Understanding why $B$ is generated by a finite number of monomials and how this affects $B_0$.

Question

Here is the question I want to understand its statement:

Let $B = \mathbb C[x_1, x_2, x_3, x_4]$ be $\mathbb Z$-graded so that each $x_i$ is homogeneous and $\deg(x_1, x_2, x_3, x_4) = (1, -2, 3, -4).$ Find $B_0.$

Hint: This ring is generated by a finite number of monomials.

My thoughts:

To find $B_0$ I have to solve this equation $$e_1 -2 e_2 + 3e_3 -4 e_4 = 0.$$

But this equation has infinitely many solutions, so I am guessing that here comes the importance of the given hint but for my life I do not understand how? could someone explain this to me ,please?

Some details of why specifically the above equation:

As I know that I am going to put a $\mathbb Z$-grading on $B$ where $B = \bigoplus_{i \in \mathbb Z} B_i$ and $x_i \in B_{\deg x_i}.$So according to the degrees we have in the question we have $x_1 \in B_1, x_2 \in B_{-2}, x_3 \in B_3, x_4 \in B_{-4}.$ And because I know that $B_i$ is the vector space with basis $\{X_1^{e_1}, X_2^{e_2}, X_3^{e_3}, X_4^{e_4}\}$ where $e_1 \deg x_1 + e_2 \deg x_2 + e_3 \deg x_3 + e_4 \deg x_4 = i.$

Answer 1

I claim $B_0$ is generated as a $\Bbb C$-algebra by the following monomials: $$1,x_1^2x_2, x_1^4x_4, x_2^3x_3^2, x_3^4x_4^3, x_1x_3x_4, x_1x_2^2x_3, x_2x_3^2x_4.$$ We'll show that given any monomial in $B_0$, it's a product of the above and a constant.

Suppose $m=x_1^ax_2^bx_3^cx_4^d$ belongs to $B_0$. This is equivalent to $a-2b+3c-4d=0$. We note that $x_1^2x_2$, $x_1^4x_4$, $x_2^3x_3^2$, and $x_3^4x_4^3$ are all in $B_0$, so if we're looking for other generators $m$ of $B_0$, it suffices to consider $m$ after factoring out as many of these monomials as possible. This gives us the following:

• If $a\geq 2$ and $b\geq 1$, then we can write $m=x_1^2x_2m'$, so we may assume that either $a<2$ or $b=0$.
• If $a\geq 4$ and $d\geq 1$, then we can write $m=x_1^4x_4m'$, so we may assume that either $a<4$ or $d=0$.
• If $b\geq 3$ and $c\geq 2$, then we can write $m=x_2^3x_3^2m'$, so we may assume $b<3$ or $c<2$.
• If $c\geq 4$ and $d\geq 3$, then we can write $m=x_3^4x_4^3m'$, so we may assume $c<4$ or $d<3$.

Now we have some case work. Suppose $m$ is a generator, and we'll go through all the combinations of the observations above.

• If $b=d=0$, then $a=c=0$, and so $m=1$.
• If $b=0$ and $a<4$, then we can apply our fourth observation: either $c<4$, and we find $(1,0,1,1),(2,0,2,2)$ as nontrivial solutions, or $d<3$ and we find the same solutions.
• If $d=0$ and $a<2$, then we can apply the third observation: either $b<3$, and we get $(1,2,1,0)$, or $c<2$ and we find the same solution.

If we assume $a<2$ and $b,d\neq 0$, we can run through the four ways to combine observations 3 and 4:

• If $b<3$ and $c<4$, then $\max(a+3c)=10$ and we have the solutions $(1,2,1,0)$, $(1,0,1,1)$, $(0,1,2,1)$, $(1,0,1,1)$, $(1,1,3,2)$, and $(1,3,3,1)$; the new only new generator we get here is $(0,1,2,1)$, as $(1,1,3,2)=(1,0,1,1)+(0,1,2,1)$ and $(1,3,3,1)=(1,0,1,1)+(0,3,2,0)$.
• If $b<3$ and $d<3$, then $\min(-2b-4d)=-12$ and all the solutions we get are things we've seen before.
• If $c<2$ and then $\max(a+3c)=4$ and we've seen all of those solutions before.

So our list of monomials from the start is a complete list of generators. There's probably a faster/easier way to do this, and I'll be glad to upvote someone who produces it.