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Community Bot
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If you throw in the residue degrees as well, then you get Zev's question. Otherwise, the converse is not true. As an example, consider a quadratic field that has an unramified everywhere $A_5$ Galois extension (see e.g. this MO threadthis MO thread). Since $A_5$ is simple, any proper intermediate extension will not be Galois over the quadratic. But the ramification indices will all be 1, since they are all 1 in the big extension.

If you throw in the residue degrees as well, then you get Zev's question. Otherwise, the converse is not true. As an example, consider a quadratic field that has an unramified everywhere $A_5$ Galois extension (see e.g. this MO thread). Since $A_5$ is simple, any proper intermediate extension will not be Galois over the quadratic. But the ramification indices will all be 1, since they are all 1 in the big extension.

If you throw in the residue degrees as well, then you get Zev's question. Otherwise, the converse is not true. As an example, consider a quadratic field that has an unramified everywhere $A_5$ Galois extension (see e.g. this MO thread). Since $A_5$ is simple, any proper intermediate extension will not be Galois over the quadratic. But the ramification indices will all be 1, since they are all 1 in the big extension.

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Alex B.
Alex B.
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If you throw in the residue degrees as well, then you get Zev's question. Otherwise, the converse is not true. As an example, consider a quadratic field that has an unramified everywhere $A_5$ Galois extension (see e.g. this MO thread). Since $A_5$ is simple, any proper intermediate extension will not be Galois over the quadratic. But the ramification indices will all be 1, since they are all 1 in the big extension.