Fulton 2.6

2.33. Factor $Y^3-2XY^2+2X^2Y+X^3$ into linear factors in $\mathbb{C}[X,Y].$

We note that factoring $F(X,Y)=Y^3-2XY^2+2X^2Y+X^3$ is the same as factoring $F(X,1)$ or $F(1,Y).$ Unfortunately, neither factors nicely, so we will simply note that $F(X,Y)=(X-Y\lambda_1)(X-Y\lambda_2)(X-Y\lambda_3),$ where $\lambda_1,\lambda_2,\lambda_3$ are the roots to $F(X,1)$ in $\mathbb{C}.$

2.34. Suppose $F,G\in k[X_1,\ldots,X_n]$ are forms of degree $r,r+1$ respectively, with no common factors ( $k$ a field). Show that $F+G$ is irreducible.

Suppose $F+G=HJ$ for some $H,J\in k[X_1,\ldots,X_n].$ By Proposition 5, $(F+G)^*=X_{n+1}F+G=H^*G^*=(HG)^*.$ Since $X_{n+1}F+G$ is degree 1 as a polynomial in $X_{n+1},$ one of $H^*,G^*$ must be of degree $0$ as a polynomial in $X_{n+1}.$ Therefore one of $H^*,G^*$ divides $F$ and $G$ and is therefore constant, since $F,G$ share no common factors. Thus one of $H,G$ are constant. Hence $F+G$ is irreducible.

2.35.* (a) Show that there are $d+1$ monomials of degree $d$ in $R[X,Y],$ and $1+2+\ldots+(d+1)=(d+1)(d+2)/2$ monomials of degree $d$ in $R[X,Y,Z].$

A monomial of degree $d$ in $R[X,Y]$ is of the form $X^iY^j,$ $i+j=d.$ There are $d+1$ options for $i$ and $j=d-i.$

A monomial of degree $d$ in $R[X,Y,Z]$ is of the form $F_iZ^j$ where $F_i$ is a monomial of degree $i$ in $R[X,Y]$ and $i+j=d.$ There are $i+1$ monomials of degree $i$ in $R[X,Y].$ Thus there are $1+2+\ldots+(d+1)=(d+1)(d+2)/2$ monomials of degree $d$ in $R[X,Y,Z].$

(b) Let $V(d,n)=\{\text{forms of degree } d \text{ in } k[X_1,\ldots,X_n]\},$ $k$ a field. Show that $V(d,n)$ is a vector space over $k,$ and the monomials of degree $d$ form a basis.

It is immediate that $V(d,n)$ is a vector space with the natural addition and scalar multiplication operations. Clearly the monomials of degree $d$ span $V(d,n)$ and are linearly indpendent over $k.$ Thus they form a basis.

(c) Let $L_1,L_2,\ldots$ and $M_1,M_2,\ldots$ be sequences of nonzero linear forms in $k[X,Y],$ and assume no $L_i=\lambda M_j,$ $\lambda\in k.$ Let $A_{ij}=L_1L_2\ldots L_iM_1M_2\ldots M_j,$ $i,j\ge 0$ ( $A_{00}=1$ ). Show that $\{A_{ij} \, | \, i+j=d\}$ forms a basis for $V(d,2).$

It suffices to show that $\{A_{ij} \, | \, i+j=d\}$ is a linearly independent set.

Suppose to the contrary that $0= \sum_{i+j=d} a_{ij}L_1\ldots L_i M_1\ldots M_j, \ a_{ij}\in k\setminus\{0\}.$ Then $L_1\ldots L_d=\sum_{i+j=d, j>0} a_{ij}L_1\ldots L_i M_1\ldots M_j.$ So $L_1\ldots L_d=M_1\sum_{i+j={d-1}} a_{ij}L_1\ldots L_i M_2\ldots M_j.$ Thus, since $L_1,\ldots, L_d, M_1$ are all linear, $M_1=\lambda L_i,$ a contradiction.

2.36. With the above notation, show that $\dim V(d,n)=\binom{d+n-1}{n-1}.$

There are $n$ variables, whose powers must add up to $d.$ This is counted by $\left(\binom{n}{d}\right)=\binom{d+n-1}{n-1}.$