# Graded commutative rings and graded and commutative rings

###### 11th February, 2022

- Graded rings
- Commutativity
- Freeness
- Circles and spheres
- The Künneth theorem
- Some useful cohomology rings

One of the many reasons that teaching is fun is because you get to look back at things that you haven’t seen in a while and try to understand them in light of what you’ve learnt in the meantime. This means that you sometimes have the unexpected joy of having to teach something that always used to confuse you, but that now seems so much more straightforward! I experienced this last year when teaching an algebraic topology course: I remember being super lost when it came to the graded ring structure of cohomology and getting very annoyed at Hatcher’s book; now I look back and realise that it’s really neat! This post has a slightly different intended audience than normal: I’m just gonna assume that you know a bit about rings in the first half; the second half is aimed for somebody who’s a reasonable way through a first course on algebraic topology (e.g. knows what the cup product in cohomology is).

# Graded rings

↟Our starting point is the omnipresent *polynomial ring*: given
a ring S, we define S[x] to be the set of polynomials in one
variable (namely x) with coefficients
in S, so elements of S[x] are of the form
f = s_n x^n + s_{n-1} x^{n-1} + \ldots + s_1 x + s_0
where the s_i are elements of
R. (More generally, we have the
*polynomial ring in n
indeterminates*: S[x_1,x_2,\ldots,x_n], which is entirely
analogous, but with more than one variable; this is important, but we
don’t need to worry about it too much for now).

These rings are *really* nice, in lots of ways that we won’t
talk about today, but one *particularly* nice thing that they
have is a *grading*, given by the *degree*. Recall that
the degree of a *monomial* is defined to be the “total power”,
i.e.
\deg(x^n) = n
(we say *total* power to deal with the case of multiple
variables, e.g. \deg(x^my^n)=m+n). The
degree behaves nicely under multiplication
\deg(x^m\cdot x^n) = m+n
and we have a *unique* way of writing *any*
polynomial f\in S[x] as a sum of
monomials, basically by definition of what it means to be a polynomial:
f = \sum_{i=0}^n r_i x^i.

So let’s do what we always do and generalise this to an abstract structure!

A ring R is a *graded ring*
if there exist abelian groups (R_d)_{d\in\mathbb{N}}, where R_d\subseteq R for all d\in\mathbb{N}, such that

- R\cong\bigoplus_{d=0}^\infty R_d;
- R_d R_e\subseteq R_{d+e}.

Given r\in R\setminus\{0\}, if there
exists some d\in\mathbb{N} such that
r\in R_d, then we say that r is *homogeneous of degree d*, and we write |r|=d.

The prototypical example is the thing we started with: the polynomial ring R=S[x] is a graded ring with R_d\coloneqq\{sx^d \mid s\in S\}; the more general polynomial ring R=S[x_1,\ldots,x_n] in n variables is a graded ring with R_d\coloneqq\{s x_1^{m_1}x_2^{m_2}\cdots x_n^{m_n}\mid s\in S, m_1+m_2+\ldots+m_n=d\}.

# Commutativity

↟Now let’s talk about something confusing: what is a *graded
commutative ring*? Or should we say *commutative graded
ring*? Or should these two things be different?

Well, it makes sense that a graded commutative ring would just be a
commutative ring that is graded, i.e. we parse it as “graded
(commutative ring)”. Annoyingly, however, this is *not* the way
that most algebraists or geometers will parse this! If you want to talk
about commutative rings that are graded, then your best bet is really to
just say “a commutative ring that is graded”, but if you want to be
snappier, then I would advise that you say *commutative graded
ring*. Why am I making such a point out of this? What do people mean
when they actually say “graded commutative ring” then?

The answer lies in “bracketing” the adjectives in a different way, namely: “(graded commutative) ring”. But this just prompts the question: what does it mean for a ring to be “graded commutative”?

A ring R is *graded
commutative* if R is a
*graded* ring R=\bigoplus_{d=0}^\infty
R_d such that
rs = (-1)^{|r||s|}sr
for all homogeneous elements r,s\in
R.

