What facts in commutative algebra fail miserably for simplicial commutative rings, even up to homotopy? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:15:12Z http://mathoverflow.net/feeds/question/45273 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/45273/what-facts-in-commutative-algebra-fail-miserably-for-simplicial-commutative-rings What facts in commutative algebra fail miserably for simplicial commutative rings, even up to homotopy? Harry Gindi 2010-11-08T07:03:15Z 2012-05-31T23:33:10Z <p>Simplicial commutative rings are very easy to describe. They're just commutative monoids in the monoidal category of simplicial abelian groups. However, I just realized that a priori, it's not clear that even some of the simplest facts we can prove for ordinary commutative rings (in particular those that depend integrally on the axiom of choice, or even those that depend on the law of the excluded middle) will hold for simplicial commutative rings. However, we have at least one saving grace. That is, the interesting parts of simplicial commutative algebra come from considering things up to homotopy. </p> <p>So, for example, as far as it makes sense, can we prove that every simplicial ideal of a simplicial commutative ring is weakly equivalent to one contained in a maximal simplicial ideal? Perhaps a better way to state this would be something like, "every noncontractible simplicial commutative ring admits at least one surjective map to a simplicial commutative ring that's weakly equivalent to a simplicial field", or some variation on where the homotopy equivalence appears. Given that the axiom of choice does not necessarily hold in $sSet$, it doesn't seem reasonable to think that the ordinary theorem will hold.</p> <p>Is there a version of the Hilbert basis theorem that holds up to isomorphism? How about weak equivalence?</p> <p>What other well-known theorems will fail, even up to homotopy?</p> http://mathoverflow.net/questions/45273/what-facts-in-commutative-algebra-fail-miserably-for-simplicial-commutative-rings/45278#45278 Answer by Neil Strickland for What facts in commutative algebra fail miserably for simplicial commutative rings, even up to homotopy? Neil Strickland 2010-11-08T09:00:25Z 2010-11-08T09:43:41Z <p>Given a simplicial ring $A_\bullet$, it is standard that $\pi_0(A)$ is a ring and $\pi_n(A)$ is a module over it, so $A_\bullet$ is weakly contractible iff $\pi_0(A)=0$. Also, $\pi_0(A)$ is just the cokernel of $d_0-d_1:A_1\to A_0$. We can take any chain of face maps $A_n\to A_0$ and compose with the projection to $\pi_0(A)$, and this will be independent of which chain we chose. These maps $A_n\to\pi_0(A)$ give a surjective map from $A_\bullet$ to the constant simplicial ring $\pi_0(A)$. (This is just the simplest piece of the Postnikov tower.) If $\pi_0(A)\neq 0$ we can then compose with a surjective map to a constant simplicial field.</p> <p>Note also that all simplicial fields are constant. This is just because the composite $A_0\xrightarrow{s_0^n}A_n\xrightarrow{d_0^n}A_0$ is the identity, so $d_0^n$ is surjective, but all field homomorphisms are injective, so $d_0^n$ is an isomorphism.</p> <p>I think it also works out that a simplicial ring is contractible as a simplicial group iff it is weakly contractible (because you can find an element $x\in A_1$ with $d_0(x)-d_1(x)=1$, and then shuffle-multiplication by $x$ should give a contraction). It would be stronger to say that there is a contraction by ring maps. I don't immediately see the full picture about that.</p> http://mathoverflow.net/questions/45273/what-facts-in-commutative-algebra-fail-miserably-for-simplicial-commutative-rings/45304#45304 Answer by Tyler Lawson for What facts in commutative algebra fail miserably for simplicial commutative rings, even up to homotopy? Tyler Lawson 2010-11-08T13:57:25Z 2010-11-08T13:57:25Z <p>Most of the things that stop working are things related to procedures in commutative algebra that don't preserve exactness. The tensor product of simplicial modules always has to be the derived tensor product in order to be meaningful, etc.</p> <p>One of the sticky points is that "free" is not the same as "polynomial" in higher degrees except in characteristic zero, and this makes building simplicial rings from a generators-and-relations perspective quite difficult.</p> <p>For example, if R is an ordinary ring viewed as a constant simplicial commutative ring (with homotopy R, concentrated in degree 0), then you can take the free simplicial R-algebra on a class x in degree n; let me call it A. If n=0, the homotopy groups of A are just the polynomial algebra R[x] concentrated in degree zero. If n=1, the homotopy groups of A are an exterior algebra over R on a generator in degree 1. If n=2, the homotopy groups of A are a <em>divided power algebra</em> on a over R on a generator in degree 2 (a simplicial commutative ring always has a divided power structure on its higher homotopy groups). If n>3 and 2R = 0, then the answer is always a countable tensor product of exterior algebras over R. If n > 3 and pR = 0, you get algebras whose structure is simply complicated (because it involves iterated Tor starting with a divided power algebra).</p> <p>Similar remarks apply to trying to construct an ideal generated by a collection of elements in homotopy groups which are anything other than a regular sequence in <code>$\pi_0$</code> - you are unlikely to get the "quotient" you expect.</p> http://mathoverflow.net/questions/45273/what-facts-in-commutative-algebra-fail-miserably-for-simplicial-commutative-rings/98526#98526 Answer by temp for What facts in commutative algebra fail miserably for simplicial commutative rings, even up to homotopy? temp 2012-05-31T23:33:10Z 2012-05-31T23:33:10Z <p>Maybe it is too late to answer this, but here is a classical "epic fail" which I'm sure you've already seen in disguise:</p> <p>Even for the constant ring $\mathbb{Z}$, thought of as a simplicial ring, short exact sequences of free modules of finite rank (i.e. free abelian group of finite rank on each degree) don't split.</p> <p>Simplicial free abelian groups are just complexes of abelian groups. Short exact sequence of complexes produces a long exact sequence, and if the boundary map is non-zero then it certainly won't split.</p> <p>I don't have a more conceptual way to "explain" this, but for me it is like such failure are caused by some non-abelian input, (e.g. those coming from a topological space or homotopy groups.)</p>