学习代数的笔记

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 \(\exists b \in R, b \neq 0\), 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 \(I\).