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Show that every ideal that is a proper subset of a field is maximal

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Final answer:

A proper ideal in a field is maximal because a field only has trivial ideals (the set containing only the zero element and the field itself) due to every non-zero element having a multiplicative inverse in the field, which leads to any proper ideal being incapable of being properly contained in any larger ideal and thus being maximal.

Step-by-step explanation:

To show that every ideal that is a proper subset of a field is maximal, we must first understand what these terms mean in the context of ring theory, a branch of abstract algebra. An ideal in ring theory is a special subset of a ring that absorbs multiplication by elements of the ring. A field is a type of ring where every non-zero element has a multiplicative inverse. A proper ideal is an ideal that is not equal to the entire ring itself, and a maximal ideal is an ideal such that there are no other ideals that contain it, other than the ring itself.

Now, let's prove that every proper ideal in a field is maximal. Let F be a field, and let I be a proper ideal of F. Assume there is an ideal J such that I is a subset of J, and J is a subset of F. Since F is a field, every non-zero element of F must have a multiplicative inverse in F, which means the only ideals possible are the trivial ideal {0} and the entire field F itself.

This is because any non-zero element a in F that belongs to an ideal would also require its inverse a-1 to belong to the ideal, which would give us a * a-1 = 1 in the ideal, and by the ideal property, all of F would then be included in the ideal. Thus, our assumption that there exists a non-trivial ideal J that contains I must be incorrect, meaning I is maximal.

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