1.2 Irreducible Algebraic Sets

Definition

A (closed) algebraic subset of $k^n$ is defined as $V:=\{z\in k^n:f_1(z)=\cdots =f_m(z)=0\}$ with $f_i\in k[x_1,\cdots ,x_n]$.

Let $I=(f_1,\cdots ,f_n)$ be the ideal generated by $f_i$. Then $V=V(I)=\{z\in k^n:f(z)=0,\forall f\in I\}$. Note that $k(x_1,\cdots ,x_n)$ is Noetherian, then any ideal $I$ is finitely generated and $I=(f_1,\cdots ,f_{\ell })$. It deduces that $V(I)=\{z\in k^n:f_1(z)=\cdots =f_{\ell }(z)=0\}$.

Remark. Note that $V((0))=k[x_1,\cdots ,x_n]$ and $V(k[x_1,\cdots ,x_n])=\varnothing$.

Definition

Let $V\subseteq k^n$ be an algebraic set. Define $I(V)=\{f\in k[x_1,\cdots ,x_n]:f(V)=0\}$.

Facts.

• $I(V)\subseteq k[x_1,\cdots ,x_n]$ is an ideal.
• $V(I(V))=V$.

Hilbert's Nullstellensatz

$I(V(I))=\sqrt{I}:=\{f\in k[x_1,\cdots ,x_n]:\text{mbox}thereexistsm\text{mbox},suchthat{f}^m\in I\}$.

\begin{proof} "→" If $f\in \sqrt{I}$, then $f^m\in I$ and $f^m(z)=0$ for all $z\in V(I)$. Thus $f(z)=0$ and so $f\in I(V(I))$.

"←" Let $I=(f_1,\cdots ,f_n)$. For $z$ with $f_1(z)=\cdots =f_n(z)=0$ and $g$ with $g(z)=0$, we aim to show there exists $m$ such that $g^m\in I$.

Define an ideal $B=I\cdot k[x_1,\cdots ,x_n,x_{n+1}]+(1-g\cdot x_{n+1})$.

Step 1. We claim that $B=k[x_1,\cdots ,x_{n+1}]$. Otherwise, if $B⊊k[x_1,\cdots ,x_{n+1}]$, then $B\subseteq M$ for some maximal ideal $M$ and $M=(x_1-a_1,\cdots ,x_{n+1}-a_{n+1})$ by weak Nullstellensatz. Since $f\in B\subseteq M$, we have $f_i(a_1,\cdots ,a_m)=0$ for $1⩽i⩽n$. It deduces that $g(a_1,\cdots ,a_n)=0$. Since $1-g\cdot x_{n+1}\in B$, then $1-g(a_1,\cdots ,a_n)\cdot a_{n+1}=1$, which is a contradiction. Therefore, $B=k[x_1,\cdots ,x_{n+1}]$.

Step 2. As $B=k[x_1,\cdots ,x_{n+1}]$, we have

$$1=\sum _{i=1}^n{h}_i(x_1,\cdots ,x_n,x_{n+1})f_i(x_1,\cdots ,x_n)+h_{n+1}(x_1,\cdots ,x_{n+1})(1-g{x}_{n+1}).$$

Let $x_{n+1}=1/g$, then

$$1=\sum _{i=1}^n{h}_i(x_1,\cdots ,x_n,1/g)f_i(x_1,\cdots ,x_n).$$

There exists $m$ large enough such that $g^m{h}_i(x_1,\cdots ,x_n,1/g)\in (f_1,\cdots ,f_m)$ and then

$$g^m=\sum g^m{h}_i(x_1,\cdots ,x_n,1/g)f_i(x_1,\cdots ,x_n)\in (f_1,\cdots ,f_m).$$

Now we finish the proof. \end{proof}

Corollary

There exists a 1-1 correspondence

$$\{V\subseteq k^n:V\text{mbox}isanalgebraicset\}\leftrightarrow \{I\subseteq k[x_1,\cdots ,x_n]:I=\sqrt{I}\}.$$

Proposition

The followings hold.

