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The reasonable meaning following example (1) seems to be that $E \otimes_k F$ is a field. If so, then it is isomorphic to every compositum. If not, then there exists a compositum within which they are not linearly disjoint.

I am not (yet) getting voter support, but I stand my ground! :-)

First, clearly if $E \otimes_k F$ is a field, then it is isomorphic to every compositum.

Second, if $E \otimes_k F$ is not a field, then there exists a compositum in which $E$ and $F$ are not linearly disjoint. It has a non-trivial quotient field, and that field can serve as a compositum. As Pete Clark points out, there is a difference between the case that $E \otimes_k F$ is an integral domain and the case that it has zero divisors. (And Pete is right that I forgot about this distinction.) In the former case, there exists a compositum in which they are linearly disjoint, namely the fraction field of $E \otimes_k F$. In the latter case, $E$ and $F$ are not linearly disjoint in any compositum.

If $E$ and $F$ are both transcendental extensions, then there are two different criteria: Weakly linearly disjoint, when $E \otimes_k F$ is an integral domain, and strongly linearly disjoint, when it is a field. Which you think is the more important condition is up to you. In Andrew's examples, $E$ and $F$ aren't both transcendental, so the distinction is moot.

(I needed to think about this issue in this paper.)

Actually, the previous isn't the whole story. If $E$ and $F$ are both transcendental, then they are extensions of purely transcendental extensions $E'$ and $F'$. $E'$ and $F'$ are only weakly linearly disjoint, and therefore $E$ and $F$ are too. So the distinction is always moot. Pete and Andrew's intuition was more correct all along. The correct statement is that when $E$ and $F$ are both transcendental, linearly disjoint extensions have different behavior.

The reasonable meaning following example (1) seems to be that $E \otimes_k F$ is a field. If so, then it is isomorphic to every compositum. If not, then there exists a compositum within which they are not linearly disjoint.

I am not (yet) getting voter support, but I stand my ground! :-)

First, clearly if $E \otimes_k F$ is a field, then it is isomorphic to every compositum.

Second, if $E \otimes_k F$ is not a field, then there exists a compositum in which $E$ and $F$ are not linearly disjoint. It has a non-trivial quotient field, and that field can serve as a compositum. As Pete Clark points out, there is a difference between the case that $E \otimes_k F$ is an integral domain and the case that it has zero divisors. (And Pete is right that I forgot about this distinction.) In the former case, there exists a compositum in which they are linearly disjoint, namely the fraction field of $E \otimes_k F$. In the latter case, $E$ and $F$ are not linearly disjoint in any compositum.

If $E$ and $F$ are both transcendental extensions, then there are two different criteria: Weakly linearly disjoint, when $E \otimes_k F$ is an integral domain, and strongly linearly disjoint, when it is a field. Which you think is the more important condition is up to you. In Andrew's examples, $E$ and $F$ aren't both transcendental, so the distinction is moot.

(I needed to think about this issue in Finite, connected, semisimple, rigid tensor categories are linear.)

Actually, the previous isn't the whole story. If $E$ and $F$ are both transcendental, then they are extensions of purely transcendental extensions $E'$ and $F'$. $E'$ and $F'$ are only weakly linearly disjoint, and therefore $E$ and $F$ are too. So the distinction is always moot. Pete and Andrew's intuition was more correct all along. The correct statement is that when $E$ and $F$ are both transcendental, linearly disjoint extensions have different behavior.

The reasonable meaning following example (1) seems to be that $E \otimes_k F$ is a field. If so, then it is isomorphic to every compositum. If not, then there exists a compositum within which they are not linearly disjoint.

I am not (yet) getting voter support, but I stand my ground! :-)

First, clearly if $E \otimes_k F$ is a field, then it is isomorphic to every compositum.

Second, if $E \otimes_k F$ is not a field, then there exists a compositum in which $E$ and $F$ are not linearly disjoint. It has a non-trivial quotient field, and that field can serve as a compositum. As Pete Clark points out, there is a difference between the case that $E \otimes_k F$ is an integral domain and the case that it has zero divisors. (And Pete is right that I forgot about this distinction.) In the former case, there exists a compositum in which they are linearly independent, namely the fraction field of $E \otimes_k F$. In the latter case, $E$ and $F$ are not linearly independent in any compositum.

If $E$ and $F$ are both transcendental extensions, then there are two different criteria: Weakly linearly independent, when $E \otimes_k F$ is an integral domain, and strongly linearly independent, when it is a field. Which you think is the more important condition is up to you. In Andrew's examples, $E$ and $F$ aren't both transcendental, so the distinction is moot.

(I needed to think about this issue in this paper.)

