In the pieces so far, mean field returned a transition. But mean field is an approximation, made by replacing the neighbours with an average. Whether it is right can only be checked against a case that can be solved without approximating at all.
The easiest such case is one dimension. Lay the spins in a single line and tie only neighbouring pairs. In that shape the whole thing can be carried through without a single approximation.
The figure is that chain. So long as the temperature is not 0, somewhere along it a spin has turned over. That one fact, that something somewhere turns over, settles the answer for this piece.
The sum runs over 2^N arrangements, which is no number to lay out honestly. But in one dimension a spin is tied only to its left and its right, so they can be cleared away one at a time from an end.
Write the weight e^{\beta J s s'} between a neighbouring pair as a 2 \times 2 array. This is called the transfer matrix, and the partition function is the trace of its N th power: Z = \mathrm{Tr}\, T^N.
The eigenvalues of that matrix are \lambda_+ = 2\cosh\beta J and \lambda_- = 2\sinh\beta J, giving Z = \lambda_+^N + \lambda_-^N. For large N only the larger of the two survives.
Taking the free energy from what is left, Z = (2\cosh\beta J)^N, there is no kink anywhere in temperature. \cosh is smooth, and differentiable at every \beta. Where there is no kink there is no transition.
Working out the magnetisation gives 0 at every temperature, unless a field is applied from outside. The flat line in the figure is that, with the mean-field curve laid over it for comparison.
Mean field returned a transition and the exact calculation did not. It is the approximation that was wrong. So what did mean field fail to count?
Turn one place over in the middle of an ordered chain and one boundary appears. The energy goes up by 2J, because one tie is broken where the boundary sits. Up to here mean field would say the same.
But that boundary may be put anywhere along the chain. There are N places to put it, so the entropy goes up by k\log N. It is this count of places that mean field never took.
The change in free energy is 2J - kT\log N. Make N large and the second term wins at any temperature, however low. A boundary always goes in, and the ordered state cannot hold. The crossing of the two curves in the figure is the size at which boundaries begin to enter. For infinite N that temperature drops all the way to 0.
In two dimensions a boundary is not a point but a line. Making a boundary of length L costs 2JL, which grows in proportion to the length.
The count of ways grows too. A boundary branches a few ways at every step, so the entropy is about kL\log 3. Energy and entropy are therefore both proportional to L: which of them wins is not settled by the size of the system, only by the temperature.
So in two dimensions, below some temperature boundaries stop going in, and there is a transition. What separated one dimension from two was whether the cost of a boundary grows with the size of the system. Mean field failed in one dimension because it never counted these boundaries once. That is what was meant in the previous piece by saying that dimension settles whether fluctuation matters.