---- > [!definition] Definition. ([[projective variety]]) > Let $k$ be an [[algebraically closed]] [[field]] and $\mathbb{P}^{n}$ denote $k$-[[projective space]]. Consider the [[polynomial 4|polynomial ring]] $k[T_{0},\dots T_{n}]$, [[graded ring|graded by degree]]. If $I$ is an [[homogeneous ideal|homogeneous]] [[ideal]] of this [[ring]], the set $V(I)=\{ [z_{0}:\cdots: z_{n}] \in \mathbb{P}^{n} : f(z_{0},\dots,z_{n})=0 \ \fa \text{homogeneous} \ f \in I \}$ is called a **projective variety** $X \subset \mathbb{P}^{n}$.[^1] > This notion is analogous to that of [[algebraic set]]. A projective variety is irreducible iff $I(X):=\text{ideal generated by homogeneous polynomials vanishing on $X$}$ is prime, as one might hope in analogue to [[affine variety]]. In such a case, can define[^2] $k(X):=\frac{\{ f / g: f,g \text{ homogeneous of same degree}, g \notin I(X) \}}{\left( \frac{f_{1}}{g_{1}} \sim \frac{f_{2}}{g_{2}} \iff f_{1}g_{2}-g_{2}g_{1} \in I(X) \right)}.$ > >We say $h \in k(X)$ is **regular at $p \in X$** if it can be written as a quotient $f / g$ with $f,g$ [[homogeneous polynomial|homogeneous]] of the same degree, $g(p) \neq 0$. Note that the only everywhere-regular rational functions on $X$ are the constant functions.[^3] > The **projective Nullstellensatz** (an easy deduction from the [[Nullstellensatz|(affine) Nullstellensatz]]) says if $\sqrt{ I } \neq \langle X_{0},X_{1}, \dots, X_{n} \rangle$ (the [[irrelevant ideal]]), then $I(X)=\sqrt{ I }$. , ---- #### [^1]: Note that we need homogeneity for this notion to even be well defined. If $\lambda \in k^{*}$, $f(\lambda z_{0},\dots,\lambda z_{n})=\lambda^{\text{deg }f}f(z_{0},\dots,z_{n})$; we can't pull out $\lambda^{\text{deg }f}$ (it isn't even defined) unless $f$ is [[homogeneous polynomial|homogeneous]]. [^2]: Need irreducibility, for without it $\sim$ would fail to be [[equivalence relation|transitive]]. [^3]: Indeed, suppose $h$ is regular at every $p \in X$; $p$, have $h= f / g$ in reduced form, i.e., can assume we've factored/canceled out common roots so $f,g$ don't share any roots. Then $h$ is not regular at the roots of $g$, so $g$ must be constant. Hence $f$ is too. Now something with sheaf (will return after class) ---- #### 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 > ```