Maximal ideals

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 29 December 2011

Maximal ideals

Let $R$ be a commutative ring. A maximal ideal is an ideal $M$ such that $R/M$ is a field.

For each $a\in ℂ$ let $$\begin{array}{cccc}{\mathrm{ev}}_a:& ℂ\left[x\right]& \to & ℂ\\ & f\left(x\right)& \mapsto & f\left(a\right)\end{array}$$ be the evaluation homomorphism.

The map $$\begin{array}{ccc}ℂ& \to & \left\{maximal\; ideals\; ofℂ\right[x\left]\right\}\\ a& \mapsto & \ker \left({\mathrm{ev}}_a\right)=(x-a)\end{array}$$ is a bijection.

For each $a=\left(a_1,...,a_n\right)\in {ℂ}^n$ let $$\begin{array}{ccc}ℂ [x_1,...,x_n] & \to & {ℂ}^n\\ f\left(x_1,...,x_n\right)& \mapsto & f\left(a_1,...,a_n\right)\end{array}$$ be the evaluation homomorphism.

(Weak Nullstellensatz.) The map $$\begin{array}{ccc}{ℂ}^n& \to & \left\{maximal\; ideals\; ofℂ [x_1,...,x_n] \right\}\\ a& \mapsto & \ker \left({\mathrm{ev}}_a\right)=\left(x_1-a_1,...,x_n-a_n\right)\end{array}$$ is a bijection.

		Proof.
	$\Rightarrow )$ If $f\in \ker \left({\mathrm{ev}}_a\right)$ then use Taylor's theorem to write $$f\left(x\right)=f\left(a\right)+\sum _{k_1,...,k_n\in {\mathbb{Z}}_{\ge 0}}{c}_{k_1,...,k_n}(x_1-a_1{)}^{k_1}\cdots (x_n-a_n{)}^{k_n},$$ where, for each term in the sum, some $k_i\ne 0.$ This shows that $\ker \left({\mathrm{ev}}_a\right)\subseteq \left(x_1-a_1,...,x_n-a_n\right).$ Thus, $\ker \left({\mathrm{ev}}_a\right)=\left(x_1-a_1,...,x_n-a_n\right).$ $\Leftarrow )$ Let $M$ be a maximal ideal and let $𝔽=ℂ [x_1,...,x_n} /M.$ Let $\pi$ be the composition $$\pi :ℂ\left[x_1\right]\hookrightarrow ℂ [x_1,...,x_n] \to 𝔽.$$ If $f\left(x_1\right)\in \ker \pi$ then $f\left(x_1\right)=(x_1-c_1)g\left(x_1\right)$ for some $c_1\in ℂ$ and $g\left(x_1\right)\in ℂ\left[x_1\right].$ Since $\pi \left(f\right(x_1\left)\right)=0$ and $ℂ\left[x_1\right]$ is an integral domain either $\pi (x_1-c_1)=0$ or $\pi \left(g\right(x_1\left)\right)=0.$ Thus, by induction on the degree of $f\left(x_1\right),$ $\ker \pi =0$ or $x_1-a_1\in \ker \pi$ for some $a_1\in ℂ.$ If $\ker \pi =0$ then $ℂ\left[x\right]\hookrightarrow 𝔽$ and $\pi$ can be extended to a map $$\pi :ℂ\left(x_1\right)\hookrightarrow 𝔽,$$ where $ℂ\left(x_1\right)$ is the field of fractions of $ℂ\left[x_1\right].$ The field $𝔽=ℂ [x_1,...,x_n] /M$ has a countable basis but the $ℂ-$dimension of $ℂ\left(x_1\right)$ is uncountable since the functions $$\frac{1}{x-\alpha },\alpha \in ℂ,are\; linearly\; independent.$$ This is a contradiction. So $\ker \pi \ne 0.$ So $M$ contains $(x_1-a_1).$ Similarly, $M$ contains $(x_i-a_i)$ for complex numbers $a_1,...,a_n\in ℂ.$ Thus $M\supseteq \left(x_1-a_1,...,x_n-a_n\right).$ $\square$

A variety is a set $V\subseteq {ℂ}^n$ such that $$V= {a\in {ℂ}^n|f_1\left(a\right)=0,...,f_r\left(a\right)=0} ,$$ for some finite set of polynomials $f_1,...,f_r\in ℂ [x_1,...,x_n] .$

Let $f_1,...,f_r\in ℂ [x_1,...,x_n]$ and let $V= {a\in {ℂ}^n|f_1\left(a\right)=0,...,f_r\left(a\right)=0} .$ The map $$\begin{array}{ccc}V& \to & \{maximal\; ideals\; ofℂ [x_1,...,x_n] / ⟨f_1,...,f_r⟩ \}\\ a& \mapsto & \ker \left({\mathrm{ev}}_a\right)+ ⟨f_1,...,f_r⟩ \end{array}$$ is a bijection.

Notes and References

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References

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