Algebra Qualifying Exam Solutions
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Let M be the ideal 〈2, x〉. Note that M is maximal, as R/M is the field with 2 elements. Also note thatM is (0), soM is nilpotent. Thus, R is a local ring and any ideal of R other than R itself must be contained inM. So assume that I is not R itself (listed above as 〈1〉). We now consider the possibility that I 6⊆ M. Note that M/M∈ is a R/M vector space of dimension 2. If I/(I ∩M) contains any of the three nonzero vectors in M/M, then I contains 2x. Explicitly, 2 ·x = x · 2 = x(x+ 2). So, if I/(I ∩M) is nonzero, then I contains 2x. As M = 〈4, 2x, x〉 = 〈2x〉, this shows that, if I 6⊆ M, then I ⊇ M, and so such an I is determined by its image in M/M. The three 1-dimensional subspaces of M/M correspond to 〈x〉, 〈2〉 and 〈x+ 2〉; the whole 2-dimensional vector space gives 〈2, x〉.
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تاریخ انتشار 2011