Actually, the previous isn't the whole story. If $E$ and $F$ are both transcendental, then they are extensions of purely transcendental extensions $E'$ and $F'$. $E'$ and $F'$ are only weakly linearly independent, and therefore $E$ and $F$ are too. So the distinction is always moot. Pete and Andrew's intuition was more correct all along. The correct statement is that when $E$ and $F$ are both transcendental, linearly independent extensions have different behavior.

The reasonable meaning following example (1) seems to be that $E \otimes_k F$ is a field. If so, then it is isomorphic to every compositum. If not, then there exists a compositum within which they are not linearly disjoint.

I am not (yet) getting voter support, but I stand my ground! :-)

First, clearly if $E \otimes_k F$ is a field, then it is isomorphic to every compositum.

Second, if $E \otimes_k F$ is not a field, then there exists a compositum in which $E$ and $F$ are not linearly disjoint. It has a non-trivial quotient field, and that field can serve as a compositum. As Pete Clark points out, there is a difference between the case that $E \otimes_k F$ is an integral domain and the case that it has zero divisors. (And Pete is right that I forgot about this distinction.) In the former case, there exists a compositum in which they are linearly disjoint, namely the fraction field of $E \otimes_k F$. In the latter case, $E$ and $F$ are not linearly disjoint in any compositum.

If $E$ and $F$ are both transcendental extensions, then there are two different criteria: Weakly linearly disjoint, when $E \otimes_k F$ is an integral domain, and strongly linearly disjoint, when it is a field. Which you think is the more important condition is up to you. In Andrew's examples, $E$ and $F$ aren't both transcendental, so the distinction is moot.

(I needed to think about this issue in this paper.)

Actually, the previous isn't the whole story. If $E$ and $F$ are both transcendental, then they are extensions of purely transcendental extensions $E'$ and $F'$. $E'$ and $F'$ are only weakly linearly disjoint, and therefore $E$ and $F$ are too. So the distinction is always moot. Pete and Andrew's intuition was more correct all along. The correct statement is that when $E$ and $F$ are both transcendental, linearly disjoint extensions have different behavior.

The reasonable meaning following example (1) seems to be that $E \otimes_k F$ is a field. If so, then it is isomorphic to every compositum. If not, then there exists a compositum within which they are not linearly disjoint.

I am not (yet) getting voter support, but I stand my ground! :-)

First, clearly if $E \otimes_k F$ is a field, then it is isomorphic to every compositum.

Second, if $E \otimes_k F$ is not a field, then there exists a compositum in which $E$ and $F$ are not linearly disjoint. It has a non-trivial quotient field, and that field can serve as a compositum. As Pete Clark points out, there is a difference between the case that $E \otimes_k F$ is an integral domain and the case that it has zero divisors. (And Pete is right that I forgot about this distinction.) In the former case, there exists a compositum in which they are linearly independent, namely the fraction field of $E \otimes_k F$. In the latter case, $E$ and $F$ are not linearly independent in any compositum.

If $E$ and $F$ are both transcendental extensions, then there are two different criteria: Weakly linearly independent, when $E \otimes_k F$ is an integral domain, and strongly linearly independent, when it is a field. Which you think is the more important condition is up to you. In Andrew's examples, $E$ and $F$ aren't both transcendental, so the distinction is moot.

(I actually needed to think about this issue in this paper.)

The reasonable meaning following example (1) seems to be that $E \otimes_k F$ is a field. If so, then it is isomorphic to every compositum. If not, then there exists a compositum within which they are not linearly disjoint.

I am not (yet) getting voter support, but I stand my ground! :-)

First, clearly if $E \otimes_k F$ is a field, then it is isomorphic to every compositum.

Second, if $E \otimes_k F$ is not a field, then there exists a compositum in which $E$ and $F$ are not linearly disjoint. It has a non-trivial quotient field, and that field can serve as a compositum. As Pete Clark points out, there is a difference between the case that $E \otimes_k F$ is an integral domain and the case that it has zero divisors. (And Pete is right that I forgot about this distinction.) In the former case, there exists a compositum in which they are linearly independent, namely the fraction field of $E \otimes_k F$. In the latter case, $E$ and $F$ are not linearly independent in any compositum.

If $E$ and $F$ are both transcendental extensions, then there are two different criteria: Weakly linearly independent, when $E \otimes_k F$ is an integral domain, and strongly linearly independent, when it is a field. Which you think is the more important condition is up to you. In Andrew's examples, $E$ and $F$ aren't both transcendental, so the distinction is moot.

(I needed to think about this issue in this paper.)

Actually, the previous isn't the whole story. If $E$ and $F$ are both transcendental, then they are extensions of purely transcendental extensions $E'$ and $F'$. $E'$ and $F'$ are only weakly linearly independent, and therefore $E$ and $F$ are too. So the distinction is always moot. Pete and Andrew's intuition was more correct all along. The correct statement is that when $E$ and $F$ are both transcendental, linearly independent extensions have different behavior.