So this is a slightly odd definition: graded commutativity is like
commutativity, but with a possible minus sign, depending on the degree
of the (homogeneous) elements.^{1} To understand this
better, let’s turn back to our good old friend S[x]. If S
is commutative, then S[x] is clearly
commutative, but is it graded commutative? Shockingly, no! Indeed, we
are asking if the following equality holds
x^2 = x\cdot x \overset{?}{=} (-1)^{|x||x|} x\cdot x = -x^2.
We see that this only happens in two cases:

- if 2=0; or
- if x^2=0.

The first one can happen (if S is a
field of characteristic 2, for example,
e.g. \mathbb{Z}/2\mathbb{Z}), but the
second one cannot, by the very definition of S[x]. *But* this second case does
happen in the *ring of dual numbers* S[x]/(x^2).

Going back to the polynomial ring S[x], we could also do something entirely
different and *define* x to be
of *degree 2*. That is, we are
giving S[x] a *different* graded
ring structure: S[x]\cong\bigoplus_{d=0}^\infty R_d where
R_d =
\begin{cases}
\{sx^d \mid s\in S\} &\text{if }d\text{ is even;}
\\0 &\text{if }d\text{ is odd.}
\end{cases}
(After a little bit of thought, you can see that this is “the
same as” simply looking at S[y^2],
where y is an indeterminate of degree
1 again). Then we have that
(-1)^{|x||x|} x\cdot x = x^2
and so this graded ring *is* graded commutative.

A nice little summary table would be helpful right about now (and,
just in this table, we’re going to make the assumption that
**S is commutative**).

Graded ring | Commutative? | Graded commutative? |
---|---|---|

S[x] | ✅ | ❌ (unless 2=0 in S) |

S[x]/(x^2) | ✅ | ✅ |

S[x^2] | ✅ | ✅ |

Let’s look at one more example that we’re going to need when it comes
to doing some algebraic topology: the *exterior algebra* \Lambda_S[\alpha_1,\ldots,\alpha_n] of a ring
S can be defined as the graded ring
\Lambda_S[\alpha_1,\ldots,\alpha_n] \coloneqq
S[\alpha_1,\ldots,\alpha_n]/(\alpha_i^2,\alpha_i\alpha_j+\alpha_j\alpha_i)_{1\leq
i,j\leq n}
where each \alpha_i is of
degree 1 (although, again, we can
modify this if we want to). Note that \Lambda_S[\alpha]\cong S[\alpha]/(\alpha^2),
i.e. *the exterior algebra of S in
one variable is exactly the ring of dual numbers*.

The exterior algebra is commutative (if S is), but is it graded commutative? Well it’s enough to check the condition on the generators \alpha_i, but we see that \alpha_i\alpha_j = -\alpha_j\alpha_i = (-1)^{|\alpha_i||\alpha_j|}\alpha_j\alpha_i where the first equality is exactly by the definition of \Lambda_S[\alpha_1,\ldots,\alpha_n] (and note that, if i=j, then everything is zero, and we definitely have that 0=0).

Back to our table:

Graded ring | Commutative? | Graded commutative? |
---|---|---|

S[x] | ✅ | ❌ (unless 2=0 in S) |

S[x^2] | ✅ | ✅ |

\Lambda_S[\alpha_1,\ldots,\alpha_n] | ❌ (unless n=1) | ✅ |

(we’ve removed the row for S[x]/(x^2), since this is just a specific example of the exterior algebra \Lambda_S[\alpha_1,\ldots,\alpha_n] where n=1.)

# Freeness

↟Unrelated to the property of (graded) commutativity is that of being
*free*. For *commutative* rings, being free basically
means being isomorphic to a polynomial ring in finitely many variables.
So S[x] is free (as a commutative
ring), as is S[x^2], but S[x]/(x^2) is *not* free (as a
commutative ring) — the latter has a nilpotent element (i.e. some r such that r^2=0, namely r=x) and free commutative rings never have
nilpotent elements, so it cannot be isomorphic to a free commutative
ring.^{2}

But note that I’ve been very careful to say “free *as a
commutative ring*”, and this is important: \Lambda_S[\alpha] is **not** a
free *commutative ring*, but it **is** a free
*graded commutative ring*. What do I mean by this? I mean that,
if we just use the fact that \Lambda_S[\alpha] is a graded commutative
ring, *forgetting all about how it’s actually defined*, then we
can recover the fact that \alpha_i^2=0
and \alpha_i\alpha_j=-\alpha_j\alpha_i
*without having to ask for it* (…most of the time).

