# proof of the weak Nullstellensatz

Let $K$ be an algebraically closed field, let $n\ge 0$, and let $I$ be
an ideal in the polynomial ring^{} $K[{x}_{1},\mathrm{\dots},{x}_{n}]$. Suppose $I$ is
strictly smaller than $K[{x}_{1},\mathrm{\dots},{x}_{n}]$. Then $I$ is contained in a
maximal ideal^{} $M$ of $K[{x}_{1},\mathrm{\dots},{x}_{n}]$ (note that we don’t have to
accept Zorn’s lemma to find such an $M$, since $K[{x}_{1},\mathrm{\dots},{x}_{n}]$ is
Noetherian^{} by Hilbert’s basis theorem), and the quotient ring^{}

$$L=K[{x}_{1},\mathrm{\dots},{x}_{n}]/M$$ |

is a field. We view $K$ as a subfield^{} of $L$ via the natural
homomorphism^{} $K\hookrightarrow L$, and we denote the images of
${x}_{1},\mathrm{\dots},{x}_{n}$ in $L$ by ${\overline{x}}_{1},\mathrm{\dots},{\overline{x}}_{n}$. Let
$\{{t}_{1},\mathrm{\dots},{t}_{m}\}$ be a transcendence basis of $L$ over $K$; it is
finite since $L$ is finitely generated^{} as a $K$-algebra^{}. Now $L$ is
an algebraic extension^{} of $K({t}_{1},\mathrm{\dots},{t}_{m})$. By multiplying the
minimal polynomial of ${\overline{x}}_{i}$ over $K({t}_{1},\mathrm{\dots},{t}_{m})$ by a
suitable element of $K[{t}_{1},\mathrm{\dots},{t}_{m}]$ for each $i$, we obtain
non-zero polynomials^{} ${f}_{i}\in K[{t}_{1},\mathrm{\dots},{t}_{m}][X]$ with the
property that ${f}_{i}({\overline{x}}_{i})=0$ in $L$:

$${f}_{i}={c}_{i,0}+{c}_{i,1}X+\mathrm{\cdots}+{c}_{i,{d}_{i}}{X}^{{d}_{i}}\mathit{\hspace{1em}\hspace{1em}}(1\le i\le n)$$ |

for certain integers ${d}_{i}>0$ and polynomials ${c}_{i,j}\in K[{t}_{1},\mathrm{\dots},{t}_{m}]$ with ${c}_{i,{d}_{i}}\ne 0$. Since $K$ is algebraically
closed^{} (hence infinite), we can choose ${u}_{1},\mathrm{\dots},{u}_{n}\in K$ such that
${c}_{i,{d}_{i}}({u}_{1},\mathrm{\dots},{u}_{m})\ne 0$ for all $i$. We define a homomorphism^{}

$$\varphi :K[{t}_{1},\mathrm{\dots},{t}_{m}]\u27f6K$$ |

by taking $\varphi $ to be the identity^{} on $K$ and sending ${t}_{j}$ to ${u}_{j}$.
Let $N$ be the kernel of this homomorphism. Then $\varphi $ can be
extended to the localization^{} $K{[{t}_{1},\mathrm{\dots},{t}_{m}]}_{N}$ of
$K[{t}_{1},\mathrm{\dots},{t}_{m}]$. Since ${c}_{i,{d}_{i}}\notin N$ for all $i$, the
${\overline{x}}_{i}$ are integral over this ring. Since $K$ is algebraically
closed, the extension theorem for ring homomorphisms implies that
$\varphi $ can be extended to a homomorphism

$$\varphi :(K{[{t}_{1},\mathrm{\dots},{t}_{m}]}_{N})[{\overline{x}}_{1},\mathrm{\dots},{\overline{x}}_{n}]=L\u27f6K.$$ |

Because $L$ is an extension field^{} of $K$ and $\varphi $ is the identity on
$K$, we see that $\varphi $ is actually an isomorphism^{}, that $m=0$, and
that $N$ is the zero ideal^{} of $K$. Now let ${a}_{1}=\varphi ({\overline{x}}_{1}),\mathrm{\dots},{a}_{n}=\varphi ({\overline{x}}_{n})$. Then for all polynomials $f$ in the
ideal $I$ we started with, the fact that $f\in M$ implies

$$f({a}_{1},\mathrm{\dots},{a}_{n})=\varphi (f({x}_{1},\mathrm{\dots},{x}_{n})+M)=0.$$ |

We conclude that the zero set^{} $V(I)$ of $I$ is not empty.

Title | proof of the weak Nullstellensatz |
---|---|

Canonical name | ProofOfTheWeakNullstellensatz |

Date of creation | 2013-03-22 15:27:43 |

Last modified on | 2013-03-22 15:27:43 |

Owner | pbruin (1001) |

Last modified by | pbruin (1001) |

Numerical id | 4 |

Author | pbruin (1001) |

Entry type | Proof |

Classification | msc 13A10 |

Classification | msc 13A15 |