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mathwonk

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Here is an attempt at some more insight on this question. Suppose we think of power series over a field, of form a + bx + cx^2 +.....+ex^n +......

The ones that vanish at the origin form the maximal ideal M with zero constant coefficient, so they start at the linear term, i.e. look like bx + cx^2 +........ And the ones that not only vanish, but have derivative zero also at the origin, start at the quadratic term, so look like cx^2 + dx^3 +........., and thus form the ideal M^2.

Similarly those that start at the term ex^n+......., form the ideal M^n, of functions that vanish along with the first n-1 derivatives, at the origin. Since a function can have as many vanishing derivatives as desired, and still not be identically zero, the infinite descending sequence of ideals M, M^2, M^3,..... never terminates, so the ring of power series at the origin is not artinian, and in particular the submodule M is not artinian.

But if we take the quotient, say M/M^n, we are looking only at series that are truncated by throwing away all terms after the x^n term, hence there isn't much left, and this quotient is an n-1 dimensional vector space / the field of coefficients, hence it is artinian.

It's like looking at an infinite decimal, which goes on forever, and even if it only starts at the nth decimal place, still goes on forever. But if we only truncate it at the nth decimal place, like a hand held calculator does, then we get something finite. Its the same idea as this problem, start with something infinite, throw away some smaller infinite part at the tail end, and be left only with something finite.

Well forgive my rambling, but these abstract ideas are really inspired by very concrete objects from the world of calculus and infinite series, and that's where the intuition comes from for them.

In fact this is probably the example that might be most familiar to people with mainly calculus experience. I.e. let R = k[[x1,...,xn]] be the ring of power series in n variables at the origin, let M be the ideal of those power series with zero constant term, and consider the quotient M/M^2. These are power series starting at the linear term, but not having terms after that, hence consists of homogeneous linear polynomials. This is exactly the cotangent space of n - space, an n dimensional vector space, hence it is artinian as a module over R, where multiplication by a power series is defined just by multiplying by the constant term.

................

One way to look at a ring being artinian is from the point of view of dimension. A polynomial ring like k[x1,...,xn], over a field, has dimension n, corresponding to the fact that it represents functions on the space k^n which has dimension n. The dimension of the space is reflected in the existence of chains of subspaces like the origin, the x1 axis, the (x1,x2) plane, the (x1,x2,x3) space,..... that have length n+1, but no longer such chains exist.

Correspondingly the ring k[x1,...,xn] has chains of prime ideals corresponding to the ideals of functions that vanish identically on these subspaces, i.e. (x1,...,xn), (x2,...,xn), (x3,...,xn),....(0), which has length n+1 also, but no longer chains of prime ideals exist.

Now these chains of prime ideals are finite, but by taking powers of them you get infinite chains of non prime ideals, at least when there are some variables, i.e. when the polynomial ring has positive dimension, and these show that the ring is not artinian.

If the ring is artinian, you don't have these infinite chains of powers of prime ideals, so the ring must have had dimension zero. If you have a prime ideal P in the ring k[x1,...,xn] and we call the "height" of P the length of a chain of smaller prime ideals, then the height of P plus the dimension of the quotiwnt ring k[x1,...,xn]/P equals n, the dimension of k[x1,...,xn]. So if we want to get a zero dimensional quotient, we just need to mod out by a maximal height prime ideal, i.e. by maximal ideal. In this case that will give us a field, which is artinian.

More generally, a commutative artinian ring, without nilpotent elements, is just a product of fields, so these examples of (commutative) artinian rings being either fields, or of form R/m^r, for some maximal ideal m, are pretty much all there is.

To try to avoid confusion, none of the concrete examples I have given follows the general model I gave first, of constructing examples via taking a product AxN, forcing the quotient AxN/N ≈ A, to be of predetermined form. Of course using that method one can construct many more examples, but I think less interesting ones.

The ones that vanish at the origin form the maximal ideal M with zero constant coefficient, so they start at the linear term, i.e. look like bx + cx^2 +........ And the ones that not only vanish, but have derivative zero also at the origin, start at the quadratic term, so look like cx^2 + dx^3 +........., and thus form the ideal M^2.

Similarly those that start at the term ex^n+......., form the ideal M^n, of functions that vanish along with the first n-1 derivatives, at the origin. Since a function can have as many vanishing derivatives as desired, and still not be identically zero, the infinite descending sequence of ideals M, M^2, M^3,..... never terminates, so the ring of power series at the origin is not artinian, and in particular the submodule M is not artinian.

But if we take the quotient, say M/M^n, we are looking only at series that are truncated by throwing away all terms after the x^n term, hence there isn't much left, and this quotient is an n-1 dimensional vector space / the field of coefficients, hence it is artinian.

It's like looking at an infinite decimal, which goes on forever, and even if it only starts at the nth decimal place, still goes on forever. But if we only truncate it at the nth decimal place, like a hand held calculator does, then we get something finite. Its the same idea as this problem, start with something infinite, throw away some smaller infinite part at the tail end, and be left only with something finite.

Well forgive my rambling, but these abstract ideas are really inspired by very concrete objects from the world of calculus and infinite series, and that's where the intuition comes from for them.

In fact this is probably the example that might be most familiar to people with mainly calculus experience. I.e. let R = k[[x1,...,xn]] be the ring of power series in n variables at the origin, let M be the ideal of those power series with zero constant term, and consider the quotient M/M^2. These are power series starting at the linear term, but not having terms after that, hence consists of homogeneous linear polynomials. This is exactly the cotangent space of n - space, an n dimensional vector space, hence it is artinian as a module over R, where multiplication by a power series is defined just by multiplying by the constant term.

................

One way to look at a ring being artinian is from the point of view of dimension. A polynomial ring like k[x1,...,xn], over a field, has dimension n, corresponding to the fact that it represents functions on the space k^n which has dimension n. The dimension of the space is reflected in the existence of chains of subspaces like the origin, the x1 axis, the (x1,x2) plane, the (x1,x2,x3) space,..... that have length n+1, but no longer such chains exist.

Correspondingly the ring k[x1,...,xn] has chains of prime ideals corresponding to the ideals of functions that vanish identically on these subspaces, i.e. (x1,...,xn), (x2,...,xn), (x3,...,xn),....(0), which has length n+1 also, but no longer chains of prime ideals exist.

Now these chains of prime ideals are finite, but by taking powers of them you get infinite chains of non prime ideals, at least when there are some variables, i.e. when the polynomial ring has positive dimension, and these show that the ring is not artinian.

If the ring is artinian, you don't have these infinite chains of powers of prime ideals, so the ring must have had dimension zero. If you have a prime ideal P in the ring k[x1,...,xn] and we call the "height" of P the length of a chain of smaller prime ideals, then the height of P plus the dimension of the quotiwnt ring k[x1,...,xn]/P equals n, the dimension of k[x1,...,xn]. So if we want to get a zero dimensional quotient, we just need to mod out by a maximal height prime ideal, i.e. by maximal ideal. In this case that will give us a field, which is artinian.

More generally, a commutative artinian ring, without nilpotent elements, is just a product of fields, so these examples of (commutative) artinian rings being either fields, or of form R/m^r, for some maximal ideal m, are pretty much all there is.

To try to avoid confusion, none of the concrete examples I have given follows the general model I gave first, of constructing examples via taking a product AxN, forcing the quotient AxN/N ≈ A, to be of predetermined form. Of course using that method one can construct many more examples, but I think less interesting ones.

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