• If $I\subseteq J$, then $V(I)\supseteq V(J)$;
• If $V_1\subseteq V_2$, then $I(V_1)\supseteq I(V_2)$;
• If $V({\sum }_{\alpha }{I}_{\alpha })={\cap }_{\alpha }V(I_{\alpha })$;
• $V(I\cap J)=V(I)\cup V(J)$.

\begin{proof} i)-iii) are easy to proof.

iv) Since $I\cap J\subseteq I,J$, we have $V(I\cap J)\supseteq V(I)\cup V(J)$. If $z\notin V(I)\cup V(J)$, then there exists $f\in I,g\in J$ such that $f(z)\ne 0$ and $g(z)\ne 0$. It deduces that $0\ne f(z)g(z)$ with $fg\in V\cap J$ and so $z\in V(I\cap J)$. Thus $V(I\cap J)\subseteq V(I)\cup V(J)$. \end{proof}

Remark.

• Here is an example to show $V({\cap }_{\alpha \in \Lambda }{I}_{\alpha })\ne {\cup }_{\alpha \in \Lambda }V(I_{\alpha })$. Note that $I(\mathbb{Q})=\{f\in \mathbb{C}[x]:f(\mathbb{Q})=0\}=\{0\}$ and $V(I(\mathbb{Q}))={\cup }_{q\in \mathbb{Q}}V(I(q))$. If $V(I(\mathbb{Q}))$ is closed, then $\mathbb{Q}=V(I(\mathbb{Q}))=V((0))=\mathbb{C}$, which is a contradiction.
• Close set are “measure zero”: for any $f\in I\setminus \{0\}$, we have $V\subseteq V((f))=\{z:f(z)=0\}$.

Definition

A closed algebraic set is irreducible if it is NOT the union of $2$ strictly smaller closed algebraic sets.

Example. Notice that $V(xy)=\{xy=0\}\subseteq {\mathbb{C}}^2=\{x=0\}\cup \{y=0\}=V((x))\cup V((y))$, so $V((xy))$ is not irreducible.

Explanation with ideals.

• Here are some facts:
  • $\sqrt{I}=I$ and $\sqrt{J}=J$, then $\sqrt{I\cap J}=I\cap J$;
  • All prime ideals $P$ satisfying $\sqrt{P}=P$;
  • If $P$ is a prime ideal and $P=I\cap J$, then $P=I$ or $J$. (Otherwise, take $x\in I\setminus P$ and $y\in J\setminus P$, then $xy\in P$ leading to a contradiction.)
• Assume that $I,J_1,J_2$ are radical ideals. Then $V(I)=V(J_1)\cup V(J_2)=V(J_1\cap J_2)$ iff $I=J_1\cap J_2$. Hence,
  • $V(I)$ is irreducible iff for $I=J_1\cap J_2$ there is $I=J_i$ for some $i$.
  • $V(I)$ is not irreducible iff there exists $I\ne J_i$ such that $I=J_1\cap J_2$.
• Primary decomposition: If $I\subseteq k[x_1,\cdots ,x_n]$ is an ideal with $I=\sqrt{I}$, there exists unique $\{P_i:P_i\text{mbox}isaprimeideal,1⩽i⩽m\}$ such that $I=P_1\cap P_2\cap \cdots \cap P_m$.
• With the uniqueness of primary decomposition, we have the following proposition.

