# An “independence” condition on two algebraic elements over $K$.

Let $K$ be a field and let $a,b\in \overline K$ be algebraic elements.

I've stumbled upon a certain condition on $a,b$, which I feel could be considered an "independence" condition. I would like to know more about it.

Let's say that $a,b$ are weakly independent when $$\deg_K(a)=\deg_{K(b)}(a)$$

and $$\deg_K(b)=\deg_{K(a)}(b).$$

A condition weaker than algebraic independence implies this condition:

Fact. Suppose that for $g\in K[x,y],$ we have that $g(a,b)=0$ implies $g\in K[x]$ or $g\in K[y].$ Then $a,b$ are weakly independent.

Proof. Suppose $$\deg_K(a)>\deg_{K(b)}(a)$$ ($\geq$ always holds). Let $f(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_0$ be the minimal (monic) polynomial of $a$ over $K(b).$ We have $$f\in K(b)[x]\setminus K[x]$$ because monic minimal polynomials are unique. For each $i=0,1,\ldots,n-1,$ there exists $g_i\in K[x]$ such that $g_i(b)=a_i,$ and a least one $g_i$ isn't constant because otherwise $f\in K[x].$ Thus $$f=x^n+g_{n-1}(b)x^{n-1}+\ldots+g_0(b).$$

Let $g\in K[x,y]$ be defined by $$g(x,y)=x^n+g_{n-1}(y)x^{n-1}++\ldots+g_0(y).$$ Clearly $g(a,b)=0.$ But also, since $g_j(y)=b_my^m+\ldots+b_0$ is non-constant, there is $1\leq k\leq m$ such that $b_k\neq 0.$ Therefore, the $x^jy^k$-coefficient of $g$ is non-zero, and so $g\not\in K[x]$. It is clear that $g\not\in K[y].$ The symmetric case is proved symmetrically. $\square$

The converse doesn't hold. For example, $\sqrt 2,\sqrt 3$ are weakly independent over $\mathbb Q$ but $g(x,y)=x^2y-2y$ annihilates $(\sqrt 2,\sqrt 3)$. Something weaker does hold, but I won't post it here because I don't understand it very well and I don't want to make this question too long.

I would like to know if this "weak independence" has any real name and if it's equivalent to anything interesting. I've been having different ideas as to what it could be equivalent to but nothing seemed to work. Most of my ideas have been somewhere around the fact above.

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Is this equivalent to $K(a)$ and $K(b)$ being linearly disjoint over $K$? Two fields $L,L'$ (between a base field $K$ and a larger "playground" field, e.g. $\overline{K}$ are said to be linearly disjoint, iff elements of $L'$ are linearly independent over $L$ whenever they are so over $K$, iff elements of $L$ are linearly independents over $L'$ whenever they are so over $K$. – Jyrki Lahtonen Jun 20 '12 at 15:22
As it happens, our seminar once needed a refresher, so I scribbled down a few pages worth of basics. Does this fit? – Jyrki Lahtonen Jun 20 '12 at 15:28
I don't understand your polynomial condition. If $a,b$ are algebraic, and $g(x,y)$ is $p(x)+q(y)$ where $p(a)=q(b)=0$, then $g(a,b)=0$, and $g$ is in neither $K[x]$ nor $K[y]$, so the hypothesis of your "Fact" never applies. – Gerry Myerson Jun 21 '12 at 0:47
The polynomial condition can only hold if $a$ and $b$ are both transcendental. If $a$ is algebraic, then any polynomial in $K[x,y]$ that has the minimal polynomial of $a$ as a factor will necessarily evaluate to $0$, even if it does involve $y$. – Arturo Magidin Jun 21 '12 at 14:22
@GerryMyerson Well, I did say I didn't understand it very well. :-) Thank you. I won't delete it because I think there should be some real condition around this, and I'm not sure how to show what I kind of condition I'm looking for other than by giving an example of a reasoning I tried. – user23211 Jun 21 '12 at 14:39

As Jyrki notes, your condition is a special case of linear disjointness. E.g., this is taken from Lang's Algebra, Section VIII.3:

Definition. Let $K$ and $L$ be extensions of $k$, contained in some common algebraically closed field $\Omega$ that contains $k$. We say that $K$ is linearly disjoint from $L$ over $k$ if every finite set of elements of $K$ that is linearly independent over $k$ is also linearly independent over $L$.

