# Can a principal ideal generated by a polynomial with unit leading coefficient contain nonzero elements of lesser degree?

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)?

• Are you considering left ideals or right ideals or two-sided ideals? – Angina Seng Jul 19 '20 at 1:40
• @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 – GMarks2000 Jul 19 '20 at 1:47
• $b(X+a)-(X+a)b=ba-ab$? – Angina Seng Jul 19 '20 at 1:47
• 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! – GMarks2000 Jul 19 '20 at 1:52
• 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). – tkf Jul 19 '20 at 3:41

## 1 Answer

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$$(f)$$.

• 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. – tkf Jul 19 '20 at 3:19
• 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. – Mark Sapir Jul 19 '20 at 3:26
• 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. – tkf Jul 19 '20 at 3:32
• 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. – Mark Sapir Jul 19 '20 at 3:42