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The field norm and trace exist when $K$ is a finite algebraic extension of $F$. In this case, an element $\alpha \in K$ can be interpreted as an $F$-linear map on $K$ by multiplication. The field norm is just the determinant of $\alpha$ as a linear map, while the trace is the trace of $\alpha$ as a linear map. This yields an evident generalization: Norm and trace are part of a family of nice maps, namely the coefficients of the characteristic polynomial of $\alpha$.

Since Zev asks for a uniqueness theorem in the comments, here is one that shows both the merits and limitations of the characteristic polynomial as an answer.

For simplicity let $F$ have characteristic 0. Let $K$ be a field extension of degree $n$ which is generic in the sense that the Galois group is $S_n$. Then any Galois-invariant polynomial in $\alpha \in K$ and its Galois conjugates, is a symmetric polynomial. The theorem is that the algebra of symmetric polynomials is generated by elementary symmetric polynomials, which are exactly the coefficients of the characteristic polynomial of $\alpha$. (This is using the fact that the eigenvalues of $\alpha$ as a map are itself and its Galois conjugates.) In particular, the trace is the unique linear such map up to a scalar; and any multiplicative polynomial of this type is a power of the norm. You can also describe the norm as the last Galois-invariant polynomial (the one of degree $n$) that provides new information.

But if the Galois group is smaller, then the ring of invariant polynomials in $\alpha$ and its Galois conjugates is larger, and any of these other invariant polynomials is also "nice".

Well, the original question was open-ended. I think that this answer does fit one interpretation of the question, but maybe it is too standard and maybe there are also other interesting answers.

The field norm and trace exist when $K$ is a finite algebraic extension of $F$. In this case, an element $\alpha \in K$ can be interpreted as an $F$-linear map on $K$ by multiplication. The field norm is just the determinant of $\alpha$ as a linear map, while the trace is the trace of $\alpha$ as a linear map. This yields an evident generalization: Norm and trace are part of a family of nice maps, namely the coefficients of the characteristic polynomial of $\alpha$.

Since Zev asks for a uniqueness theorem in the comments, here is one that shows both the merits and limitations of the characteristic polynomial as an answer.

For simplicity let $F$ have characteristic 0. Let $K$ be a field extension of degree $n$ which is generic in the sense that the Galois group is $S_n$. Then any Galois-invariant polynomial in $\alpha \in K$ and its Galois conjugates, is a symmetric polynomial. The theorem is that the algebra of symmetric polynomials is generated by elementary symmetric polynomials, which are exactly the coefficients of the characteristic polynomial of $\alpha$. (This is using the fact that the eigenvalues of $\alpha$ as a map are itself and its Galois conjugates.) In particular, the trace is the unique linear such map up to a scalar; and any multiplicative polynomial of this type is a power of the norm. You can also describe the norm as the last Galois-invariant polynomial (the one of degree $n$) that provides new information.

But if the Galois group is smaller, then the ring of invariant polynomials in $\alpha$ and its Galois conjugates is larger, and any of these other invariant polynomials is also "nice". These extras are somewhat hidden by the fact that, for any Galois group, the trace is still the only linear example and the norm is still the only multiplicative example.

Well, the original question was open-ended. I think that this answer does fit one interpretation of the question, but maybe it is too standard and maybe there are also other interesting answers.

The field norm and trace exist when $K$ is a finite algebraic extension of $F$. In this case, an element $\alpha \in K$ can be interpreted as an $F$-linear map on $K$ by multiplication. The field norm is just the determinant of $\alpha$ as a linear map, while the trace is the trace of $\alpha$ as a linear map. This yields an evident generalization: Norm and trace are part of a family of nice maps, namely the coefficients of the characteristic polynomial of $\alpha$.

Since Zev asks for a uniqueness theorem in the comments, here is one that shows both the merits and limitations of the characteristic polynomial as an answer.

For simplicity let $F$ have characteristic 0. Let $K$ be a field extension of degree $n$ which is generic in the sense that the Galois group is $S_n$. Then any Galois-invariant of degree $n$ from $K$ to $F$ which is a polynomial in $\alpha \in K$ and its Galois conjugates, is a symmetric polynomial. The theorem is that the algebra of symmetric polynomials is generated by elementary symmetric polynomials, which are exactly the coefficients of the characteristic polynomial of $\alpha$. (This is using the fact that the eigenvalues of $\alpha$ as a map are itself and its Galois conjugates.) In particular, the trace is the unique linear such map up to a scalar; and any multiplicative polynomial of this type is a power of the norm. You can also describe the norm as the last Galois-invariant polynomial (the one of degree $n$) that provides new information.

