Ring theory by Rachel Quinlan

By Rachel Quinlan

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After reordering the gj (x) we can suppose that for i = 1, . . , r fi (x) = ui (gi (x) where ui is a non-zero rational number. However since fi (x) and gi (x) are both primitive polynomials in Z[x], we must have ui = ±1 for each i, so fi (x) and gi (x) are associates not only in Q[x] but in Z[x]. Thus Z[x] is a UFD.

Suppose that M is a maximal ideal of R and let a+M be a non-zero element of R/M. We need to show the existence of b + m ∈ R/M with (a + M)(b + M) = 1 + M. This means ab + M = 1 + M, or ab − 1 ∈ M. So we need to show that there exists b ∈ R for which ab − 1 ∈ M. Let M denote the set of elements of R of the form ar + s, for some r ∈ R and s ∈ M. Then M is an ideal of R (check), and M properly contains M since a ∈ M and a ∈ M. Then M = R since M is a maximal ideal of R. In particular then 1 ∈ M and 1 = ab + m for some b ∈ R and m ∈ M.

N − 1 (this is guaranteed by the division algorithm in Z). 3. Let F be a field and let I be an ideal in F[x]. 3. If g(x) ∈ F[x] then the coset g(x) + I contains all those polynomials that differ from g(x) by a multiple of f(x). If F is infinite then the number of cosets of I in F[x] is infinite but each has exactly one representative of degree less than that of f(x). Q UESTION FOR THE S EMINAR: Why is this? g. F = Z/pZ for some prime p), then the number of cosets of I in F[x]is finite. 2 Let a and b be elements of a ring R in which I is a two-sided ideal.

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