Wikipedia define the graded ( ring, module, vector space, ...) as here

I noted that in the rings and modules it required the condition of inclusion but in the vector spaces it did not. just the direct sum condition ..why ??

Any hint ?


We would impose a condition like $F_i V_j \subseteq V_{i+j}$ for all $i,j$, if the vector space were graded ($V = \bigoplus V_j$) and also the field were graded: $F = \bigoplus F_i$. That is the case for graded modules over graded rings, or simply for graded rings (which are, equivalently, graded modules over themselves). So it is reasonable to expect a similar condition for graded vector spaces over fields with gradings.

And yet, even though this seems reasonable, Wikipedia lists no such condition. Why is this? It turns out that the reason there is no such condition is because the field $F$ is always trivially graded, i.e., $F = F_0$, and $F V_j = F_0 V_j = V_{0+j} = V_j$ for all $j$. So there are no $F_i$ that need to have any condition imposed—except $i=0$, and we impose the condition that for every $j$, $V_j$ is a sub-vector space, including that it is closed under scalar products, i.e., $F V_j \subseteq V_j$. That is pretty much the same as $F_0 V_j \subseteq V_{0+j}$. (There is also a requirement for each $V_j$ to be closed under addition.)

Okay, that explains why there is no condition listed on Wikipedia, but it raises a new question: why is a field with a grading always trivially graded, i.e., graded in degree $0$? Here, let me assume that "graded" means $\mathbb{Z}$-graded (as opposed to some other semigroup). Let $F$ be a field and suppose that $F = \bigoplus_{i \in \mathbb{Z}} F_i$, each $F_i \subseteq F$ an abelian group, such that $F_i F_j \subseteq F_{i+j}$.

A first claim is that $1 \in F_0$. Left as an exercise, but hint: Say $1 = \sum f_i$, $i \in F_i$. Each $f_i = f_i 1 = \sum_j f_i f_j$, where $f_i f_j \in F_{i+j}$; but $f_i \in F_i$...

Next claim, every unit is homogeneous. Again exercise, with hint: Say $0 \neq u = \sum f_i$ with inverse $u^{-1} = \sum g_j$. In $1 = u u^{-1} = \sum_{i,j} f_i g_j$ consider the maximum and minimum of $i+j$...

Now finally every nonzero element in the field is homogeneous. But if $F$ has even two degrees occurring, say $f_{i_1}$ and $f_{i_2}$ are nonzero elements of degrees $i_1 \neq i_2$, then consider their sum; it is nonzero (why?) and nonhomogeneous, but every nonzero element is supposed to be homogeneous.

So if $F$ is a field that is graded (meaning $\mathbb{Z}$-graded), then it is concentrated in a single degree, specifically degree $0$. That is why there is no need for conditions on the scalar multiplication on graded vector spaces.

In closing, (1) I haven't thought about what happens if the grading monoid is different, maybe some weird thing with lots of idempotents... I don't know. (2) Because there are no fields with nontrivial gradings, the term "graded field" has been "recycled" to mean a graded ring in which every nonzero homogeneous element is invertible (e.g., $K[x,x^{-1}]$).

  • $\begingroup$ Very nice answer .. thank you so much for your detailed explanation 👍🏼 $\endgroup$ – raindrops Jul 22 '17 at 7:42
  • 1
    $\begingroup$ If you allow gradings by other groups, you can have non-trivially graded fields. For example $\mathbb{C} = \mathbb{R} + i\mathbb{R}$ is $\mathbb{Z}/2\mathbb{Z}$-graded. More generally: If $K$ is a field containing $n$-th roots of unity and $L$ is the splitting field of $X^n - a$, then $\operatorname{Gal}(L/K) = \mathbb{Z}/m\mathbb{Z}$ for some $m$ dividing $n$ and this action induces a $\mathbb{Z}/m\mathbb{Z}$-grading on $L$. $\endgroup$ – Elvorfirilmathredia Sep 27 '17 at 11:54

Any inclusion V -> W of vector spaces creates a direct sum decomposition $W \cong V \oplus V^{\perp}$.

A graded vector space $W = V_1 \oplus V_2$ would have graded part ($\le 2$) equal to W not $V_2$.

  • $\begingroup$ I ask why there is no condition like this for example: $ F_i V_j \subseteq V_{i+j} $ where $V$ is a vector space over field $F$ ?? $\endgroup$ – raindrops Jul 21 '17 at 23:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.