Could someone explain the steps to finding the generator of the prime ideal

Question

Question:

Answer:

So I understand part a and the fact that because it's euclidean it must be a PID but how do you get (-1,4) are the minimal co-ords is it just trail and error? and the fact that 3+4i is the generator you get.

Answer 1

First, a small correction: the $x$ and $y$ we're looking for should minimize the given expression with the constraint that $(x,y) \neq (0,0)$. Of course the expression is actually minimized at $(0,0)$, but this does not give a generator of the ideal; we want a non-zero element of $I$ with minimal Euclidean norm.

With that out of the way, note that $$(25x + 7y)^2 + y^2 = 625 x^2 + 350 xy + 50y^2 = 25(25x^2 + 14xy + 2y^2),$$ so it suffices to minimize $f(x,y) := 25x^2 + 14xy + 2y^2$.

Note that $z = 25x^2 + 14xy + 2y^2$ defines a paraboloid in $\mathbb{R}^3$. Thus, it should be minimized over $\mathbb{Z}^2 \setminus \{(0,0)\}$ at a point near the minimum in $\mathbb{R}^2$. As previously mentioned, $f$ attains a minimum at $(0,0)$, so we are looking for an integer point "near the origin".

To make this precise, we begin by finding an upper bound for the minimum value. Simply note that $f(0,1) = 2$, so we want a point $(x,y) \in \mathbb{Z}^2 \setminus \{(0,0)\}$ such that $f(x,y) \leq 2$ in particular. Indeed, if $(0,1)$ is not the desired point, we must be looking for a point $(x,y) \in \mathbb{Z}^2$ with $f(x,y) = 1$ (since the paraboloid only attains the value $0$ at the origin). So let's suppose for now that such a point exists, and if we reach a contradiction we'll conclude that in fact $(0,1)$ is the desired point.

For any fixed $x = x_0$, by basic calculus, the function $y \mapsto f(x_0, y)$ is minimized at $y = \frac{-7}{2}x_0$ with minimum value $x_0^2/2$. Thus $f(x,y) \geq x^2/2$ for all $(x,y) \in \mathbb{Z}^2$. In particular, our desired point $(x,y)$ satisfies $1 = f(x,y) \geq x^2/2$, so $x \in \{-1,0,1\}$.

Likewise, for any fixed $y = y_0$, the function $x \mapsto f(x,y_0)$ is minimized at $x = \frac{-7}{25} y_0$ with minimum value $y_0^2/25$. Thus $f(x,y) \geq y^2/25$ for all $(x,y) \in \mathbb{Z}^2$. In particular, our desired point $(x,y)$ satisfied $1 = f(x,y) \geq y^2/25$, so $y \in \{-5, \dots, 5\}$.

We now know that if any point $(x,y) \in \mathbb{Z}^2 \setminus \{(0,0)\}$ satisfies $f(x,y) = 1$, then it lies in the finite set $\{-1,0,1\} \times \{-5, \dots, 5\}$. We now plug in all $33$ elements of this set into the function $f$, and find that $(-1,4)$ has the desired property!

I'm sure this isn't the most efficient approach, it's just what I came up with first.

Now that we have $x = -1, y = 4$ we plug this in to get the generator $$25(-1) + (7+i)4 = 3+4i.$$ (your question about this step belies some confusion about the basic setup of the solution. please try to re-explain to yourself where the $3+4i$ came from!)