## Maximal ideals

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 $ev a : ℂ [ x ] → ℂ f ( x ) ↦ f ( a )$ be the evaluation homomorphism.

The map $ℂ → { maximal ideals of ℂ [ x ] } a ↦ ker ( ev a ) = ( x - a )$ is a bijection.

For each $a=\left({a}_{1},...,{a}_{n}\right)\in {ℂ}^{n}$ let $ℂ x 1 ... x n → ℂ n f x 1 ... x n ↦ f a 1 ... a n$ be the evaluation homomorphism.

(Weak Nullstellensatz.) The map $ℂ n → { maximal ideals of ℂ x 1 ... x n } a ↦ ker ( ev a ) = x 1 - a 1 ... x n - a n$ is a bijection.

 Proof. $⇒\right)$ If $f\in \mathrm{ker}\left({\mathrm{ev}}_{a}\right)$ then use Taylor's theorem to write $f ( x ) = f ( a ) + ∑ k 1 , ... , k n ∈ ℤ ≥ 0 c k 1 , ... , k n ( x 1 - a 1 ) k 1 ⋯ ( x n - a n ) k n ,$ where, for each term in the sum, some ${k}_{i}\ne 0.$ This shows that $\mathrm{ker}\left({\mathrm{ev}}_{a}\right)\subseteq \left({x}_{1}-{a}_{1},...,{x}_{n}-{a}_{n}\right).$ Thus, $\mathrm{ker}\left({\mathrm{ev}}_{a}\right)=\left({x}_{1}-{a}_{1},...,{x}_{n}-{a}_{n}\right).$ $⇐\right)$ Let $M$ be a maximal ideal and let $𝔽=ℂ [{x}_{1},...,{x}_{n}} /M.$ Let $\pi$ be the composition $π : ℂ [ x 1 ] ↪ ℂ x 1 ... x n → 𝔽 .$ If $f\left({x}_{1}\right)\in \mathrm{ker}\pi$ then $f\left({x}_{1}\right)=\left({x}_{1}-{c}_{1}\right)g\left({x}_{1}\right)$ for some ${c}_{1}\in ℂ$ and $g\left({x}_{1}\right)\in ℂ\left[{x}_{1}\right].$ Since $\pi \left(f\left({x}_{1}\right)\right)=0$ and $ℂ\left[{x}_{1}\right]$ is an integral domain either $\pi \left({x}_{1}-{c}_{1}\right)=0$ or $\pi \left(g\left({x}_{1}\right)\right)=0.$ Thus, by induction on the degree of $f\left({x}_{1}\right),$ $\mathrm{ker}\pi =0$ or ${x}_{1}-{a}_{1}\in \mathrm{ker}\pi$ for some ${a}_{1}\in ℂ.$ If $\mathrm{ker}\pi =0$ then $ℂ\left[x\right]↪𝔽$ and $\pi$ can be extended to a map $π : ℂ ( x 1 ) ↪ 𝔽 ,$ 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 $1 x-α , α ∈ ℂ , are linearly independent.$ This is a contradiction. So $\mathrm{ker}\pi \ne 0.$ So $M$ contains $\left({x}_{1}-{a}_{1}\right).$ Similarly, $M$ contains $\left({x}_{i}-{a}_{i}\right)$ 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 $V = a ∈ ℂ n f 1 ( a ) = 0 , ... , f r ( a ) = 0 ,$ 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 $V → { maximal ideals of ℂ x 1 ... x n / f 1 ... f r } a ↦ ker ( ev a ) + f 1 ... f r$ is a bijection.

## Notes and References

Where are these from?

References?