User talk:Marty/UCSC Math 203 Fall 2008

Is $x^2 + 3 = 7$? Marty 15:46, 29 September 2008 (PDT)

geometric series formula in A x ?
does the following work for $f \in A x $ in part (i) or exercise 5?

$g = \frac{ 1 }{f} = \frac{1}{1-(1-f)} = \sum_{n=0}^\infty (1-f)^n \in A x $

I'm having trouble seeing the implicit use of $a_0$ as a unit here (if it's there). The formula should work for any nonzero $f \in Z x $, but no $f$ should be invertible, so what's going on here? Alex B 17:47, 5 October 2008 (PDT)
 * I don't think think that this will work directly for all $f$ in the power series ring. But, something like this should work if you first multiply $f$ by the multiplicative inverse of $a_0$ in $A$.  Then, you get a power series $a_0^{-1} f$ with zeroeth coefficient $1$.  It follows that $1 - a_0^{-1} f \in (X)$.  Then, you can prove that the formula $\sum_{n=0}^\infty (1 - a_0^{-1} f)^n$ makes sense. Marty 19:25, 5 October 2008 (PDT)

idempotent-like elements splitting rings
A few days ago we talked about idempotents splitting a ring into two parts. If we have a ring $A$ and an idempotent-like element that satisfies $x^3 = x \neq 0,1$, would we be able to split $A$ into 3 parts? one part for each factor of $x^3-x$? For example

$\phi: A \to (x^2-1)A \times (x^2-x)A \times (x^2+x)A$

for $((x^2-1)a,(x^2-x)b,(x^2+x)c)$ in the RHS. We should get that $\phi( (x^2-1)a + (x^2-x)b + (x^2+x)c ) = ((x^2-1)a,(x^2-x)b,(x^2+x)c)$, so the map is onto. Also injective because $\phi^{-1}(0,0,0) = (x^3-x) = (0)$. So I guess these are isomorphic.

Seems like this should work for any $x^k = x$, but for larger $k$ we start to see imaginary roots of $x^k-x$, such as when $k=5$. So if $x^k-x$ can't split completely in $A$, then it seems neither can $A$.

So I guess, what I'm wondering is what happens if $x^k-x$ can't split completely in $A$, and can this even happen? And does anything odd happen when $k$ is composite. Alex B 23:44, 12 October 2008 (PDT)
 * Its a good question. Here are two important examples, which might help with intuition.  Consider first, the ring $\RR[X] / $.  Then, since the polynomial $X^3 - X$ splits into three irreducible linear factors, we obtain an isomorphism (the Chinese remainder theorem):

$$\RR[X] /  \equiv \RR[X] /  \times \RR[X] /  \times \RR[X] / .$$
 * Here, as you suggest, the ring splits into a direct product of three fields, as you expect.
 * But, such a phenomenon does not always work in a straightforward way. Consider, for example, the ring $A = \RR[X] / $.  Then, splitting the polynomial into irreducible factors yields an isomorphism:

$$A \equiv \RR[X] /  \times \RR[X] /  \times \RR[X] /  \times \RR[X] / .$$ This splits $A$ as the product of four fields; three are isomorphic to $\RR$, and the last is isomorphic to $\CC$ (which certainly does not split further!).
 * Although $Spec(A)$ has four points, the "base change" $Spec(A) \times_{Spec(\RR)} Spec(\CC)$, which corresponds to the tensor product $A_\CC = A \otimes_\RR \CC$, has five points. An algebraic geometer would say that the scheme $Spec(A)$ has five geometric points, referring to points after base change to an algebraically closed field.
 * Hope this helps. It is a subtle question!  Marty 23:56, 12 October 2008 (PDT)