That is, if we know that \Lambda_S[\alpha_1,\ldots,\alpha_n] is graded
commutative, then we know that
\alpha_i\cdot\alpha_i =
(-1)^{|\alpha_i||\alpha_i|}\alpha_i\cdot\alpha_i =
-\alpha_i\cdot\alpha_i
which rearranges to give
2\alpha_i^2=0.
But then, if we can *divide by 2* (i.e. if 2 is an invertible element) in our ring S, then we see that \alpha^2 *must be* equal to zero,
automatically! (The other relation that we need to impose, that \alpha_i\alpha_j=-\alpha_j\alpha_i is exactly
the definition of being graded commutative (since all our \alpha_i are of degree 1), so we don’t even need to worry about
enforcing that one separately!)

Of course, I’m going to put this in the table (and I’ll write “gc-ring” to mean “graded commutative ring”).

Graded ring | Commutative? | Graded commutative? | Free (as a gc-ring) |
---|---|---|---|

S[x] | ✅ | ❌ (unless 2=0 in S) |
✅ |

S[x^d] | ✅ | ❌ (unless 2 divides d) |
✅ |

\Lambda_S[\alpha_1,\ldots,\alpha_n] | ❌ (unless n=1) | ✅ | ❌ (unless 2 is invertible in S) |

By now you have probably noticed that all of the difficulties and
subtleties come from how the number 2
behaves (except for the exterior algebra only being commutative when in
one variable, but this isn’t a “graded” property, so we’ll ignore that
one), and this is a very common thing to happen.^{3}

# Circles and spheres

↟Before diving into the applications, we just need one more abstract
remark about graded rings: the tensor product R\otimes S of two graded rings has a natural
graded ring structure by setting
\deg(r\otimes s) \coloneqq \deg(r)+\deg(s)
and defining the multiplication to have a sign that “makes
things work nicely”:
(r\otimes s)(r'\otimes s') \coloneqq
(-1)^{|r'||s|}(rr')\otimes(ss')
(note that the exponent uses the two “inner” terms, r' and s, **not** r and s, nor
r' and s' (nor r and s', for that matter)).

What does this all have to do with algebraic topology? Hopefully
you’ve already seen the ring structure on cohomology (of a manifold),
given by the cup product. In fact, the cup product lets us assemble the
cohomology groups into a *commutative graded ring* (this is
indeed one justification for there being a minus sign in the definition
of the cup product: it ensures graded commutativity!).

*In what follows, we say “space” to mean “CW complex that is nice
enough to satisfy whatever hypotheses the theorem in question might
need”.*

Here’s a fun fact that we won’t prove:

Let X and Y be spaces. If Y is such that \operatorname{H}^\bullet(Y) is *free*
(as a graded commutative ring) and *finitely generated*, then
\operatorname{H}^\bullet(X\times Y) \cong
\operatorname{H}^\bullet(X)\otimes\operatorname{H}^\bullet(Y)
(where \cong means “isomorphic
*as graded rings*”).

Using this, we can look at two examples.

We know that \operatorname{H}^n(S^1) = \begin{cases} \mathbb{Z} &\text{if }n=0,1\text{;} \\0 &\text{otherwise.} \end{cases} If we denote by \alpha the generator of \operatorname{H}^1(S^1), then \alpha^2 \coloneqq \alpha\smile\alpha is an element of \operatorname{H}^2(S^1), but this group is zero, and so it must be the case that \alpha^2=0. Thus \operatorname{H}^\bullet(S^1) \cong \mathbb{Z}\langle1\rangle \oplus \mathbb{Z}\langle\alpha\rangle/(\alpha^2) = \Lambda_{\mathbb{Z}}[\alpha] (where here the angled brackets mean “the abelian group generated by these elements”).

