Problem Statement: Let I be an integral domain. Show that I is a field if and only if I has no nontrivial ideals.

Proof: First assume that I is a field and let J be an ideal of I. If 1\in J then it follows that J=I since x1=x\in J for every x\in I. Suppose then that 1\not\in J and let j\in J, then there exists a j^{-1}\in I since by assumption I is a field. But since J is an ideal it follows that jj^{-1}\in J, but since 1\not\in J this is only possible if j=0 and so it follows that J=\{0\}. Thus the only ideals of I are the trivial ideals.

Now assume that the only ideals of I are J=\{0\} and I. Fix a non-zero x\in I and define \phi_{x}:I\rightarrow I by \phi_{x}(y)=xy. \phi_{x} is a homomorphism with \ker\phi_{x}=J=\{0\} since I is an integral domain and so we have no zero divisors. So we have that \phi_{x} is one to one and thus \phi_{x} is onto since it maps I to itself. Now, since 1\in I it follows that there exists y\in I such that \phi_{x}(y)=1\in I. By definition of \phi_{x}(y) this implies that xy=1 and so y=x^{-1}. Thus every non-zero element in I is a unit and so I is a field.


Reflection: This technique of using a homomorphism to find an inverse is a really helpful tool. This only works since we can force \ker\phi_{x}=\{0\} which is due to the fact that I is an integral domain.

This entry was posted in Algebra, Field, Homomorphism, Ideal, Integral Domain. Bookmark the permalink.

1 Response to Suzie-Q

  1. j2kun says:

    Great technique! You’re quite inspired.

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