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The task is to determine if $\mathbb R[x]$, which represents the set of all polynomials with real coefficients, is a field. My response is that it is not, since I feel that not all of these polynomials have a multiplicative inverse.

If I am correct, how would I show this in detail?

Kenny Wong
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Analysis15
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    You need three things basically: (1) To know what the degree of a polynomial is (2) To be convinced that the degree of a product $f(x)g(x)$ is exactly the sum of the degree of $f$ plus the degree of $g$ (3) To know what it means for $x$ to have an inverse: it means there exists a polynomial $f(x)$ such that $xf(x) = 1$. Can you put the pieces together? (hint: compare the degree of $xf(x)$ with the degree of $1$, which should be equal) – Daniel Apr 02 '17 at 07:50

1 Answers1

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What is the inverse of the polynomial $X$?

Suppose $P(X)$ were its inverse; say $P(X) = a_n X^n + Q(X)$, where all the powers of $X$ in $Q$ are of order less than $n$, and where $a_n \not = 0$. Then $1=X P(X) = a_n X^{n+1} + X Q(X)$ where $X Q(X)$ has no $X^{n+1}$ coefficient, so the $X^{n+1}$ coefficient of $X P(X)$ cannot be $0$, so $X P(X)$ cannot be $1$.

Dietrich Burde
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Patrick Stevens
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  • Your explanation is kind of confusing for me, as I don't understand your logic in showing that the polynomial x has no inverse. – Analysis15 Apr 02 '17 at 07:44
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    @Analysis15 If $;XP(X)=1;$ , it must be that all the coefficients of positive powers of $;X;$ vanish. Yet in this product, one already has $;a_nX^{n+1};$ , and the basic assumption (that Patrick perhaps forgot or maybe he left for you to complete) is that $;a_n\neq0;$ . Since $;\deg(XQ(X))\le n;$ , the monomial $;a_nX^{n+1};$ cannot vanish in the product $;XP(X);$ , contradiction...+1 – DonAntonio Apr 02 '17 at 08:01
  • I just forgot to say that $a_n \not = 0$; thanks for that. – Patrick Stevens Apr 02 '17 at 09:37