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May 13th, 2017, 06:04 PM   #1
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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?
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June 1st, 2017, 01:05 AM   #2
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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 $\displaystyle R[x]$
$\displaystyle ax\cdot bx = axbx= a(xb)x=a(bx)x=(ab)(xx)=abx^2.$
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June 19th, 2017, 05:54 AM   #3
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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?
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June 20th, 2017, 02:30 AM   #4
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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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