Suppose can be generated by a single element. That means that there exist such thatandbut since so it is impossible.

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- January 16th 2009, 06:12 AM #1

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- January 16th 2009, 06:36 AM #2

- January 16th 2009, 05:07 PM #3

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you also need to show that: : since the constant term of every polynomial in is a multiple of 2, we have and thus is a proper ideal of

similarly you can easily prove a general result which is quite useful: if is a non-zero ring, then the polynomial ring is a PID if and only if is a field.

- January 19th 2009, 08:01 AM #4

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- January 19th 2009, 03:41 PM #5

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if is a field, we know that will be a PID. that's a well-known fact. conversely, suppose is a PID. then since is a subring of it has to be an integral domain. we'll prove that

every non-zero element of is a unit, which means is a field. so let let since is a PID, there exists such that hence must

be a__unit element of__because so i.e. there exists such that now put to get i.e. is a unit element of