英国代考| MATH0021 Commutative Algebra交换式代数

Now suppose that $t>1$, and let $\sigma: R^{t} \rightarrow R$ be “projection to the last term ${ }^{n}$, that is,
$$\sigma\left(r_{1} e_{1}+\cdots+r_{t} e_{t}\right)=r_{t}$$
Put $I=\operatorname{Im}(\sigma)$. Since $\sigma$ is an $R$-module homomorphism, $I$ is an ideal of $R$.
If $I=0$, then $K \subseteq R^{t-1}$, so we are finished by our induction hypothesis.
So suppose that $I=R a$ is not 0 . By definition of $I$, there is an element $w$ in $K$ with $\sigma(w)=a$, that is,
$$w=w_{1} e_{1}+\cdots+w_{t-1} e_{t-1}+a e_{t},$$
where $e_{1}, \ldots, e_{t}$ is the standard basis of $R^{t}$. For any other element
$$x=x_{1} e_{1}+\cdots+x_{t-1} e_{t-1}+x_{t} e_{t} \in K,$$

we have $x_{t}=b a$ for some $b$ in $R$. Then
$$x-b w \in \operatorname{Ker}(\sigma) \cap K .$$
Write $L=\operatorname{Ker}(\sigma) \cap K$. We have
$$x=(x-b w)+b w \in L+R w$$
which gives
$$K=L+R w .$$
If $x \in L \cap R w$, then $x=b w$ with $b a=0$. Thus $b=0$ and so $x=0$. This establishes that
$$K=L \oplus R w .$$
Now $L$ is a submodule of $R^{t-1}$, so the induction hypothesis tells us that $L$ is free of rank $h$ say, with $h \leq t-1$. Let $\left{c_{1}, \ldots, c_{h}\right}$ be a basis of $L$. Since $R w$ has basis the single element $w$ (and has rank 1 ), $K$ is free with basis $\left{c_{1}, \ldots, c_{h}, w\right}$ and rank $s=h+1 \leq t$.

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