Why do we specify the degree of elements in algebra? In algebra (especially algebraic topology where I have seen it) when we have some sort of ring like $R[x]$ we are quick to specify that $x$ has degree $n$. I understand the importance of this but isn't this just the same as the ring $R[x^n]$?
I would find the later notion more compact and clear given that in early mathematical education $R[x]$ is always assumed to have $x$ in degree 1. In fact then it seems to me that only a single definition of $R[x]$ needs to be given.
I am interested in the historic or good reasons for defining it above. This notion is really common in cohomology for example.
 A: Recall the graded commutativity of the cup product: $xy = (-1)^{(\deg x) (\deg y)} yx$ for homogeneous elements $x,y$. Applied to $x = y$, you get $x^2 = (-1)^{\deg x} x^2$. If $x$ has odd degree (and you're not working in characteristic $2$...) then this implies that $x^2 = 0$ and thus $x^n = 0$ for all $n \ge 2$. But you want to define the generator as $x^n$ where $x$ has degree $1$, and $1$ is odd, so you would have $x^n = 0$ automatically... The problem with your notation is that it suggests that the generator of the ring is a power of some element of degree one, and that's problematic.
One other obvious reason. When you're dealing with cohomology, then a cohomology class of degree $n$ on a space $X$ is something that eats continuous maps $\sigma : \Delta^n \to X$ and spits a number (+ cocycle condition). The degree is intrinsic to the class, you cannot forget it. The cohomology ring is not a subring of $R[x]$ generated by $x^n$ (even if we forget that that's problematic, see above). It has an element in degree $n$, and there's no way to see it as something other than in degree $n$. So it doesn't feel natural at all to introduce some artificial class of degree $1$ and then decree that $x$ is actually $x^n$. The class $x$ exists on its own right, and it's of degree $n$, and it happens to generate the ring.
Moreover it's possible for $x$ to have degree zero (hence the whole ring is concentrated in degree zero). There's no way to represent this correctly with your point of view: $x^0 = 1$ so that doesn't work. Or you could say that "$x$ has degree zero", but your goal is to precisely avoid that. Even "crazier", you could have some $x$ of negative degree (not when dealing with usual cohomology of course, but it's not the only setting where graded rings appear).
And you're not accounting for relations or more complicated rings. What do you do if you find a cohomology ring that looks like $R[x,y,z] / (x^3 = y^2 + 3z^7)$, say with $\deg x = 14$, $\deg y = 7$ and $\deg z = 2$? Then it's just a pain. You need to rewrite all your expressions in terms of dummy variables such that $x = (...)^{14}$, $y = (...)^7$, $z = (...)^2$, do your computations with that, and in the end remember that you need to divide any exponent of $x$ by $14$ and so on. It's just not worth it (and of course impossible, because of my point in the first paragraph – graded commutativity would go berserk).