M/N+N=M?
My colleague suggested to me that given M a module and N a submodule that perhaps it is not true that M/N+N=M where were mean the isomorphic copy of M/N in M. We are not requiring a direct sum here. Is there an obvious counter example to this? Shawn O&#39;hare 12:59, 19 October 2008 (PDT)
 * Often, there is not an "isomorphic copy of M/N in M". For example, consider when $R = \ZZ$, $M = \ZZ$, and $N = 2 \ZZ$.  Then $M/N = \ZZ / 2 \ZZ$, which certainly does not embed in $M$.  Marty 13:38, 19 October 2008 (PDT)


 * Yes, I'm not sure what I was thinking. Shortly afterward I realized you could also consider Q/Z.  Q/Z is torsion, but Q is not. Shawn O&#39;hare 18:08, 19 October 2008 (PDT)

LaTeX
So is there an actual way to have TeX typesetting show up on the Wiki? At first I thought maybe we were all using the LaTeX markup instead of Wiki math markup, butthe standard wiki math environment $$ $$ produces an error. Shawn O&#39;hare 18:08, 19 October 2008 (PDT)


 * In your user preferences there is a "math" tab, that lets you specify by what method you want your TeX markup to be displayed/translated. I use jsMath with unicode fonts, but the bulletin board font (i.e. /mathbb{Z} ) gives me an error.


 * To type in TeX, just use the regular dollar signs (or double dollar signs for equation lines) around the math parts. Click edit on the threads above to see examples. Alex B 18:39, 19 October 2008 (PDT)


 * Alex is correct. I set it up so that regular dollar signs work.  For mathbb fonts, which are not supported as directly with jsMath/Mediawiki, I put in some special commands.  For $\NN$, $\ZZ$, $\QQ$, $\RR$, $\CC$, $\FF$, edit this thread to see how its done. To get math to show up correctly, you might need to click on the jsMath control panel (a rectangle at the bottom-right of the window), and select unicode fonts from the options menu.  Or, you can download the jsMath font package to get optimum appearance.  Marty 20:43, 19 October 2008 (PDT)


 * Odd. None of my browsers seem to want to render the math, under any of the options in the Math panel.  (Even after clearing cache).  Shawn O&#39;Hare 21:52, 19 October 2008 (PDT)
 * Did you try clicking on the jsMath control panel? It's accessible through a small rectangle on the bottom-right of the window.  It's not related to the math section of the wiki preferences.  I've tested under IE, Firefox, Chrome, and Safari.  Marty 21:55, 19 October 2008 (PDT)


 * There is an option in that control panel to show font warnings. Check that option and try reloading, it should tell you what fonts you're missing at the top of the page and give you a link. If it's not showing up, here's a link to some of the fonts you may need.


 * http://www.math.union.edu/~dpvc/jsMath/download/jsMath-fonts.html
 * http://www.math.union.edu/~dpvc/jsMath/download/extra-fonts/welcome.html


 * Alex B 22:17, 19 October 2008 (PDT)

$f \otimes 1$ when $f$ is a natural inclusion map
I wanted to explore question 2 from the homework a bit more. (Please bear any LaTeX mistakes. I still can't get the fonts to render for the life of me). Suppose that $A$ is a ring, $I$ an ideal of the ring. Then let $f: I \to A$ be the natural inclusion map, that is for $a \in I$ we have $f(a)=a \in A$. The question: How can $f \otimes 1: I \otimes M \to A \otimes M$ possibly fail to be injective? Do we not have $a \otimes m=f(a) \otimes m=[f \otimes 1](a \otimes m)=[f \otimes 1](b \otimes m)=f(b) \otimes m=b \otimes m$? Naively it seems to me that if for $x \neq y$ for $x, y \in A$ but $x \otimes m=y \otimes m$, the "collapse of elements" in $A \otimes M$ also occurs in $I \otimes M$, so injectivity is preserved. This is essentially because $f \otimes 1$ is really just the restriction of $1 \otimes 1$.

