Context
Let $\theta$ be an algebraic number of degree $n$ over $\mathbb{Q}$, and let $K=\mathbb{Q}[\theta]$. Then we have field isomorphisms $\sigma_1,\ldots\sigma_n$ such that $\sigma_1(\theta)=\theta$, and the other $n-1$ send $\theta$ to each of its algebraic conjugates (which may or may not belong to $K$). Applying these maps to the entire field $K$, we get its $n$ different embeddings in $\mathbb{C}$.
Definitions
For an element $\alpha\in K$, define $S(\alpha)=\prod_{i=2}^n\sigma_i(\alpha)$, the product of all of $\alpha$'s algebraic conjugates, excuding itself. Equivalently, define $S(\alpha)$ by the conditions $S(0)=0$, and $\alpha\cdot S(\alpha)=N_K(\alpha)$, where the right hand side is the field norm of $\alpha$ in $K$.
Now, let $\mathcal{O}$ be the ring of integers of $K$ (the elements satisfying monic integer polynomials), and let $I\subseteq\mathcal{O}$ be an ideal of that ring.
A non-zero number $\alpha\in I$ is a "weakly minimal element" of $I$ if the only $\beta\in I$ satisfying $|\sigma_i(\beta)|<|\sigma_i(\alpha)|$ for every $i=1,\ldots,n$ is the trivial element $\beta=0$.
A non-zero number $\alpha\in I$ is a "strongly minimal element" of $I$ if the only $\beta\in I$ satisfying $|\beta|<|\alpha|$ and $|S(\beta)|<|S(\alpha)|$ is the trivial element $\beta=0$.
My question
Glancing at the contrapositives of these statements, it is clear that strongly minimal implies weakly minimal, which is why I chose the names that way. Is the converse true, and if not, what is a counterexample?
My thoughts
If $[K:\mathbb{Q}]=2$, then the two definitions coincide, and there's nothing to talk about. If $[K:\mathbb{Q}]=3$, and $K$ is real, with its other two embeddings complex conjugates, then the definitions still coincide, because $|\sigma_2(\beta)|=|\sigma_3(\beta)|$ for all $\beta\in K$.
I suspect the perspective I'm taking might be limiting my thinking here, because I'm thinking of $K$ as a subset of $\mathbb{C}$, as opposed to thinking of it as the quotient $\mathbb{Q}[x]/(p)$, where $p$ is the minimal polynomial of $\theta$. The latter is an abstract field, and no particular embedding enjoys a privileged spot, so I think that's probably the perspective I should adopt, but I don't know how :/
Any help is greatly appreciated.
So your definition only really makes sense assuming you have fixed an embedding $\sigma_1$ of $K \hookrightarrow \mathbf{C}$, so assume this.
Let $K = \mathbf{Q}[x]/(x^3 + 3x^2 - 12x + 6)$, which is a totally real field with three embeddings and $\mathcal{O}_K = \mathbf{Z}[x]$. Let $\sigma_1$ denote the embedding such that $$x = 1.805602\ldots$$ Let $I = (2,x)$.
Give the three conjugates $\{1.805\ldots,0.613\ldots,-5.418\ldots\}$ of $x$, it is a simple enough exercise in linear algebra to see that the only elements of $\mathcal{O}_K$ all of whose conjugates are strictly less than $2$ are $0,1,-1$. Hence $2$ is a weakly minimal element of $I$, because $I$ is a proper ideal.
On the other hand, $2$ is not strongly minimal, because
$$1.805602\ldots = |x| < 2, \ |S(x)| = \frac{6}{|x|} = 3.32299\ldots < 4 = S(2).$$
How to find this: If $I = (\alpha)$ is a principal ideal, then it is easy to see that $\alpha$ will be both weakly and strongly minimal. So imagine that $I = (2,x)$ and one wants $2$ to be weakly minimal but not strongly minimal. You need to take at least a cubic field. The conditions on $x$ to realize $2$ as not strongly minimal are $|x| < 2$ and $N(x)/|x| < 4$ so $2 > |x| > N(x)/4$. That meant that $N(x) < 8$ is small. In order for $(2,x)$ to be a proper ideal, $N(x)$ should be even. So take $|N(x)| = 6$ and $f(x) = x^3 + a x^2 + b x + 6$. In order for there to be a root slightly less than $2$, you want $f(2)$ to be small. It is always even if $f(0)$ is even, so set $f(2) = 2$, or $b = -2(3+a)$. Then play around with some small examples, in particular making sure that the ideal $I = (2,x)$ is not principal.