Proposition

• $V(I)\subseteq k^n$ is irreducible with $I=\sqrt{I}$ iff $I$ is a prime ideal;
• $V(I)=V_1\cup \cdots \cup V_m$ is uniquely decomposed into irreducible sets with $V_i\ne V_j$.

\begin{proof} If $I=P_1\cap \cdots \cap P_m$ is uniquely decomposed, then $V(I)=V(P_1)\cup \cdots \cup V(P_m)$. If $V(I)$ is irreducible, then note that $V(I)=V(P_1)\cup V(P_2\cap \cdots \cap P_m)$ and there exists $V(I)=V(P_j)$ such that $I=P_j$. If $V(I)=V(P_1)\cup \cdots \cup V(P_m)=V(Q_1)\cup \cdots \cup V(Q_{\ell })$, then $P_1\cap \cdots \cap P_m=Q_1\cap \cdots \cap Q_{\ell }$ and $\{P_1,\cdots ,P_m\}=\{Q_1,\cdots ,Q_{\ell }\}$ by uniqueness of primary decomposition. \end{proof}

Projective Algebraic Set

Definition

Define projective space ${\mathbb{P}}^n(k)=\{(z_1,\cdots ,z_n)\in k^{n+1}\setminus \{0\}/\sim \}=\{[z_0:\cdots :z_n]:z_0,\cdots ,z_n\text{mbox}arenotallzero\}$, where $(z_0,\cdots ,z_n)\sim (u_0,\cdots ,u_n)$ if there exists $\lambda \in k^{\ast }=k\setminus \{0\}$ such that $(z_0,\cdots ,z_n)=(\lambda u_0,\cdots ,\lambda u_n)$ and $[z_0,\cdots ,z_n]$ are homogeneous coordinate.

Definition

For homogeneous polynomials $f_1,\cdots ,f_m$, define $V=\{\overline{x}\in {\mathbb{P}}^n(k):f_1(\overline{x})=\cdots =f_m(\overline{x})=0\}$ and define $(f_1,\cdots ,f_m)$ as the ideal generated by homogeneous polynomial.

Conversely, for any homogeneous ideal $I$, define $V(I)=\{\overline{x}\in {\mathbb{P}}^n(k):\forall f\in I,f(\overline{x})=0\}$.

Fact. It has same propositions as ^487780 and ^894d70.

Theorem

There is a $1$-$1$ correspondence

$$\{V\subseteq {\mathbb{P}}^n(k):V\text{mbox}isanalgebraicset\}\leftrightarrow \{\text{mbox}homogeneousidealI\text{mbox}suchthatI=\sqrt{I}\text{mbox}except(x_0,\cdots ,x_n)\}.$$

Proposition

• Any algebraic set $V\subseteq {\mathbb{P}}^n(k)$, we have a unique composition $V=V_1\cup \cdots \cup V_m$ with irreducible $V_i$.
• $\{V\text{mbox}irreducible\}\leftrightarrow \{\text{mbox}homogeneousprimeidealexcept(x_0,\cdots ,x_n)\}$.

Remark. Note that $U_i=\{[z_0,\cdots ,z_i,\cdots ,z_n]\in {\mathbb{P}}^k:z_i\ne 0\}=\{[z_0/z_i:\cdots :1:\cdots :z_n/z_i]\in {\mathbb{P}}^k\}$, and there is a $1$-$1$ correspondence $U_i\leftrightarrow k^n$. Then we can get a homogeneous polynomial from a polynomial, and get a projective algebraic set from an affine algebraic set. For example, $V=\{x_1{x}_2+x_0^2=0\}\subseteq {\mathbb{P}}^2$, and take $u=x_1/x_0$ and $v=x_2/x_0$. Then $V{|}_{U_0}\simeq \{uv+1=0\}$.

Algebraic Sets vs. Varieties: Two Conventions (Concise)

Convention 1 (Common): Variety = Irreducible Algebraic Set

• Algebraic Set: Zero locus of polynomials (can be reducible).
• Variety: Irreducible algebraic set.

Convention 2 (Less Common): Variety = Algebraic Set

• Variety: Zero locus of polynomials (can be reducible or irreducible).
• Irreducible Variety: Used specifically for irreducible cases.

Key Point for Convention 2: Under this convention, “algebraic set” and “variety” are synonymous.