Fulton 1.5

1.23. Give an example of a collection of ideals $\mathcal{P}$ ideals in a Noetherian ring such that no maximal member of $\mathcal{P}$ is a maximal ideal.

Take $\mathcal{P}=\{(x^n) \ | \ n > 0\}$ and note all ideals in $\mathcal{P}$ are properly contained in the maximal ideal $(x,2).$

1.24. Show that every proper ideal in a Noetherian ring is contained in a maximal ideal.

Let $P$ be a prime ideal in a Noetherian ring $R.$ Let $\mathcal{P}=\{I\subset R \ | \ I \text{ is an ideal, } P\subset I\}$ The maximal element of $\mathcal{P}$ is a maximal ideal of $R.$

Let $I$ be a non-zero prime ideal of $R[X].$

Suppose $J=I\cap R\ne \{0\}.$ Then $R[X]/I\cong (R[X]/J[X])/(I/J[X])\cong(R/J)[X]/(I/(J[X])).$ Since $R/J$ is a field, $(R/J)[X]$ is a Euclidean Domain. Thus $I/(J[X])=(F(X)+J[X])$ for some $F\in R[X],$ of minumum degree, such that $F+J[X]$ is irreducible over $(R/J)[X].$ Therefore $I=(J, F(X)).$

Suppose $I\cap R=\{0\}.$ Let $F\in I$ be of minimum degree. Suppose $G\in I$ such that $\text{gcd}(F,G)=1.$ By Gauss’ Lemma $\text{gcd}(F,G)=1$ in $K[X]$ where $K$ is the field of fractions of $R.$ Since $K[X]$ is a Euclidean domain, there exist $A,B\in K[X]$ such that $A(X)F(X)+B(X)G(X)=1\in K[X].$ If we let $\alpha\in R\setminus\{0\}$ be a common denominator for the coefficients of $A,B$ then $\alpha A(X)F(X)+\alpha B(X)G(X)=\alpha.$ Therefore $\alpha\in I,$ a contradiction. Thus $F\mid G$ for all $G\in I.$ So $I=(F).$

1.25. (a) Show that $V(Y-X^2)\subset\mathbb{A}^2(\mathbb{C})$ is irreducible, in fact, $I(V(Y-X^2))=(Y-X^2).$

Note $Y-X^2$ is irreducible in $\mathbb{C}(Y)[X].$ Thus by Gauss’ lemma $Y-X^2$ is reducible in $\mathbb{C}[Y][X]=\mathbb{C}[X,Y].$

(b) Decompose $V(Y^4-X^2,Y^4-X^2Y^2+XY^2-X^3)\subset\mathbb{A}^2(\mathbb{C})$ into irreducible components.

Let $V=V(Y^4-X^2,Y^4-X^2Y^2+XY^2-X^3).$ Observe $Y^4-X^2=(Y^2-X)(Y^2+X)$ and $Y^4-X^2Y^2+XY^2-X^3=(Y^2-X^2)(Y^2+X) =(Y-X)(Y+X)(Y^2+X).$ Thus $V = V(Y^2-X,(Y-X)(Y+X))\cup V(Y^2+X).$ Observe if $Y-X=0$ or $Y+X=0$ then $Y^2=X^2.$ So $X^2-X=0$ and $X=0$ or $1.$ $\begin{aligned}[t] V(Y^2-X,(Y-X)(Y+X))&=V(X,Y)\cup V(X-1,Y^2-1)\\ &=V(X,Y)\cup V(X-1,Y-1)\cup V(X-1,Y+1). \end{aligned}$ By lemma 2, these are each irreducible varieties. Thus $V=V(X,Y)\cup V(X-1,Y-1)\cup V(X-1,Y+1)\cup V(Y^2+X).$

1.26. Show that $F=Y^2+X^2(X-1)^2\in\mathbb{R}[X,Y]$ is an irreducible polynomial, but $V(F)$ is reducible.

Observe $Y^2+X^2(X-1)^2$ is irreducible in $k(Y)[X]$ and thus irreducible in $k[X,Y].$ However, $V(F)=\{(0,0),(1,0)\}=V(X, Y)\cup V(X-1, Y).$

1.27. Let $V,W$ be algebraic sets in $\mathbb{A}^n(k)$ with $V\subset W.$ Show that each irreducible component of $V$ is contained in some irreducible component of $W.$

Let $V=V_1\cup\ldots\cup V_n$ and $W=W_1\cup\ldots\cup W_m$ be the decomposition of each into irreducible algebraic sets. Let $i\in\{1,\ldots,n\}.$ Suppose $V_i \not\subset W_j$ for any $j\in\{1,\ldots,m\}.$ Then since $V\subset W$ it must be that $V_i\subset \left(\bigcup_{k} W_{j_k}\right).$ But then $(V_i\cap W_{j_1}),\ldots,(V_i\cap W_{j_k}$ is a decomposition of $V_i$ into proper algebraic subsets, a contradiction.

1.28. If $V=V_1\cup\ldots\cup V_r$ is the decomposition of an algebraic set into irreducible components, show that $V_i\not\subset\bigcup_j\ne i V_j.$

Suppose $V_i\subset V_j$ for some $j\ne i.$ Then $V_i\cap V_j$ and $V_i\cap\left(\bigcup_{k\ne j}V_k\right)$ is a decomposition of $V_i$ into proper algebraic subsets, a contradiction.

1.29.* Show that $\mathbb{A}^n(k)$ is irreducible if $k$ is infinite.

Suppose $k$ is an infinite field. Suppose $\mathbb{A}^k=V_1\cup V_2.$ Then $V_1\subset V(F)$ and $V_2\subset V(G)$ for some nonzero $F,G\in k[X_1,\ldots,X_n].$ So $\mathbb{A}^k= V(F)\cup V(G)=V(FG).$ By Problem 1.4, $FG=0,$ thus $F=0$ or $G=0,$ a contradiction.