Return to Answer

The following answer was communicated to me by Keith Conrad:

See:

M. Isaacs, Degree of sums in a separable field extension, Proc. AMS 25 (1970), 638--641.

http://alpha.math.uga.edu/~pete/Isaacs70.pdf

Isaacs shows: when $K$ has characteristic $0$ and $[K(a):K]$ and $[K(b):K]$ are relatively prime, then $K(a,b)$ = $K(a+b)$, which answers the students question in the affirmative. His proof shows the same conclusion holds under the weaker assumption that

$[K(a,b):K] = [K(a):K][K(b):K]$.

since Isaacs uses the relative primality assumption on the degrees only to get that degree formula above, which can occur even in cases where the degrees of $K(a)$ and $K(b)$ over $K$ are not relatively prime.

The following answer was communicated to me by Keith Conrad:

See:

M. Isaacs, Degree of sums in a separable field extension, Proc. AMS 25 (1970), 638--641.

http://alpha.math.uga.edu/~pete/Isaacs70.pdf

Isaacs shows: when $K$ has characteristic $0$ and $[K(a):K]$ and $[K(b):K]$ are relatively prime, then $K(a,b)$ = $K(a+b)$, which answers the students question in the affirmative. His proof shows the same conclusion holds under the weaker assumption that

$[K(a,b):K] = [K(a):K][K(b):K]$.

since Isaacs uses the relative primality assumption on the degrees only to get that degree formula above, which can occur even in cases where the degrees of $K(a)$ and $K(b)$ over $K$ are not relatively prime.

The following answer was communicated to me by Keith Conrad:

See:

M. Isaacs, Degree of sums in a separable field extension, Proc. AMS 25 (1970), 638--641.

http://math.uga.edu/~pete/Isaacs70.pdf

Isaacs shows: when $K$ has characteristic $0$ and $[K(a):K]$ and $[K(b):K]$ are relatively prime, then $K(a,b)$ = $K(a+b)$, which answers the students question in the affirmative. His proof shows the same conclusion holds under the weaker assumption that

$[K(a,b):K] = [K(a):K][K(b):K]$.

since Isaacs uses the relative primality assumption on the degrees only to get that degree formula above, which can occur even in cases where the degrees of $K(a)$ and $K(b)$ over $K$ are not relatively prime.

The following answer was communicated to me by Keith Conrad:

See:

M. Isaacs, Degree of sums in a separable field extension, Proc. AMS 25 (1970), 638--641.

http://alpha.math.uga.edu/~pete/Isaacs70.pdf

Isaacs shows: when $K$ has characteristic $0$ and $[K(a):K]$ and $[K(b):K]$ are relatively prime, then $K(a,b)$ = $K(a+b)$, which answers the students question in the affirmative. His proof shows the same conclusion holds under the weaker assumption that

$[K(a,b):K] = [K(a):K][K(b):K]$.

since Isaacs uses the relative primality assumption on the degrees only to get that degree formula above, which can occur even in cases where the degrees of $K(a)$ and $K(b)$ over $K$ are not relatively prime.

The following answer was communicated to me by Keith Conrad:

See:

M. Isaacs, Degree of sums in a separable field extension, Proc. AMS 25 (1970), 638--641.

http://math.uga.edu/~pete/Isaacs70.pdf

Isaacs shows: when $K$ has characteristic $0$ and $[K(a):K]$ and $[K(b):K]$ are relatively prime, then $K(a,b)$ = $K(a+b)$, which answers the students question in the affirmative. His proof shows the same conclusion holds under the weaker assumption that

$[K(a+b):K] = [K(a):K][K(b):K]$.

since Isaacs uses the relative primality assumption on the degrees only to get that degree formula above, which can occur even in cases where the degrees of $K(a)$ and $K(b)$ over $K$ are not relatively prime.

The following answer was communicated to me by Keith Conrad:

See:

M. Isaacs, Degree of sums in a separable field extension, Proc. AMS 25 (1970), 638--641.

http://math.uga.edu/~pete/Isaacs70.pdf

Isaacs shows: when $K$ has characteristic $0$ and $[K(a):K]$ and $[K(b):K]$ are relatively prime, then $K(a,b)$ = $K(a+b)$, which answers the students question in the affirmative. His proof shows the same conclusion holds under the weaker assumption that

$[K(a,b):K] = [K(a):K][K(b):K]$.

since Isaacs uses the relative primality assumption on the degrees only to get that degree formula above, which can occur even in cases where the degrees of $K(a)$ and $K(b)$ over $K$ are not relatively prime.