### 澳洲代写｜MATH6216｜Advanced Topics in Algebra 代数高级专题 澳洲国立大学

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This course introduces students to key concepts and techniques in advanced algebra. Topics will be taken from contemporary research areas in Algebra.

Possible topics include:

Algebraic number theory, Analytic number theory, Algebraic geometry and scheme theory, Sheaf theory, Derived and Triangulated categories, Algebraic curves and Riemann surfaces

## Advanced Topics in Algebra 代数高级专题问题集

$\sharp$ EXERCISE. Let $K$ be a field and let $a_1, \ldots, a_n \in K$. Show that the ideal
$$\left(X_1-a_1, \ldots, X_n-a_n\right)$$
of the ring $K\left[X_1, \ldots, X_n\right]$ (of polynomials with coefficients in $K$ in indeterminates $\left.X_1, \ldots, X_n\right)$ is maximal.

The concept of maximal ideal in a commutative ring immediately leads to the very important idea of the Jacobson radical of such a ring.

Definition. Let $R$ be a commutative ring. We define the Jacobson radical of $R$, sometimes denoted by $\operatorname{Jac}(R)$, to be the intersection of all the maximal ideals of $R$.

Thus $\operatorname{Jac}(R)$ is an ideal of $R$ : even in the case when $R$ is trivial, our convention concerning the intersection of the empty family of ideals of a commutative ring means that $\operatorname{Jac}(R)=R$.

Note that when $R$ is quasi-local, $\operatorname{Jac}(R)$ is the unique maximal ideal of $R$.

We can provide a characterization of the Jacobson radical of a commutative ring.

Let $R$ be a commutative ring, and let $r \in R$. Then $r \in$ $\mathrm{Jac}(R)$ if and only if, for every $a \in R$, the element $1-r a$ is a unit of $R$.
Proof. ( $\Rightarrow$ ) Suppose that $r \in \operatorname{Jac}(R)$. Suppose that, for some $a \in R$, it is the case that $1-r a$ is not a unit of $R$. Then, by 3.11, there exists a maximal ideal $M$ of $R$ such that $1-r a \in M$. But $r \in M$ by definition of $\operatorname{Jac}(R)$, and so
$$1=(1-r a)+r a \in M$$
$(\Leftarrow)$ Suppose that, for each $a \in R$, it is the case that $1-r a$ is a unit of $R$. Let $M$ be a maximal ideal of $R$ : we shall show that $r \in M$. If this were not the case, then we should have
$$M \subset M+R r \subseteq R$$
Hence, by the maximality of $M$, we deduce that $M+R r=R$, so that there exist $b \in M$ and $a \in R$ with $b+a r=1$. Hence $1-r a \in M$, and so cannot be a unit of $R$. This contradiction shows that $r \in M$, as claimed. As this is true for each maximal ideal of $R$, we have $r \in \mathrm{Jac}(R)$.

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