My Math Forum Construction of a polynomial rings

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 May 13th, 2017, 05:04 PM #1 Newbie   Joined: Apr 2017 From: Europe Posts: 13 Thanks: 1 Construction of a polynomial rings Can anyone explain why we need to construct a polynomial ring necessarily over a commutative ring with an identity? What would happen if we tried to make one over a noncommutative ring without identity?
 June 1st, 2017, 12:05 AM #2 Newbie   Joined: May 2017 From: Russia Posts: 18 Thanks: 0 We can construct a polynomial ring over a non-commutative ring without unity. But its variables must commute with all the elements of the ring. Otherwise, the product of two polynomials might be not a polynomial. For example, in $\displaystyle R[x]$ $\displaystyle ax\cdot bx = axbx= a(xb)x=a(bx)x=(ab)(xx)=abx^2.$
June 19th, 2017, 04:54 AM   #3
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Joined: Apr 2017
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Quote:
 Otherwise, the product of two polynomials might be not a polynomial.
Can you give a particular example of that property?

 June 20th, 2017, 01:30 AM #4 Newbie   Joined: May 2017 From: Russia Posts: 18 Thanks: 0 Let $\displaystyle R$ be a ring, and $\displaystyle a,b,c\in R$. And let $\displaystyle x$ be a variable. We can construct polynomials in $\displaystyle x$: $\displaystyle ax^2+bx,\ \ cx^2.$ If $\displaystyle x$ didn't commute with $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$, the product of the polynomials would be $\displaystyle (ax^2+bx)\cdot cx^2=ax^2cx^2+bxcx^2.$ We could not rewrite the result in the form $\displaystyle a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$ where $\displaystyle a_i\in R.$

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