Algebraically closed proof.

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

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

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

• What if $\deg t < 0$? – mathmusic Feb 24 at 3:03
• Do I have redundant steps in my proof?If so, can you point them out ? – mathmusic Feb 24 at 3:11
• I feel I found two contradictions the one of the degrees and the one of being integral domain, do I really need both? – mathmusic Feb 24 at 3:14
• 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? – mathmusic Feb 24 at 3:16
• If the degree is negative the same argument goes through, just in the reverse order. – Pedro Tamaroff Feb 24 at 11:57