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I'm working through a proof of Amitsur's Theorem, that if $\mathbb{K}$ is a field and $A$ a $\mathbb{K}$-algebra with $\dim_{\mathbb{K}}A<|\mathbb{K}|$, then $J(A)$ (the Jacobson radical) is nil (all it's elements are nilpotent).

We let $a\in J(A)$. In the final step of the proof, we arrive at $$ 0 = a^k\underbrace{(b_0 + b_1a + ... + b_m a^m)}_{=: p(a)},\quad b_i\in\mathbb{K}, b_0\neq 0 $$

My prof says $p(a)$ is invertible "since each $b_i$ is a unit", but I fail to see how this is true.

Is it true, in general, that if each $u_i$ is a unit of a ring $R$, and each $u_i$ commutes with $a$ along with every other $u_j$, then $u_0 + u_1a + \cdots + u_n a^n$ is also a unit? I know that if $u\in R^\times$, $a\in R$ is nilpotent, and $ua = au$, then $u + a\in R$, but this cannot be applied in our case as we cannot assume a priori that $a$ is nilpotent (this is what we are trying to show). If the general statement is false, what could be "special" about our $p(a)$ above which guarantees it's a unit?

user3002473
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    Would the downvoter care to explain what they disapprove of? – user3002473 Dec 11 '20 at 20:24
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    I don't think the downvote is fair. I upvoted to cancel out the downvote. This is a fine question. – morrowmh Dec 11 '20 at 20:37
  • @MichaelMorrow Thank you! I know the question is strange and niche, but I didn't think it was ill-posed. – user3002473 Dec 11 '20 at 20:44
  • See the Theorem in the linked dupe. – Bill Dubuque Dec 12 '20 at 19:58
  • @BillDubuque I think that's a little far-reaching to say it's a duplicate. My concern about whether or not $p(a)$ is invertible is, certainly, a simple corollary of the result of the theorem in the first linked answer. But to the untrained eye (myself included), these aren't duplicates. Perhaps it will be useful for some other student, new to Jacobson radicals, in the future? – user3002473 Dec 12 '20 at 20:31
  • See abstract duplicates. It is an immediate corollary of the linked theoren, so, as the dupe link claims, "this question is already answered there". – Bill Dubuque Dec 13 '20 at 02:38

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I see what the issue is. The statement in the question title is not true in general (although I suppose it might be true if $a$ is nilpotent). Instead, the reason $p(a)$ is invertible is because $a\in J(A)$.

Recall that if $a\in J(A)$ then $1 - ra$ is invertible for all $r\in A$. From this we can deduce that $b_0 + ra$ is invertible for all $b_0\in\mathbb{K}\backslash\{0\}$ and $r\in A$. This includes $r = b_1 + b_2a + \cdots + b_m a^{m-1}$, hence $b_0 + (b_1 + \cdots + b_m a^{m-1})a = p(a)$ is invertible.

user3002473
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