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I am trying to find the generators of the ideal $(3)$ in the ring of integers of $mathbbQ[sqrt-83]$ the ring of integers is $mathbbZleft[frac1+sqrt-832right]$ I evaluated the Minkowski constant and it is $2/pi sqrt83 sim 5.8;$ then I checked the minimal polynomial of $x^2-x+84/2, $ which is reducible module $3,$ hence the ideal $(3)$ is generated by two elements, right? Did I miss something? I want to say the ring of integers is not a UFD.

You seem to have touched upon several different ideas here.

Generators of the ideal $(3)$. Usually, when you talk about the generators of $(3)$, you mean its generators as an abelian group.

Define $theta := tfrac 1 2 (1 + sqrt-83)$. We know that the ring of integers $mathbb Z[theta]$ is generated by $1, theta $ as an abelian group, so $(3)$ is generated by $3, 3theta $ as an abelian group.

But having read the remainder of your question, it looks like this is not what you're after! What you're really after are the generators of the ideal class group for $mathbb Z[theta]$...

Minkowski constant. The fact that the Minkowski constant is $5.8$ implies that the ideal class group is generated by prime ideals that are factors of $(2)$ or $(3)$.

Dedekind's criterion. Dedekind's criterion is a way of factorising $(3)$ as a product of primes.

As you pointed out, we have the factorisation
$$ x^2 - x + frac 842 equiv x(x-1) mod 3,$$

so Dedekind's criterion says that
$$ (3) = (3, theta)(3, theta - 1)$$
is the prime factorisation of $(3)$ in $mathbb Z[theta]$.

Whether $mathbb Z[theta]$ is a UFD. This is equivalent to asking whether the ideal class group is trivial, i.e. whether all ideals are principal.

Why don't we check whether $(3, theta)$ is principal? The answer must be "no". Note that $(3, theta)$ has norm $3$. If it was principal, then there would exist $x, y in mathbb Z$ such that $(3, theta) = (x + ytheta)$. But then
$$ 3 = N(3, theta) = N(x + ytheta) = (x + tfrac 1 2 y)^2 + 83(tfrac 1 2 y)^2,$$
which is impossible.

So $mathbb Z[theta]$ is not a principal ideal domain, and hence, is not a unique factorisation domain.

The minimal polynomial of $alpha=frac1+sqrt-832$ is $f = x^2 - x + 84/4 = x^2 - x + 21$. Modulo $3$ this factors as $$f equiv x^2 - x = x(x-1) pmod 3,$$
so by the Kummer-Dedekind theorem the ideal $(3)$ factors into primes in $mathbbZbigg[frac1+sqrt-832bigg]$ as $(3) = (3, alpha)(3, alpha-1)$.

The ideal $(3)$ is principal, generated by $3$. You can show that the prime factors are not principal, e.g. using the field norm:

If $(3,alpha) = (beta)$ then $N(beta)$ divides both $N(3) = 3^2$ and $N(alpha) = f(0) = 21 = 3cdot 7$, so $N(beta) = 3$.

If we write $beta = a+balpha$ then $$N(beta)=(a+balpha)(a+b(1-alpha)) = ldots = a^2+ab + 21b^2.$$

The equation $a^2+ab + 21b^2 = 3$ is an ellipse in the $(a,b)$-plane without integral points, so there is no such $beta$.