Maybe I need to explore a bit more about non-flat modules. What am I missing? Shawn O&#39;Hare 21:19, 20 October 2008 (PDT)


 * Consider the following: $A = \ZZ$, $I = 2 \ZZ$, and $M = \ZZ / 2 \ZZ$.  Then, there is a short exact sequence:

$$0 \rightarrow I \rightarrow A \rightarrow A/I \rightarrow 0.$$
 * Observe that tensoring with $M$ yields:

$$I \otimes_A M \rightarrow A \otimes_A M \rightarrow A/I \otimes_A M \rightarrow 0.$$
 * There exist isomorphisms: $I \otimes_A M \cong M$, $A \otimes_A M \cong M$, $A/I \otimes_A M \cong M$, for which the above short exact sequence looks like:

$$M \rightarrow M \rightarrow M \rightarrow 0,$$
 * and the first arrow is "multiplication by two", i.e., the zero map, which is certainly not injective. Marty 22:12, 20 October 2008 (PDT)


 * In light of this, the issue then is actually choosing the appropriate sequence? In truth, choosing the appropriate maps.  For instance, rather than "multiplication by 2" we can simply use the natural embedding of $2 \ZZ$ into $Z$, this just being $f(2n)=2n$.  Then injectivity would seem to hold upon pain of contradiction.


 * These oddities seem to suggest to me that there are fundamental differences between maps which identify honest subspaces and maps which isomorphically embed objects. As per Marty's example, $2 \ZZ$ can be isomorphically embedded in $\ZZ$ via "multiplication by 2" yet this embedding does not behave the same in our short exact sequence as the map $f(2n)=2n$, which I think of as the "subspace identification" map.  Shawn O&#39;Hare 22:57, 20 October 2008 (PDT)  (PS:  I apologize if you're getting spammed by emails saying I am making edits to the page.  I have been hasty twice now in the last few minutes and realized I made a spelling mistake or grammar error after submitting.  This, in spite of using the preview mode a few times.  Oh well.)

Hom from $I\otimes M$ to $IM$
I had trouble showing explicitly that it was an $A$-module homomorphism, particularly that $$[f \otimes 1](a_1 \otimes m_1 + a_2 \otimes m_2 ) = [f\otimes 1](a_1 \otimes m_1) + [f\otimes 1](a_2 \otimes m_2) $$ because (I think) we aren't guaranteed that $a_1$ and $a_2$ have a common divisor in $I$ (like we do in $A$). But the text after Exercise 2.15 on page 27 suggests that the tensor of two $A$-module homomorphisms is again an $A$-module homomorphism (which I don't see) but should prove that $f\otimes 1$ is an $A$-module homomorphism, and also injective by your argument. Alex B 21:55, 20 October 2008 (PDT)

(edit: moved this to seperate thread, because I realized this wasn't what that thread was about)Alex B 13:54, 21 October 2008 (PDT)

Localization and Absolute flatness
I seem to have shown something that isn't true in 10.(ii) and can't seem to spot my mistake, here's my proof:

problem: $A$ is absolutely flat $\iff$ $A_M$ is a field for each maximal ideal $M$

proof: "$\Longrightarrow$": $A_M$ absolutely flat is implied by part (i). From there we use the fact that $A_M$ is a local ring with maximal ideal $S^{-1}M$ to show by contradiction that $S^{-1}M$ must be the zero ideal.

We say $(e)$ is some principal ideal generated by a nontrivial idempotent, so it must be contained in $S^{-1}M$. Similarly $(1-e)$ is another principal ideal, so must also be contained in $S^{-1}M$ as well. This is a contradiction, as $S^{-1}M$ cannot contain both $e$ and $1-e$. So we conclude that $S^{-1}M$ cannot contain any principal ideals, hence must be the zero ideal.

This shows that $A_M$ is a field.
 * This is basically correct, though it could be stated better. Marty 21:48, 26 October 2008 (PDT)

"$\Longleftarrow$": Here we say $A_M$ is absolutely flat because $(1)$ and $(0)$ are the only ideals of a field, and both are generated by idempotents.

