我会在这里记录一下自己学习代数时遇到的各种问题。长期更新（也许吧）
环
McCoy定理的证明
在学代数的时候发现了一个有趣的定理。
Theorem Let (R) be a commutative ring, and let (f(x)) be a zero-divisor in (R[x]). Then there (exists b in R, b eq 0), such that (f(x)b = 0)
Proof Let $f(x) = a_{d}x^{d} + ... + a_(0) $, nad let (g(x) = b_{e}x^{e} + ... + b_{0}) be a nonzero polynomial of minimal degree (e) such that (f(x)g(x) = 0). We know (a_{d}b_{e} = 0), so (deg(a_{d}g(x))<e). But (f(x)a_{d}g(x)=f(x)g(x)a_{d} = 0) since (R[x]) is a commutative ring as (R). (a_{d}g(x)) must be (0) because of minimality of degree of (g(x)). So, (a_{d}g(x) = 0).
Follow the distribution of ring, we have $$f(x)g(x) = a_{d}x^{d}g(x) + (a_{d-1}x^{d-1} + ... + a_{0})g(x) = (a_{d-1}x^{d-1} + ... + a_{0})g(x) = 0$$ By induction, we have (a_{d-i}g(x) = 0) for all (0 leq i leq d). Then we know (a_{d-i}b_{e} = 0) for all (0 leq i leq d), so (f(x)b_{e} = 0).
这个证明的关键一步在于由(f(x)a_{d}g(x)=f(x)g(x)a_{d} = 0)进而得到(a_{d}g(x)=0)。
when ideals of polynomials ring of an algebraically closed field is maxmimal?
Theorem Let (k) be an algebraically closed field, and let $I subseteq k[x] $ be an ideal. Then (I) is maximal if and only if $I = (x-c) $ for some $c in k $
Proof One direction is trivially easy. For every commutative ring R, (R[x]/(x-c) cong R). Field implies commutative ring, so (k[x]/(x - c) cong k) is a field, then ((x-c)) is a maximal ideal.
For the other direction, we know that (k[x]) is PID for k is field. Suppose a maximal ideal I of (k[x]), we know (I = (f(x))) for a polynomial (f(x)). (f(x)) cannot be constant, otherwise (J) will the whole (k[x]) contra the hypothesis that (k[x]/I) is a field.
Because (k) is algebraically closed, there exists (c in k) such that (f(c) = 0). By long division, we have (f(x) = g(x)(x-c) + r), which implies that (r=0) and (f(x)=g(x)(x-c)) for some polynomial (g(x)). Then we have (f(x) in (x-c)). So ((f(x)) = I subseteq (x-c)). Then (I = (x-c)) for the maximal property of (I).