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May 13th, 2017, 05:04 PM  #1 
Newbie Joined: Apr 2017 From: Europe Posts: 15 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: 19 Thanks: 1 
We can construct a polynomial ring over a noncommutative 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  
Newbie Joined: Apr 2017 From: Europe Posts: 15 Thanks: 1  Quote:
 
June 20th, 2017, 01:30 AM  #4 
Newbie Joined: May 2017 From: Russia Posts: 19 Thanks: 1 
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 $\displaystyle a_nx^n+a_{n1}x^{n1}+\ldots+a_1x+a_0$ where $\displaystyle a_i\in R.$ 

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construction, polynomial, rings 
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