But if the Galois group is smaller, then the ring of invariant polynomials in $\alpha$ and its Galois conjugates is larger, and any of these other invariant polynomials is also "nice".

Well, the original question was open-ended. I think that this answer does fit one interpretation of the question, but maybe it is too standard and maybe there are also other interesting answers.

The field norm and trace exist when $K$ is a finite algebraic extension of $F$. In this case, an element $\alpha \in K$ can be interpreted as an $F$-linear map on $K$ by multiplication. The field norm is just the determinant of $\alpha$ as a linear map, while the trace is the trace of $\alpha$ as a linear map. This yields an evident generalization: Norm and trace are part of a family of nice maps, namely the coefficients of the characteristic polynomial of $\alpha$.

Since Zev asks for a uniqueness theorem in the comments, here is one that shows both the merits and limitations of the characteristic polynomial as an answer.

For simplicity let $F$ have characteristic 0. Let $K$ be a field extension of degree $n$ which is generic in the sense that the Galois group is $S_n$. Then any Galois-invariant polynomial in $\alpha \in K$ and its Galois conjugates, is a symmetric polynomial. The theorem is that the algebra of symmetric polynomials is generated by elementary symmetric polynomials, which are exactly the coefficients of the characteristic polynomial of $\alpha$. (This is using the fact that the eigenvalues of $\alpha$ as a map are itself and its Galois conjugates.) In particular, the trace is the unique linear such map up to a scalar; and any multiplicative polynomial of this type is a power of the norm. You can also describe the norm as the last Galois-invariant polynomial (the one of degree $n$) that provides new information.

But if the Galois group is smaller, then the ring of invariant polynomials in $\alpha$ and its Galois conjugates is larger, and any of these other invariant polynomials is also "nice".

Well, the original question was open-ended. I think that this answer does fit one interpretation of the question, but maybe it is too standard and maybe there are also other interesting answers.

The field norm and trace exist when $K$ is a finite algebraic extension of $F$. In this case, an element $\alpha \in K$ can be interpreted as an $F$-linear map on $K$ by multiplication. The field norm is just the determinant of $\alpha$ as a linear map, while the trace is the trace of $\alpha$ as a linear map. This yields an evident generalization: Norm and trace are part of a family of nice maps, namely the coefficients of the characteristic polynomial of $\alpha$.

The field norm and trace exist when $K$ is a finite algebraic extension of $F$. In this case, an element $\alpha \in K$ can be interpreted as an $F$-linear map on $K$ by multiplication. The field norm is just the determinant of $\alpha$ as a linear map, while the trace is the trace of $\alpha$ as a linear map. This yields an evident generalization: Norm and trace are part of a family of nice maps, namely the coefficients of the characteristic polynomial of $\alpha$.

Since Zev asks for a uniqueness theorem in the comments, here is one that shows both the merits and limitations of the characteristic polynomial as an answer.

For simplicity let $F$ have characteristic 0. Let $K$ be a field extension of degree $n$ which is generic in the sense that the Galois group is $S_n$. Then any Galois-invariant of degree $n$ from $K$ to $F$ which is a polynomial in $\alpha \in K$ and its Galois conjugates, is a symmetric polynomial. The theorem is that the algebra of symmetric polynomials is generated by elementary symmetric polynomials, which are exactly the coefficients of the characteristic polynomial of $\alpha$. (This is using the fact that the eigenvalues of $\alpha$ as a map are itself and its Galois conjugates.) In particular, the trace is the unique linear such map up to a scalar; and any multiplicative polynomial of this type is a power of the norm. You can also describe the norm as the last Galois-invariant polynomial (the one of degree $n$) that provides new information.

But if the Galois group is smaller, then the ring of invariant polynomials in $\alpha$ and its Galois conjugates is larger, and any of these other invariant polynomials is also "nice".

Well, the original question was open-ended. I think that this answer does fit one interpretation of the question, but maybe it is too standard and maybe there are also other interesting answers.