Rings and Fields: Understanding Polynomials as a Commutative Ring with Unity

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Homework Statement



is the set of all polynomials a ring,and a fieldd.Is is commutative and does it have unity

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The Attempt at a Solution



now if we add or multiply any polynomials we get a polynomial. So it is a ring, but i am not sure what the multiplicative inverse is or whether every non zero element has an inverse to constitute a field.
 
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Multiplication is just ordinary multiplication of polynomials and the identity is the "constant" polynomial p(x)= 1 for all x. Does there exist a polynomial p(x) such that (x-1)(p(x))= 1? Think about the degree of p and what multiplication of polynomials does to degrees.
 
i am thinking that we do not have multiplicative inverses because there is no polynomial that would give one, we would need to use a rational function, is this a correct assesmnent?
 
Yes, but to give a complete answer, you need to say why "no polynomial would give one" (I presume you mean there is no polynomial, p(x), such that (x-1)p(x)= 1.)

Again, think of the "degree" of polynomials. The degree of x- 1 is 1. What could the degree of (x-1)p(x) be? Could it be equal to 0, the degree of the constant polynomial, 1?
 
ok thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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