Show that the A-modules $\displaystyle A^{2010}$ and $\displaystyle A^{2012}$ cannot be isomorphic for any commutative ring with 1.
Could someone please show me step by step how to solve it?
consider the map f:A^{2012}--->A^{2010} given by:
f((a_{1},a_{2},....,a_{2010},a_{2011},a_{2012})) = (a_{1},a_{2},....,a_{2010}).
show f is an A-linear map. note that (0,....,0,1) is a non-zero element of the kernel.
but this is an endomorphism right? (i. e homomorphism between two modules), so I need to show that this endomorphism is injective and surjective and thus is an isomorphism, but how to show that?
It would have been surjective if the kernel was trivial but as you wrote the kernel is is not trivial...
And to show that f is A-linear I should show smth like that:
$\displaystyle f(am+bn)=af(m)+bf(n)$, $\displaystyle \forall a,b\in A, \forall m,n\in A^{2012}$?
thanks in advance
ah of course....I mixed that up
I do have one question more, how do you know that $\displaystyle kerf=(0,.......,0,1)$?
Kernel by def is a subset of the domain (so we want to list all elements from $\displaystyle A^{2012}$ that will give us 0 in $\displaystyle A^{2010}$) right?
I don't understand your answer. You seem to be implying the "truth" of the following: If M1 and M2 are two A modules with K non-zero, and M1/K isomorphic to M2, then M1 is not isomorphic to M2.
To see this is false, let M1 be the cross product of A indexed by the naturals, M2 the cross product of A indexed by the integers greater than or equal to 1. Let K=(A, 0, 0, ...). Then clearly M1/K is isomorphic to M2. But also M1 is isomorphic to M2 via the function g where g(f)(i) = f(i-1) for all i >= 1.
Hello,
The standard way to prove this is as follows. By Krull's theorem there exists a maximal ideal $\displaystyle \mathfrak{m}$ for $\displaystyle A$. Note then that $\displaystyle A/\mathfrak{m}$ is a field, and the isomorphism $\displaystyle A^{2010}\xrightarrow{\simeq} A^{2012}$ would induce a $\displaystyle A/\mathfrak{m}$ isomorphism $\displaystyle A/\mathfrak{m}\otimes_A A^{2010}\xrightarrow{\simeq} A/\mathfrak{m}\otimes_A A^{2012}$. But, it's easy to show that this says that, as $\displaystyle A/\mathfrak{m}$ modules one has that $\displaystyle (A/\mathfrak{m})^{2010}\cong(A/\mathfrak{m})^{2012}$--but since these are fields, it's standard vector space theory that this can't happen.
not quite. i'm saying commutative rings have an invariant basis number. A^{2010} and A^{2012} are free A-modules of rank 2010 and 2012, respectively.
the PROOF that commutative rings are IBN rings is essentially what drexel28 just wrote: it can be shown that if R-->S is a surjective homomorphism and S is an IBN ring, so is R. but if I is a maximal ideal of R, then R-->R/I is certainly a surjective homomorphism, and R/I is a field, and fields are certainly IBN rings (a la linear algebra).
note that krull's theorem (that drexel28 quoted) is just a fancy way of saying: the axiom of choice implies every ring has a maximal ideal. so we can always "mod our way" to the field case.
so....what have i left out? well i havent REALLY showed A^{n} is a free A-module.
but {(1,0,...,0),(0,1,0,...,0),...,(0,...,0,1)} is a basis (it's clearly a spanning set, and if an A-linear combination of these is 0, we have each coefficient = 0, by the definition of equality on the underlying cartesian product set).
your counter-example is well taken, but....we're dealing with free A-modules of finite rank. you can construct a similar example using vector spaces of infinite dimension, but that doesn't invalidate the rank-nullity theorem in the least.
I feel embarrassed now We had quite similar problem to this one on problem session where we had smth like this. If $\displaystyle \phi A^{n}\to A^{m}$ is an isomorphism of A-modules then $\displaystyle m=n$
Anyway thank you very much for the replies.
P.S Is Krull's thm the same as Zorn's lemma. We proved in class that every commutative ring with ''one'' has a maximal ideal by Zorn's lemma.