But we know that \Lambda_{\mathbb{Z}}[\alpha] is free as a graded commutative ring, and it has only one generator so it’s also finitely generated; we can apply the Lemma above to get that \operatorname{H}^\bullet(S^1\times S^1) \cong \operatorname{H}^\bullet(S^1)\otimes\operatorname{H}^\bullet(S^1) but this is exactly \Lambda_{\mathbb{Z}}[\alpha_1,\alpha_2] (where \alpha_i is the generator of the first cohomology of the i-th copy of S^1).

Finally, recall that \Lambda_{\mathbb{Z}}[\alpha_1,\alpha_2] is
graded commutative but *not* commutative (i.e. \alpha_1\alpha_2=-\alpha_2\alpha_1).

We know that
\operatorname{H}^n(S^2) =
\begin{cases}
\mathbb{Z} &\text{if }n=0,2\text{;}
\\0 &\text{otherwise.}
\end{cases}
If we denote by \beta the
generator of \operatorname{H}^2(S^2),
then
\beta^2 \coloneqq \beta\smile\beta
is an element of \operatorname{H}^4(S^2), but this group is
zero, and so it must be the case that \beta^2=0. Thus
\operatorname{H}^\bullet(S^2) \cong \mathbb{Z}\langle1\rangle \oplus
\mathbb{Z}\langle\beta\rangle/(\beta^2)
= \Lambda_{\mathbb{Z}}[\beta]
where *\beta is of degree
2*.

As in the previous example, the Lemma gives us that
\operatorname{H}^\bullet(S^2\times S^2)
\cong
\operatorname{H}^\bullet(S^2)\otimes\operatorname{H}^\bullet(S^2)
which is exactly \Lambda_{\mathbb{Z}}[\beta_1,\beta_2], but
*where \beta is of degree 2* (we’re repeating this because it’s
important).

Now, \Lambda_{\mathbb{Z}}[\beta_1,\beta_2] is graded commutative (since it’s a cohomology ring, and recall that these are always graded commutative, by the sign in the definition of the cup product), but we see that \beta_1\beta_2 = (-1)^{|\beta_1||\beta_2|} \beta_2\beta_1 = (-1)^{4}\beta_2\beta_1 = \beta_2\beta_1 and so this cohomology ring is actually commutative!

The cohomology rings of S^1\times S^1 and S^2\times S^2 are not isomorphic (because one is commutative and the other is not).

# The Künneth theorem

↟What happens to our useful lemma if the cohomology ring \operatorname{H}^\bullet(Y) is finitely
generated but *not* free? That is (by the classification of
finite groups), what if it has *torsion*? The *Künneth
theorem* tells us that this torsion is exactly the “correction term”
needed to fix the Lemma.

Let X and Y be spaces. If Y is such that \operatorname{H}^\bullet(Y) is finitely
generated, then
\operatorname{H}^\bullet(X\times Y)
\cong (\operatorname{H}^\bullet(X)\otimes\operatorname{H}^\bullet(Y))
\oplus
\operatorname{Tor}_{\bullet+1}(\operatorname{H}^\bullet(X),\operatorname{H}^\bullet(Y))
(note that this \operatorname{Tor} term is zero if \operatorname{H}^\bullet(Y) is free^{4}, so this really is a generalisation
of our previous lemma).