Although the definition is asymmetric, the condition is in fact symmetric:

Proposition. $K$ is linearly disjoint from $L$ over $k$ if and only if $L$ is linearly disjoint from $K$ over $k$.

Proof. Let $y_1,\ldots,y_n\in L$ be elements that are linearly independent over $k$, and let $$\alpha_1y_1+\cdots +\alpha_ny_n = 0\tag{1}$$ be a $K$-linear combination equal to $0$. Reordering if necessary, assume that $\alpha_1,\ldots,\alpha_r$ are linearly independent over $k$, and $\alpha_{r+1},\ldots,\alpha_n$ are $k$-linear combinations of $\alpha_1,\ldots,\alpha_r$; that is, $$\alpha_i = \sum_{j=1}^r \beta_{ij}\alpha_j,\qquad i=r+1,\ldots,n.$$

We can rewrite $(1)$ to get \begin{align*} \sum_{j=1}^r \alpha_j y_j + \sum_{i=r+1}^{n}\left(\sum_{j=1}^r \beta_{ij}\alpha_j\right)y_i &=0\\ \sum_{j=1}^r \left(y_j + \sum_{i=r+1}^n\beta_{ij}y_i\right)\alpha_j&=0 \end{align*} Since $K$ is linearly disjoint from $L$ over $k$ and $\alpha_1,\ldots,\alpha_r$ are $k$-linearly independent, it follows that for each $j=1,\ldots,r$ we have $$y_j + \sum_{i=r+1}^n\beta_{ij}y_i = 0.$$ But since the $y_1,\ldots,y_n$ are $k$-linearly independent, this cannot occur; thus, $r=0$, so that $\alpha_1=\cdots=\alpha_n=0$, as desired. $\Box$

A question was raised in comment, on whether, like linear disjointness, the condiition off being weakly independent can be made one sided. That is, suppose that $a$ and $b$ are such that $[K(a,b):K(b)] = [K(a):K]$. Does it follow that $[K(a,b):K(b)] = [K(b):K]$?

Theorem. Let $K$ be a field, and let $a$ and $b$ be elements of some overfield that contains $K$. If $[K(a,b):K(b)] = [K(a):K]$, then $[K(a,b):K(a)] = [K(b):K]$.

Proof. It suffices to show that if $1,b,b^2,\ldots,b^{m-1}$ are linearly independent over $K$, then they are linearly independent over $K(a)$. Suppose we have $$p_0(a) + p_1(a)b + \cdots +p_{m-1}(a)b^{m-1}=0,$$ where each $p_i(x)$ is a rational function on $a$; clearing denominators, we may assume that they are in fact polynomials, and that they are of degree less than $[K(a):K]$ (arbitrary degree if $a$ is transcendental). Let $n$ be the highest power of $a$ that occurs. Then we can rewrite this expression in the form $$q_0(b) + q_1(b)a + \cdots + q_{n}(b)a^{n}=0$$ where the $q_i$ are polynomials with coefficients in $K$ (namely, $q_0(x)$ has the constant coefficient of $p_i$ as the degree $i$th coefficient; $q_1(x)$ has the degree one coefficient of $p_i$ as the degree $i$th coefficient, etc). Since $1,a,a^2,\ldots,a^n$ are linearly independent over $K$, they are linearly independent over $K(b)$, so we conclude that $q_I(x)=0$ for all $i$; this yields that the $p_i$ are $0$ for all $i$ as well, which establishes the claim. $\Box$

Theorem. Let $a$ and $b$ be algebraic over $K$. Then $a$ and $b$ are weakly independent over $K$ if and only if $K(a)$ is linearly disjoint from $K(b)$ over $K$.

Proof. Let $n=[K(a):K]$ and $m=[K(b):K]$. Assume first that $K(a)$ is linearly disjoint from $K(b)$ over $K$. Since $1,a,a^2,\ldots,a^{n-1}$ are $K$-linearly independent in $K(a)$, it follows that they are $K(b)$-linearly independent, and therefore the minimal polynomial of $a$ over $K(b)$ has degree at least $n$. Since the minimal polynomial of $a$ over $K$ has degree exactly $n$, it follows that $[K(a,b):K(b)]=n$. A symmetric argument shows that $[K(a,b):K(a)]=m$, proving that $a$ and $b$ are weakly independent over $K$.

