Is the "spacetime" the same thing as the mathematical 4th dimension?
Yes, time can be treated as a fourth axis- that idea was developed by a German mathematician called Hermann Minkowski not long after Einstein published his theory of special relativity (Minkowski was Einstein's supervisor for a time).
Representing time as a fourth axis- along with the usual three spatial axes- is now standard in text books and scientific papers. I saw a quote from Einstein implying that he didn't like it at first, something along the lines of 'Now that mathematicians have got hold of relativity I'm not sure I understand it myself any more.'
The Minkowski institute has a website where you can read English translations of his papers.
Minkowski's spacetime is in some ways analogous to 3D space. For example, in 3D space there is no predefined value of 'up', so you can pick any direction you like to orient your Z axis, for example. Likewise in Minkowski's space there's no predefined direction for the T axis- if two observers are moving relative to each other then their respective T axes diverge, with the divergence increasing with their relative speed.
You can use the concept of 4 D spacetime to get a feel for things like time-dilation and length contraction in a way that's analogous to measurements in ordinary space. For example if you use the normal 'Z equals up' orientation you might tell me that a certain flagpole is a hundred feet high and a foot wide. If I have my Z axis tilted away from yours I will say that the height of the flagpole is less than a hundred feet, but it's a lot wider than a foot. Similar things happen in spacetime, where a diverging direction for the T axis would mean that observers measure different elapsed times.
However, you can't take the analogy too far, as the geometry of Minkovski space (ie the rule for calculating distances etc) isn't the same as the geometry of Euclidian space, which is what we were all used to before we were introduced to relativity. In that respect you can't really think of time as something you can treat exactly like the three spatial dimensions.
That said, it turns out that mathematically you can represent 'flat' spacetime as Euclidian if you make your fourth dimension iT (ie T multiplied by the square root of -1). I read somewhere that representing spacetime that way used to be more popular.
So the direct answer is a great NO, because spacetime is something physical which has a deep mathematical meaning.
I) Intuitive Idea
Intuitively and roughly speaking, spacetime is the "place" of all events, or the set of all events. An event is something that "happens in a time $\tau$ and takes place somewhere". You can grasp the main concept with a simple example: You have a physics test, Friday 11:00 AM, at the Physics Departament building, on floor 5. Well, if you go to the right place but wrong time you will miss the test. To access the event "test" you have to be in the right place at right time. So, you necessarily must to deal with four numbers: one for time and three for space.
Because of relativity, the time are not just a parameter, but a coordinate! In Lorentz transformations you transform time as a usual coordinate. You must consider time as just another coordinate as the usual spatial ones.
II) Dimension
The most elementary definition of dimension comes from a mathematical subject called linear algebra, which is one of the "mathematical tools" used to properly describe general relativity (GR) in mathematical terms. In GR, we basically deal with finite-dimensional vector spaces, so the concept of dimension is the most elementary one:
A dimension is the number of the basis vectors of a given vector space.
So a "mathematical 4th dimension" is just a 4-dimensional vector space.
III) Spacetime: A brief explanation
Well, here is where we use math to describe physics. The physics of spacetime was introduced in 1905 with Einstein's paper. But spacetime was born in 1906 with Minkowski's paper. Now, there are some facts that we will use to construct the proper idea of spacetime:
1) In physics we can measure lengths and time and a mathematical object which have this property of "measure" is the norm given by a inner product. In Newtonian mechanics, the norm is the Euclidian one:
$$ \|v\|^{2} := \langle v,v\rangle = \sum^{3}_{i=1}\sum^{3}_{j=1} \delta_{ij}v^{i}v^{j} \tag{1}$$
where $\delta_{ij}$ is the matrix:
$$ \delta_{ij} = \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{bmatrix} $$
In a sense, this norm together with a vector space gives the geometrical structure of Newtonian mechanics because we can calculate lengths, define vectors, calculate velocities and accelerations, and so on....
2) This norm sets what we call "Euclidean space" or "Euclidean geometry". Note that if you define another dimension, "the 4th dimension", you will just construct a 4-dimensional Euclidean space.
