Let $B = \bigoplus_{i \in \mathbb Z}B_i$ be a $\mathbb Z$-graded integral domain. Given $f \in B$ define $\deg: B \rightarrow \mathbb Z \cup \{-\infty\}$ by $$\deg(f) = \mathrm{maxSupp}{(f)} \text{ if } f \neq 0 \text{ and } \deg(0) = -\infty $$

Note that elements of $B_i$ are homogeneous and each $f \in B$ has a unique decomposition $f = \sum_{j \in \mathbb Z} f_j, f_j \in B_j.$

A function $f$ is homogeneous if it can be written as sum of monomials of different degrees.

I am trying to prove that $B_0$ is algebraically closed in $B.$

I want to show that $B_0$ is algebraically closed in $B.$ Which means that we want to show that $B_0$ is equal to its algebraic closure $\bar{B_0}$, where the algebraic closure of $B_0$ in $B$ is the subring $\bar{B_0}$ consisting of all $b\in B$ that are algebraic over $B_{0}.$ And $b\in B$ is algebraic over $B_0$ means that there exists a polynomial, say $f(b) \in B_0[x]$, i.e. with coefficients in $B_0$ such that $f(b) = 0$.

Now, assume by contradiction that $b\in B\setminus B_0$ such that $b \neq 0$ and $\deg b \neq 0$ and $b$ is algebraic over $B_0.$

Since $b$ is algebraic over $B_0,$ then there exists $f(b) \in B_0[x]$ such that $f(b) = 0$ i.e., there exists $c_0, c_1, \dots , c_t \in B_0$ with $c_t \neq 0$ such that $$f(b) = \sum_{j=0}^t c_jb^j = 0 \quad (1)$$

But since $b\in B$ then $$b=\sum_{i\in \Bbb Z}b_i \quad (2)$$ i.e. $b_i$ is the $i$-the homogeneous component of $b.$ Since $b$ has some non-zero component in a positive degree by assumption, let $m$ be the largest positive integer for which $b_m\ne0$ (if $b$ has no non-zero component in positive degree we take $m$ to be the minimum integer with $b_m\ne0$)

Now, substituting from $(2)$ into $(1)$ we get $$f(b) = \sum_{n=0}^t c_n(\sum_{i\in \Bbb Z}b_i)^n = 0 \quad (3)$$

But then the term $(b_m)^n$ has degree $mn$ and there is exactly one of it because other nonzero $b_i$ have lower degree (note that degrees of $c_n$'s are $0$ because they are in $B_0$). Now, since $B$ is an integral domain, then $c_n(b_m)^n \neq 0 $ because both of $c_n,(b_m)^n$ not equal $0.$ In particular, this means $f(b)$ can not be zero as it is not zero in the $mn$-th component which is a contradiction.

But I got a hint that I have to show that the values of $\deg(b^n), n \geq 0,$ are distinct.

So I modified the proof above (which I understand with the help of a proof given by @Leoli) to the following:

$\deg \sum_{i=0}^{n-1} c_ib^i \leq \max_{0 \leq i \leq (n-1)} \deg(c_i b^i) \leq \max_{0 \leq i \leq (n-1)} \deg(b^i) = \max_{0 \leq i \leq (n-1)} \{im \} = (n-1)m$

Where the first inequality by property $3$ of $\deg$ we proved before and the second inequality because $\deg c_i = 0$ and the equality before last because $\operatorname {maxSupp}b$ is $m.$

But $\deg c_nb^n = nm \neq 0$ because $B$ is an integral domain and $c_n \neq 0$ and $(b_m)^n \neq 0.$

Therefore, the maxSupp term in $\sum_{i=0}^{n-1} c_ib^i$ does not cancel the maxSupp term in $ c_nb^n$

But $\sum_{i=0}^{n} c_ib^i = 0,$ then $ mn = \deg c_nb^n = \deg \sum_{i=0}^{n} c_ib^i = \deg (0) = -\infty$

A contradiction.

I feel like my proof is not ordered well, could someone criticize it and give me a more elegant proof.

  • $\begingroup$ Perhaps you should define what $B_0$ and $B$ are? $\endgroup$ – Mindlack Feb 23 at 17:10
  • $\begingroup$ @Mindlack I am sorry about that I will include my definitons $\endgroup$ – mathmusic Feb 23 at 17:13

Your idea of using the degree function is perfectly good, so here I just offer you a way of writing your argument in a shorter way.

Consider an element $b\in B$ and write it in the form $b=t+s$ where $t\neq 0$ is homogeneous of degree $N$ and $s$ is a sum of homogeneous terms of lower degree. Let us call $t$ the leading term of $b$. Let us assume that $\deg t>0$.

Then for any $k$, we see that the leading term of $b^k$ is $t^k$, that is, $b^k = t^k+s'$ where $s'$ is a sum of elements of lower degree: this follows from the binomial expansion.

If $f\in B_0[x]$ is a polynomial of degree $d$ with coefficients of degree zero, then it follows from the above that the leading term of $f(b)$ is $c_dt^d$, so that if $f(b) =0$, then we must in fact have that $c_dt^d=0$: this follows by using the direct sum decomposition you wrote in your post.

Since $B$ is a domain and since $c_d\neq 0$ and $t\neq 0$, we can conclude that $c_dt^d\neq 0$, so that no element with leading term which is homogeneous of positive degree can be algebraic over $B_0$, like you claimed.

  • $\begingroup$ What if $\deg t < 0$? $\endgroup$ – mathmusic Feb 24 at 3:03
  • $\begingroup$ Do I have redundant steps in my proof?If so, can you point them out ? $\endgroup$ – mathmusic Feb 24 at 3:11
  • $\begingroup$ I feel I found two contradictions the one of the degrees and the one of being integral domain, do I really need both? $\endgroup$ – mathmusic Feb 24 at 3:14
  • $\begingroup$ did I showed in my proof above that the values of deg(b^n) are really distinct? I feel like my proof does not show this in a clear way, what is your opinion? $\endgroup$ – mathmusic Feb 24 at 3:16
  • $\begingroup$ If the degree is negative the same argument goes through, just in the reverse order. $\endgroup$ – Pedro Tamaroff Feb 24 at 11:57

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.