Conversely, assume that $a$ and $b$ are weakly independent over $K$. Let $y_1,\ldots,y_r$ be elements of $K(a)$ that are linearly independent over $K$. We can write them in terms of $1,a,\ldots,a^{n-1}$, and we get: \begin{align*} y_1 &= \alpha_{01} + \alpha_{11}a +\alpha_{21}a^2+\cdots +\alpha_{n-1,1}a^{n-1}\\ y_2 &= \alpha_{02} + \alpha_{12}a + \alpha_{22}a^2+\cdots + \alpha_{n-1,2}a^{n-1}\\ &\vdots\\ y_r &= \alpha_{0r} + \alpha_{1r}a + \alpha_{2r}a^2 + \cdots + \alpha_{n-1,r}a^{n-1}. \end{align*} Because $y_1,\ldots,y_r$ are linearly independent over $K$, any $r\times r$ subdeterminant of the $\alpha_{ij}$ will be nonzero. That is, the matrix of the $\alpha_{ij}$ has rank $r$.

Suppose that $\beta_1,\ldots,\beta_n\in K(b)$ are such that $\beta_1y_1+\cdots + \beta_ry_r = 0$. Plugging in and reordering, we get: $$0 = \sum_{i=1}^r \beta_i\alpha_{0i} + \left(\sum_{i=1}^r\beta_i\alpha_{1i}\right)a + \cdots + \left(\sum_{i=1}^r\beta_i\alpha_{n-1,i}\right)a^{n-1}.$$ Since $1,a,a^2,\ldots,a^{n-1}$ are linearly independent over $K(b)$ (because $[K(b)(a):K(b)]=n$), it follows that $$\sum_{i=1}^r \beta_i\alpha_{ji} = 0$$ for $j=0,\ldots,n-1$. Viewing this as a system of $n$ linear equations over the $\beta_i$, we have $n$ equations in $r$ unknowns; the coefficient matrix is full rank (rank $r$), and therefore the system has a unique solution, namely $\beta_1=\cdots=\beta_r=0$. Thus, $y_1,\ldots,y_r$ are linearly independent over $K(b)$, showing that $K(a)$ is linearly disjoint from $K(b)$, as claimed. $\Box$

Proposition: If $a$ is algebraic over $K$ and $b$ is transcendental over $K$, then both the following conditions hold:

1. $a$ and $b$ are weakly independent over $K$; and
2. $K(a)$ and $K(b)$ are linearly disjoint over $K$.

Proof. If $a$ is algebraic and $b$ is transcendental over $K$, then $b$ is transcendental over $K(a)$; hence $[K(a,b):K(a)] = [K(b):K]$; by previously established proposition, it follows that $[K(a,b):K(b)] = [K(a):K]$. Alternatively, the fact that $[K(a,b):K(b)]=[K(a):K]$ when $a$ is algebraic and $b$ is transcendental is well-known. Thus, $a$ and $b$ are weakly independent.

For the second part, we can proceed as above: let $y_1,\ldots,y_m$ be linearly independent elements of $K(a)$; to show that they are linearly independent over $K(b)$, write out $r_1(b)y_1 + \cdots r_m(b)y_m = 0$, where the $r_i$ are rational functions on $b$. Clearling denominators, we may assume that they are in fact polynomials; then looking at the terms of degree $n$ in $b$ we see that the coefficients of degree $n$ must be equal to $0$ (by the linear independence of $y_1,\ldots,y_m$ over $K$), so $r_1(b)=\cdots=r_m(b)=0$. Thus, $y_1,\ldots,y_m$ remain independent over $K(b)$, so $K(a)$ is linearly disjoint from $K(b)$, as claimed. $\Box$

Proposition. Let $K$ be a field, and let $a$ and $b$ be transcendental over $K$. Then the following are equivalent:

1. $a$ and $b$ are weakly independent over $K$;
2. $K(a)$ and $K(b)$ are linearly disjoint.

Proof. (2)$\implies$(1): since $1,a,a^2,\ldots,a^n$ are linearly independent over $K$ for every $n$, then they are linearly independent over $K(b)$. Thus, $[K(a,b):K(b)]\gt n$ for all $n$< so $[K(a,b):K(b)]=\infty = [K(a):K]$. Thus, $a$ and $b$ are weakly independent over $K$.

