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This came up as part of a proof I'm trying to write. Suppose $P(x)$ is a polynomial of degree $n$ over a ring $R$ with identity. If its leading coefficient is a unit (i.e. has a multiplicative inverse), can the principal (two-sided) ideal $(P)$ contain nonzero polynomials of degree less than $n$? I strongly suspect the answer is no. It's clearly impossible in the case of commutative $R$, but the noncommutative case is proving surprisingly difficult because the characterization of principal ideals isn't as nice.

Can anyone spot a proof, or if I'm mistaken, a counterexample for noncommutative $R$ (preferably as elementary as possible; I'm still fairly new to ring theory)?

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    $\begingroup$ Are you considering left ideals or right ideals or two-sided ideals? $\endgroup$ – Angina Seng Jul 19 '20 at 1:40
  • $\begingroup$ @AnginaSeng two-sided ideals. Apologies; my book simply uses "ideals" to mean "two-sided ideals", as a quick google showed me. I'll update the question $\endgroup$ – GMarks2000 Jul 19 '20 at 1:47
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    $\begingroup$ $b(X+a)-(X+a)b=ba-ab$? $\endgroup$ – Angina Seng Jul 19 '20 at 1:47
  • $\begingroup$ Oh, yes, that would be one. I'll have to think through why this proof isn't working, then, as I know the result is true independently. Thanks! $\endgroup$ – GMarks2000 Jul 19 '20 at 1:52
  • $\begingroup$ Following @JCAA 's comment on his answer, suppose that $X$ is not assumed to be central in the definition of polynomial ring. Then $(X^2-a)X-X(X^2-a)=Xa-aX$ provides a counterexample (with the obvious interpretation of degree). $\endgroup$ – tkf Jul 19 '20 at 3:41
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I guess your rings are commutative. Then the ideal $I$ generated by a monic polynomial $f(x)$ has Groebner basis $\{f(x)\}$. If a polynomial $g(x)\in I$, the highest term of $g$ must be divisible by the highest term of $f(x)$ so degree$(g)$$\ge$degree$(f)$.

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  • $\begingroup$ I think the OP was familiar with this argument already. His question was specifically about non-commutative rings and @AnginaSeng gave (in my view) the best answer possible in the comments. $\endgroup$ – tkf Jul 19 '20 at 3:19
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    $\begingroup$ Now I see that the OP does talk about non-commutative rings. But then it is not clear what a polynomial is. There are several non-equivalent definitions. Also the word "two-sided" appeared in the OP after I posted my answer. $\endgroup$ – Mark Sapir Jul 19 '20 at 3:26
  • $\begingroup$ Good point (+1): The answer in comments does not work if you do not assume the indeterminate is central. I interpreted (I think correctly) that the OP meant for it to be central, but it is not stated. $\endgroup$ – tkf Jul 19 '20 at 3:32
  • $\begingroup$ I am not sure "central indeterminate" is the most frequent assumption when people define polynomials over non-commutative rings. I would call a polynomial an element of the free $R$-algebra of rank 1. $\endgroup$ – Mark Sapir Jul 19 '20 at 3:42

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