then it is in fact a polynomial in $n-1$ variables $S_1, \ldots, S_{n-1}$ that fulfils $P(\sigma_1, \ldots, \sigma_{n-1}) = 0$. Then substituting $X_n=0$ inside this expression leads to $P(\sigma'_1, \ldots, \sigma'_{n-1}) = 0$. Hence $P=0$ according to the induction hypothesis.
Regarding the fact that $X_n=0$ sends $\sigma_i$ to $\sigma'_i$, this follows from the expression $\sigma_i = \sum_{1\leq j_1<j_2<\ldots<j_i\leq n} \prod_i X_{j_i}$, which becomes, under the substitution $X_n=0$
$\sum_{1\leq j_1<j_2<\ldots<j_i< n} \prod_i X_{j_i}= \sum_{1\leq j_1<j_2<\ldots<j_i\leq n-1} \prod_i X_{j_i}=\sigma'_i$.

 

[fresh_42]

Arghh! You are right. I now understand that the induction hypothesis does not imply immediately that $P = 0$. But this can be seen in the following way: if $P$ does not depend on $S_n$, [Edit: , i.e. $P(S_1, \ldots, S_{n})=P(S_1, \ldots, S_{n-1})$ ] then it is in fact a polynomial in $n-1$ variables $S_1, \ldots, S_{n-1}$ that fulfills [Edit: $0=P(\sigma_1, \ldots, \sigma_{n-1},\sigma_n) = P(\sigma_1, \ldots, \sigma_{n-1})$ ]. Then substituting $X_n=0$ inside this expression leads to $P(\sigma'_1, \ldots, \sigma'_{n-1}) = 0$. Hence $P=0$ according to the induction hypothesis.

I'm convinced, I don't think you'll have to explicitly mention the polynomials. But I would mention that $\psi(P(\sigma_1, \ldots, \sigma_{n}))=P(\sigma'_1, \ldots, \sigma'_{n-1},0)$ because this is the only place, where the specific form of the $\sigma_k$ is actually used, and this is essential. ($\psi$ being the ring homomorphism which substitutes $X_n=0$) So maybe at the beginning to outline the idea.

 

[coquelicot]

I'm convinced, I don't think you'll have to explicitly mention the polynomials. But I would mention that $\psi(P(\sigma_1, \ldots, \sigma_{n}))=P(\sigma'_1, \ldots, \sigma'_{n-1},0)$ because this is the only place, where the specific form of the $\sigma_k$ is actually used, and this is essential. ($\psi$ being the ring homomorphism which substitutes $X_n=0$)

Thank you again for your help fresh_42. I think the proof is now correct.

 

[mathwonk]

aren't the symbols X1,...,Xn all algebraic over the field generated by the symmetric functions of them, essentially by definition, or rather by the fact that the coefficients of the polynomial whose roots are the Xi are exactly those symmetric functions? That says that the field generated by the symmetric functions has the same transcendence degree as the one generated by the Xi, i.e. n.

 

[coquelicot]

aren't the symbols X1,...,Xn all algebraic over the field generated by the symmetric functions of them, essentially by definition, or rather by the fact that the coefficients of the polynomial whose roots are the Xi are exactly those symmetric functions? That says that the field generated by the suymmetric functions has the same transcendence degree as the one ghenertated by the Xi, i.e. n.

Yes of course. So what?

 

[mathwonk]

this implies the n symmetric functions are algebraically independent, hence claim 2 is true. i.e. they do not satisfy any non zero polynomial.

 

[coquelicot]

Yes, good remark. But this implies you know a theorem about the transcendence degree, and also this demonstration should be adapted to make it valid not only for fields K but for commutative rings A (or better Z-algebra A). I believe this is feasible because the argument is essentially universal over the field of rational numbers.

 

[fresh_42]

Yes, good remark. But this implies you know a theorem about the transcendence degree, and also this demonstration should be adapted to make it valid not only for fields K but for commutative rings A (or better Z-algebra A). I believe this is feasible because the argument is essentially universal over the field of rational numbers.

I very much think that the work for claim 2 and the proof that $\deg_K K(X_1,\ldots,X_n)=\deg_K K(\sigma_1,\ldots,\sigma_n)$ is almost the same. Somewhere has to be an argument, because although both are plausible, they aren't obvious.

 

[mathwonk]

I agree with the comments above, but I suggest that this discussion demonstrates that the fundmental facts about transcendence degree make all these problems somewhat elementary, and thus my view is that one should master that theory first, in order to understand well these questions. That is there are several basic results aboutb transcendence degree which are parallel to the basic results on linear dimension. Namely given any field extension E over F, a subset S of E exists such that S is algebraically independent over F, and E is algebraic over the field generated by S over F. Such a set is called a transcendence basis of E over F. Furthermore, any subset T of E such that E is algebraic over the field generated by T over F contains a transcendence basis. Next, any algebraically independent subset of E over F can be enlarged to a transcendence basis. Finally all transcendence bases of E over F have the same cardinality.

These results can be proved by lemmas which are exactly analogous to those used to prove the analogous reults for linear dimension, as exposed in Zariski - Samuel: Commutative Algebra, volume I. I agree that it is likely the arguments used to prove the problems solved by the OP are similar to those needed for this development, but I suggest that it will be found more insightful to prove the basic results on transcendence degree than to prove just these specialized problems. Moreover the problems will fall out, as will many other results. Good luck! I conjecture you will also be able to prove the basic tr. deg results, since you have the creativity to do these problems directly.