Now, the physical fact is: the geometry of spacetime isn't Euclidean, because we use a particular norm called the "Minkowski norm" or "Lorentz norm" on a 4-dimensional vector space. Because of this fact all the "conventional linear algebra" must to be adaptated to the Lorenztian geometry, given by the norm:
$$ \|v\|^{2} := \langle v,v\rangle = \sum^{3}_{\mu=0}\sum^{3}_{\nu=0} \eta_{\mu\nu}v^{\mu}v^{\nu} \tag{2}$$
where $\eta_{\mu\nu}$ is the matrix:
$$ \eta_{\mu\nu} = \begin{bmatrix} -1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} $$
III) Spacetime: The general picture
The matrix of inner product $(2)$ (and in general) is called the components of the metric tensor $g$. The metric tensor is (roughly speaking) a bilinear map which produces a particular scalar called a line element, which is simply the value of the norm of differential line element vectors, i.e.
$$ ds^{2}\equiv g\Bigg(dx^{\mu}\frac{\partial \vec{r}}{\partial x^{\mu}},dx^{\nu}\frac{\partial \vec{r}}{\partial x^{\nu}}\Bigg) := \|d\vec{r}\|^{2} =: \langle d\vec{r},d\vec{r}\rangle = \sum^{3}_{\mu=0}\sum^{3}_{\nu=0} g_{\mu\nu}dx^{\mu}dx^{\nu} \tag{3}$$
Now, in general metric tensors aren't easy matrices like $\delta_{ij}$ and $\eta_{\mu\nu}$. In fact the metric tensor can become a tensor field which varies through space (and then the geometry varies pointwise).
To describe this general behaviour of "a tensor field which varies through space (and then the geometry varies pointwise)" we need the manifold mathematical framework (which is beyond the scope of this answer).
$$ g_{\mu\nu} = \begin{bmatrix} g_{00}(x^{\mu})&g_{01}(x^{\mu})&g_{02}(x^{\mu})&g_{03}(x^{\mu})\\ g_{10}(x^{\mu})&g_{11}(x^{\mu})&g_{12}(x^{\mu})&g_{13}(x^{\mu})\\ g_{20}(x^{\mu})&g_{21}(x^{\mu})&g_{22}(x^{\mu})&g_{23}(x^{\mu})\\ g_{30}(x^{\mu})&g_{31}(x^{\mu})&g_{32}(x^{\mu})&g_{33}(x^{\mu})\\ \end{bmatrix} $$
With this manifold framework, we can give a sufficiently general description of spacetime:
A spacetime is a 4-dimensional manifold $\mathcal{M}$ with a pseudo-riemannian metric $g_{\mu \nu}$: $$ (\mathcal{M},g_{\mu \nu}) $$
IV) Spacetime: Merging the intuitive idea with math
So, spacetime is the stage of special relativity and general relativity. It tells you which events are in your future, in your past and the ones which you cannot access in a sufficiently small proper time (the time of the clock in your hand, the time of the observer at rest on his own reference system, in general a tetrad). Spacetime is also a geometrical 4-dimensional entity which tells you that because of the Lorentz signature you need to give spatial and temporal directions and, of course, the geometry isn't Euclidean anymore.
Essentially, the dimension of a space is the number of numbers you need to specify a point in it.
- The Earth's surface is two-dimensional, because you need to specify longitude and latitude.
- The set of possible electromagnetic field values is six-dimensional, because you need to specify $E_x$, $E_y$, $E_z$, $B_x$, $B_y$, and $B_z$.
- The set of possible $1000 \times 1000$ RGB images is $3000000$ dimensional, because you need to specify R, G, and B color values for each of $1000000$ pixels.
The point is that "dimension" mathematically need not have anything to do with space or spacetime. Asking if "the fourth dimension is time" is kind of like asking if "the derivative is force". No, the mathematical derivative is just that, mathematical. The statements only make sense if you're more specific: the time derivative of momentum is force, and the fourth dimension of spacetime is time.
Stated this way, it's also clear that "spacetime" as a concept has nothing to do with relativity. It takes four numbers to specify a position and a time in relativity, but it also did in Newton's day. The only difference is that in relativity, spacetime can also be given a nice mathematical structure as a whole, as a Lorentzian manifold, which is why we talk about it more. But structures like this are also not special to relativity, as Newtonian spacetime can be given a Newton-Cartan geometry.