(1)$\implies$(2): Since $a$ and $b$ are weakly independent and transcendental, it follows that $a$ is transcendental over $K(b)$, and $b$ is transcendental over $K(a)$. If $r_1(a),\ldots,r_n(a)$ are $K$-linearly independent rational functions on $a$, and $s_1(b),\ldots,s_n(b)$ are rational functions on $b$ such that $$s_1(b)r_1(a)+\cdots+s_n(b)r_n(b)=0,$$ then clearing denominators we obtain a polynomial expression in $a$ and $b$ equal to $0$. This implies the expression is trivial (all coefficients are $0$), which in turn yields that $s_1(b)=\cdots=s_n(b)=0$. Thus, $K(a)$ and $K(b)$ are linearly disjoint. $\Box$

The fact that linear disjointness is weaker than algebraic independence is also well known. For example, Proposition VIII.3.3 in Lang reads:

Proposition. Let $L$ be an extension of $k$, and let $\{u_1,\ldots,u_r\}$ be a set of quantities algebraically independent over $L$. Then $k(u)$ is linearly disjoint from $L$ over $k$.

On the other hand, we have the following:

Definition. We say that $K$ is free from $L$ over $k$ if every finite set of elements of $K$ algebraically independent over $k$ remains algebraically independent over $L$. If $(x)$ and $(y)$ are two sets of elements in $\Omega$, we say that they are free over $k$ (or independent over $k$) if $k(x)$ and $k(y)$ are free over $k$.

Proposition. If $K$ and $L$ are linearly disjoint over $k$, then they are free over $k$.

Proof. Let $x_1,\ldots,x_n$ be elements of $K$ algebraically independent over $k$. Suppose we have a relation $$\sum y_iM_i(x_1,\ldots,x_n)=0$$ where $M_i(x_1,\ldots,x_n)$ is a monomial in $x_1,\ldots,x_n$ and $y_i\in L$. This gives a linear relation over $L$ of the $M_i(x_1,\ldots,x_n)$, but because $x_1,\ldots,x_n$ are algebraically independent over $k$, we know that their monomials are linearly independent over $k$; and so by linear disjointness, their monomials are linearly independent over $L$. Therefore, $y_i=0$ for all $i$, so $x_1,\ldots,x_n$ are algebraically independent over $L$. $\Box$

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Thank you very much. I couldn't expect a better answer. (I corrected some minor things: I changed $x_i$ to $\alpha_i$ and then made some changes in the indices because I think as they stood, the betas had them switched.) I have a couple of questions. (1) Your first proof actually shows that $\deg_k(a)=\deg_{k(b)}(a)\iff\deg_k(b)=\deg_{k(a)}(b),$ right? Can it be proven without going through linear disjointness? (2) In the theorem, should there be an assumption that $a,b$ are algebraic? My definition of weak independence makes sense for transcendental elements I think but the proof fails. – user23211 Jun 20 '12 at 22:53
(3) Is there a "polynomial in $K[x,y]$" condition equivalent to the weak independence? I know it's not precise; I just mean something along the lines of the fact in my question. I spent some time trying to find a good condition but the ones I tried were either too strong or too weak and I'm curious if it can be done. – user23211 Jun 20 '12 at 23:20
@ymar:(2) I assumed they were algebraic in the proof. For the first part, the same argument shows that if $K(a)$ and $K(b)$ are linearly disjoint over $K$, then the degree of $a$ over $K(b)$ is no less than its degree over $K$; so in the infinite case, this proves that the dimension in both cases is infinite; that is, linearly disjoint implies weakly independent. For the other direction, we can write $y_1,\ldots,y_n$ as rational functions on $a$; there are only finitely many basis elements involved, so the argument should go through, but I don't have time to check details right now. – Arturo Magidin Jun 21 '12 at 2:50
@ymar: (1) Yes, the condition is symmetric, and you only need to assume one of them. – Arturo Magidin Jun 21 '12 at 4:34
@ymar: I was scooped, but your polynomial condition is vacuous; for example, if $g(a)=0$, then any polynomial that has $g(x)$ as a factor will evaluate to $0$ on $(a,b)$, even if it does involve $y$. So the only time your condition can be met is if $a$ and $b$ are both transcendental. – Arturo Magidin Jun 21 '12 at 14:20