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 Abstract Algebra Abstract Algebra Math Forum

 May 13th, 2017, 05:04 PM #1 Newbie   Joined: Apr 2017 From: Europe Posts: 16 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 Member   Joined: May 2017 From: Russia Posts: 34 Thanks: 5 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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 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 Member   Joined: May 2017 From: Russia Posts: 34 Thanks: 5 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 $\displaystyle a_i\in R.$ Tags construction, polynomial, rings Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post math interest22 Abstract Algebra 5 February 5th, 2014 06:30 AM taina007 Abstract Algebra 1 January 11th, 2011 03:52 AM remeday86 Abstract Algebra 4 June 28th, 2010 07:55 PM bjh5138 Abstract Algebra 0 December 4th, 2007 03:22 PM bjh5138 Abstract Algebra 1 October 30th, 2007 09:42 AM

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