If $k$ is a field , and $m=\langle (x,y) \rangle$ , then how to calculate $\dim _k (m^2/m^3)$?

Question

If $k$ is a field , and consider the maximal ideal $m=\langle (x,y) \rangle$ , then how to calculate $\dim _k (m^2/m^3)$ ? I can only come up with $\dfrac {k[x,y]/m^3}{m^2/m^3} \cong k[x,y]/m^2 $ , but I dont know how to proceed next or if this is the right way. Please help. Thanks in advance

Answer 1

You can compute the square and cube of the ideal explicitly: $(x,y)^2 = (x^2, xy,y^2)$, and $(x,y)^3 =(x^3, x^2y, xy^2, y^3)$. So you want to find the dimension of $$ (x^2, xy,y^2)/(x^3, x^2y, xy^2, y^3). $$ An arbitrary element of $(x^2, xy,y^2)$ looks like $$ x^2\sum_{v_1+v_2 = 0}^{r_1}a_{v_1,v_2} x^{v_1}y^{v_2} + xy \sum_{v_1+v_2 = 0}^{r_2}b_{v_1,v_2} x^{v_1}y^{v_2} + y^2 \sum_{v_1+v_2 = 0}^{r_3}c_{v_1,v_2} x^{v_1}y^{v_2} $$ (where all $v_i$ are nonnegative, and $a_{i,j}, b_{i,j}, c_{i,j}\in k$).

Now, you have to find representatives for nonzero elements of the quotient, or equivalently, throw out all elements of the above form which become zero in the quotient. As $x^3 = 0$, we can throw out all terms in $\sum_{v_1+v_2 = 0}^{r_1}a_{v_1,v_2} x^{v_1}y^{v_2}$ such that $v_1 > 0$, and because $x^2 y = 0$, we can throw out terms of $\sum_{v_1+v_2 = 0}^{r_1}a_{v_1,v_2} x^{v_1}y^{v_2}$ where $v_2 > 0$. Thus, we have narrowed down representatives of elements in the quotient to simply polynomials of the form $$ a_{0,0}x^2 + xy \sum_{v_1+v_2 = 0}^{r_2}b_{v_1,v_2} x^{v_1}y^{v_2} + y^2 \sum_{v_1+v_2 = 0}^{r_3}c_{v_1,v_2} x^{v_1}y^{v_2}. $$

You can treat the other terms similarly, and then you simply need to show that the elements you're left with are linearly independent, and count your degrees of freedom (i.e., if you were only left with polynomials of the form $a_{0,0}x^2$, you have one degree of freedom coming from the choice of $a_{0,0}\in k$, so the vector space would have $k$-dimension $1$).

Spoilers below:

You ought to find that you have representatives of the form $$ ax^2 + bxy + cy^2, $$ so that $\dim_k\mathfrak{m}^2/\mathfrak{m}^3\leq 3$. Moreover, if $ax^2 + bxy + cy^2$ is $0$ in the quotient, then you must have $ax^2 + bxy + cy^2\in\mathfrak{m}^3$, but elements of $\mathfrak{m}^3$ are of the form $x^3 f(x,y) + x^2yg(x,y) + xy^2 p(x,y) + y^3 q(x,y)$, with $f,g,p,q\in k[x,y]$, and such a sum has no $x^2$, $xy$, and $y^3$ terms, so that $a = b = c = 0$, and hence $\dim_k\mathfrak{m}^2/\mathfrak{m}^3= 3$.