In a future theorem we will prove that a left semisimple ring is a finite product of simple modules. Therefore, a left semisimple ring is left noetherian and left artinian.
The easy proof, which I didn't think of right away, is to write a decomposition series for M. Each simple module acts as a factor group, as we join fewer and fewer modules together to make smaller and smaller submodules. A finite decomposition series implies M is both noetherian and artinian.
Here is another proof, based on linear algebra. It's a little more involved, but I like it, so I'm leaving it up on my site.
Since M is spanned by n simple modules, select a generator for each module, and call these generators g1 through gn. (Remember that a simple module Mi is completely spanned by its generator gi.) Thus every element of M is a linear combination of these generators. In other words, x can be written as the sum of cigi, where the coefficients are drawn from R. We place the coefficients on the left, because M is a left R module.
Consider an ascending chain of length n, where the first submodule U1 is nontrivial. Let x be a nonzero element in U1, and write it as a linear combination of generators. Since every multiple of x lies in U1, every multiple of the corresponding vector c1g1+c2g2+c3g3+…+cngn lies in U1. The submodule U1 includes a "subspace" of dimension 1, spanned by x.
The next module U2 properly contains U1. It brings in another element y, which is another linear combination of generators. In fact, this vector is linearly independent from the vector x. It has to be independent, else y would lie in U1. Write the coefficients of y below the coefficients of x, building the second row of a matrix.
Let z be an element that is in U3 and not U2. This too is a linear combination of generators, and the vector associated with z is not spanned by the two vectors associated with x and y. The threee vectors, i.e. the first three rows of our matrix, are linearly independent.
Continue this process up the chain, through n submodules, building an n×n matrix of linearly independent vectors. There is no room for another vector, no room for a larger submodule Un+1, and the process stops.
That pretty much completes the proof, except for the fact that R may not be a field, or even a domain. Our reliance on linear algebra is a bit suspect. When R is not well behaved, it is possible to have more than n "vectors", where each is not spanned by the previous. Let's look at an example in one dimension. Consider the ring Z30. When this acts as its own R module, we can build a chain of three submodules, spanned by the elements 15, 3, and 1 respectively.
As it turns out, the linear algebra is valid because the modules are simple.
Return to our matrix, consisting of n linearly independent vectors, corresponding to the n elements brought in by our ascending chain. apply the process of gaussian Elimination on this matrix. So if the first two rows look like this, we multiply the first by ½ and subtract from the second, to clear column 1.
2 1 5 3 …
1 7 9 0 …
The scaling factor ½ is obtained by dividing 2 into 1. It's just what the doctor ordered, to clear the first column. But if R is not a division ring, we can't just divide coefficients. Fortunately we don't have to. The first and second entries in the leftmost column of our matrix are actually elements in the first simple module M1. Given any two nonzero elements from M1, the first spans the second. There is some multiplier in R that carries the first onto the second. Use this multiplier to scale the first row, then subtract from the second, and the result, which still lies in U2, has a zero in the first column. We can indeed perform gaussian elimination, clearing everything below the main diagonal.
Once this is done, use upward gaussian elimination to clear everything above the main diagonal. The result is a diagonal matrix. The last submodule in the chain, Un, contains an isolated entry from each of the n simple modules. These entries span their respective modules, which span M. Therefore Un = M, and the chain stops.
Our semisimple module, or ring, is both left noetherian and left artinian.