Since $S^{-1}M = 0$, there exists an $s \in S$ such that $sM = 0$. But $S$ doesn't contain any zero-divisors, so it must be the case that $M = 0$. So $(0)$ is a maximal ideal in $A$... so $A$ is a field.

It doesn't seem like it should be true that $A$ absolutely flat $\iff$ $A$ is a field... so I think I may have assumed something I shouldn't have. Alex B 21:30, 26 October 2008 (PDT)


 * So, I guess you mean that $S_M^{-1} M = 0$, where $S_M = A - M$. There is no reason for $S$ not to contain zero-divisors.  For example, consider $A = K \times K$, for any field $K$.  There are two maximal ideals, both principal, generated by $(1,0)$ and $(0,1)$, respectively.  Each element $(1,0)$ and $(0,1)$ is a zero-divisor, and each is not contained in the maximal ideal generated by the other.  Marty 21:48, 26 October 2008 (PDT)


 * Ah, I see. I hadn't considered that a zero divisor could generate the rest of the ring, but this is exactly what happens with idempotents. Thanks. Alex B 22:07, 26 October 2008 (PDT)

Going up
In problem 10, we're talking about a ring homomorphism $f:A \rightarrow B$ having the "going-up" property. I was wondering, is it assumed that $f$ is integral, or is that something we have to show??
 * It is not assumed that $f$ is integral, in problem 10. Just assume that $f$ is a ring homomorphism, and prove the equivalence of (a), (b), and (c).  (If $f$ were integral, then (b) would automatically be true, by the going-up theorem).  Marty 19:39, 7 November 2008 (PST)

Modding Out
If we have $N_1,N_2,L_r,L_{r+1}$ as submodules of $M$. Such that $L_r \subset L_{r+1}$ with $L_r = L_{r+1} ($ mod $ N_1)$ and $L_r = L_{r+1}($ mod $ N_2)$, is it necessarily true that $L_r = L_{r+1} ($ mod $ N_1 \cap N_2)$? It seems like it should be.

Here's what I have so far. Suppose there is some $x \in L_{r+1}/(N_1\cap N_2)$ such that $x \not \in L_r/(N_1 \cap N_2)$. By the equivalences above, there exist elements $y_1,y_2 \in L_r$ such that $x-y_1 \in N_1$ and $x-y_2 \in N_2$. But what I really want is a single element $y$ that belongs to both cosets $x+N_1$ and $x+N_2$ as well as $L_r$...

I was thinking of taking an element in the coset $x+N_1\cap N_2$ intersected with $L_r$, but then I would have to deal a lot of separate cases when intersections were empty. Is there a better way to go about this? Alex B 01:37, 23 November 2008 (PST)
 * I presume that you're talking about Problem 3 in A-M?
 * Here is a less direct approach, that works better.
 * If $N_1, N_2$ are submodules of a module $M$, then there is a exact sequence:

$$0 \rightarrow N_1 \cap N_2 \rightarrow M \rightarrow \frac{M}{N_1} \oplus \frac{M}{N_2}.$$
 * In particular, one can embed:

$$\frac{M}{N_1 \cap N_2} \hookrightarrow \frac{M}{N_1} \oplus \frac{M}{N_2}.$$
 * Submodules of a Noetherian or Artinian module are again Noetherian or Artinian (by Prop. 6.3, for example).
 * Does this help, without answering your specific question? Marty 11:04, 23 November 2008 (PST)


 * Yeah, I ended up scrapping the original idea and starting over later last night and ended up with something similar.


 * I used corollary 6.4 to say $M/N_1 \oplus M/N_2$ is Noetherian. Then took the homomorphism $\phi: M \to M/N_1 \oplus M/N_2$ defned as $\phi(m) = (m+N_1,m+N_2)$. Then computed the kernel of $\phi$ to be $N_1 \cap N_2$ to see that

$$ \frac{M}{N_1\cap N_2} \cong Im(\phi) \hookrightarrow \frac{M}{N_1} \oplus \frac{M}{N_2}.$$
 * Alex B 13:54, 23 November 2008 (PST)