We know that \operatorname{H}^n(\mathbb{RP}^2) = \begin{cases} \mathbb{Z} &\text{if }n=0\text{;} \\\mathbb{Z}/2\mathbb{Z} &\text{if }n=2\text{;} \\0 &\text{otherwise.} \end{cases} Doing some \operatorname{Tor} calculations (use the fact that \mathbb{Z} is free and thus flat (or even just projective) to see that \operatorname{Tor}(\mathbb{Z},-)=\operatorname{Tor}(-,\mathbb{Z})=0; use the fact that \operatorname{Tor}(A,\mathbb{Z}/m\mathbb{Z})\cong\{a\in A\mid ma=0\} to see that \operatorname{Tor}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/2\mathbb{Z})\cong\mathbb{Z}/2\mathbb{Z}), the above Corollary gives us that \operatorname{H}^n(\mathbb{RP}^2\times\mathbb{RP}^2) = \begin{cases} \mathbb{Z} &\text{if }n=0\text{;} \\(\mathbb{Z}/2\mathbb{Z})^2 &\text{if }n=2\text{;} \\\mathbb{Z}/2\mathbb{Z} &\text{if }n=3{;} \\\mathbb{Z}/2\mathbb{Z} &\text{if }n=4{;} \\0 &\text{otherwise.} \end{cases} We’ve written the case n=3 on a separate line because this is exactly the “correction term” given by the Künneth theorem, i.e. the previous Lemma would have given us everything else but would have said that \operatorname{H}^3=0; Künneth tells us otherwise.

Calculate \operatorname{H}^\bullet(\mathbb{RP}^2\times\mathbb{RP}^2;\mathbb{Q}/\mathbb{Z}).

*Hint: use the Universal Coefficient Theorem, or the long exact
sequence associated to the short exact sequence 0\to\mathbb{Z}\hookrightarrow\mathbb{Q}\twoheadrightarrow\mathbb{Q}/\mathbb{Z}\to0,
and recall that \operatorname{Tor}(A,\mathbb{Q}/\mathbb{Z})\cong\operatorname{tors}(A),
the torsion part of A.*

# Some useful cohomology rings

↟Real projective spaces have pretty nice cohomology rings, but they differ in presentation depending on the parity of the dimension:

- \operatorname{H}^\bullet(\mathbb{RP}^{2n})\cong\mathbb{Z}[\beta]/(2\beta,\beta^{n+1}), where |\beta|=2;
- \operatorname{H}^\bullet(\mathbb{RP}^{2n+1})\cong\mathbb{Z}[\beta,\varepsilon]/(2\beta,\beta^{n+1},\epsilon^2,\beta\epsilon), where |\beta|=2 and |\varepsilon|=2n+1 (here \epsilon is the generator of \operatorname{H}^{2n+1}).

What’s nice is that, if we work with \mathbb{Z}/2\mathbb{Z} coefficients (instead of \mathbb{Z}), then we can write both cases together as one thing:

- \operatorname{H}^\bullet(\mathbb{RP}^n;\mathbb{Z}/2\mathbb{Z})\cong(\mathbb{Z}/2\mathbb{Z})[\alpha]/(\alpha^{n+1}), where |\alpha|=1.

Given that \mathbb{C} is 2-dimensional over \mathbb{R}, complex projective space behaves much nicer, since we don’t need to worry about parity:

- \operatorname{H}^\bullet(\mathbb{CP}^n)\cong\mathbb{Z}[\beta]/(\beta^{n+1}), where |\beta|=2.

Finally, if you know how to define infinite dimensional projective spaces, then it’s a very lovely cheeky little fact that the cohomology rings are just given by “taking the limit n\to\infty”, i.e.

- \operatorname{H}^\bullet(\mathbb{RP}^\infty)\cong\mathbb{Z}[\alpha], where |\alpha|=1;
- \operatorname{H}^\bullet(\mathbb{CP}^\infty)\cong\mathbb{Z}[\beta], where |\beta|=2.

Nice!

Note that this condition extends to give some condition on the multiplication of arbitrary elements of R, since every element can be written (in a unique way) as a sum of homogeneous elements, by the fact that R\cong\bigoplus_{d=0}^\infty R_d.↩︎

This is very similar to the fact that a group homomorphism must send an element of order n to an element of order m

*such that m divides n*.↩︎Both number theorists and algebraic geometers get very tired of having to write “let k be a field of characteristic not equal to 2”, since 2 “behaves badly” for them; some algebraic topologists, on the other hand, get very excited when you say “calculate (co)homology with coefficients in a field of characteristic 2”, since 2 “behaves nicely” for them. Basically, 2 is a tricky number.↩︎

Or if we work over a field instead of over \mathbb{Z}…↩︎