The square of trace is not contained in the ideal generated by $A^2=0$ Let $A=\left[\begin{array}{*{20}{c}}
a&b\\
c&d
\end{array}\right]$ be a $2\times 2$ matrix over a field $k$, and $A^2=0$. Then we have four homogeneous polynomials in $a,b,c,d$:
$$
a^2+bc, \quad c(a+d),\quad b(a+d),\quad d^2+bc.
$$
Let $I$ be the ideal generated by above polynomials. Then, by Nullstellensatz, the trace of $A$, $a+d$, is contained in $\sqrt I$. I'd like to find the smallest degree $r$ such that $(a+d)^r\in I$. I've worked out that $(a+d)^3\in I$, but I can't show $(a+d)^2\in I$ or $(a+d)^2\notin I$.
Any hint please?
 A: There is a mechanical way to check whether an element of a polynomial ring belongs to an ideal with given generators. It first finds  a Groebner basis of the ideal ( wr to a certain ordering), and then performs a reduction of the element w r to the Groebner basis, with a unique remainder. The element belongs to the ideal, if and only if the remainder is $0$.
The first calculation in WA finds the Groebner basis
$$\{b c + d^2, a d^2 + d^3, a c + c d, a b + b d, a^2 - d^2\}$$
and the second calculation produces the remainder of $(a+d)^2$  wr to the above basis. The remainder is $2 a d + 2 d^2$, so not $0$ ( we assume $k$ of characteristic $0$).
Therefore, $(a+d)^2$ does not belong to the ideal.
Note that with similar calculations we check that the determinant $ a d - b c$ does not belong to the ideal, but its square $(a d - b c)^2$ does (!).
This question is also related, and there are further links there.
$\bf{Added:}$
There is a simpler solution to the problem. Say we are trying to solve the system
$$a^2 + b c = d^2 + b c = b(a+d)= c(a+d)=0$$
in a commutative ring $R$.  We can get a particular solution if we take $b=c=0$. Then the system becomes
$$a^2 = b^2 = 0$$
Now, does every solution $(a,b)$ to this latter system satisfy $(a+b)^2=0$?  We know that $(a+b)$ is nilpotent, but maybe not of order $2$. And indeed, we can see a simple example in the ring $R = k[x,y]/(x^2, y^2)$, where $2 \ne 0$ in $k$. Take $a = x$, $b=y$, and we have $(x+y)^2 = x^2+ 2 x y + y^2 = 2 x y \ne 0$.  Note also that $a d - b c = a d = x y$ again is not zero in this ring $R$, therefore $ad - b c$ does not belong to the ideal with the given generators ( if it did, then any solution of the system also satisfies  $a d - b c = 0$).
