---- > [!theorem] Theorem. ([[weak Nullstellensatz]]) > Let $k$ be a [[field]] and $\mathfrak{a}$ be an [[ideal]] of the [[polynomial 4|polynomial ring]] $k[T_{1},\dots,T_{n}]$. Certainly if $1 \in \mathfrak{a}$, then $V(\mathfrak{a})=\emptyset$.[^1] The weak Nullstellensatz asserts that this is the *only* obstruction to $V(\mathfrak{a})$ having substance: $V(\mathfrak{a}) = \emptyset \iff 1 \in \mathfrak{a}.$ ^theorem > [!proof]- Proof. ([[weak Nullstellensatz]]) > > Recall from field theory that if $k \subset L$ is a [[finite field extension]], i.e., $\text{dim}_{k}L < \infty$, then there is an [[injection|injective]] $k$-[[algebra]] [[algebra homomorphism|homomorphism]] $\varphi_{L}:L \to \Omega$: ![[Pasted image 20250514203242.png]] > That is, the algebraic closure $\Omega$ contains a copy of each finite extension of $k$ in a compatible way. > > --- > > > Assume $1 \not \in \mathfrak{a}$; take [[maximal ideal|then]] $\mathfrak{m} \in \text{mSpec }k[T_{1},\dots,T_{n}]$ such that $\mathfrak{a} \subset \mathfrak{m}$. $k[T_{1},\dots,T_{n}] / \mathfrak{m}$ is a [[field]], [[subalgebra generated by a subset|finitely generated]] as a $k$-[[algebra]] as a matter of definition. By [[Zariski's Lemma]], it is a [[finite algebra|finite]] $k$-[[algebra]]. Then the field theory fact above produces a $k$-[[algebra]] homomorphism $k[T_{1},\dots,T_{n}] / \mathfrak{m} \to \Omega$ that is injective. Precomposition with the [[quotient ring|quotient map]] gives a $k$-[[algebra]] homomorphism $\varphi: k[T_{1},\dots,T_{n}] \to \Omega, \operatorname{ker }\varphi=\mathfrak{m}.$ > This morphism is uniquely determined by the images of $T_{1},\dots,T_{n}$. More explicitly, we have $\varphi(f)=f(\boldsymbol x)$, where $\boldsymbol x=\big( \varphi(T_{1}),\dots,\varphi(T_{n}) \big)$. Thus, for all $f \in \mathfrak{a}$, we have $\varphi(f)=f(\boldsymbol x)$ > and also $\varphi(f)=0 \ (\text{because }\mathfrak{a} \subset \mathfrak{m}=\operatorname{ker }\varphi)$ > and so $f(\boldsymbol x)=0$, witnessing $\boldsymbol x \in V(\mathfrak{a})$. Thus $V(\mathfrak{a}) \neq \emptyset$. > ---- #### [^1]: If $1 \in \mathfrak{a}$ then $\mathfrak{a}$ is just *all* polynomials — surely they cannot all share a zero. Concretely, $V(\mathfrak{a})=V(\langle 1 \rangle)=V(1)=\emptyset$. ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```