# Why can not the Polynomial Ring be a Field?

I'm presently researching Polynomial Rings, yet I can not identify why they are Rings, not Fields. In the definition of a Field, a Set constructs a Commutative Group with Addition and also Multiplication. This indicates an inverted numerous for every single Element in the Set.

Guide does not specify on this, nonetheless. I do not recognize why a Polynomial Ring could not have an inverted multiplicative for every single component (at the very least in the entire numbers, and also it's currently considered that it has a neutral component). Could someone please clarify why this can not be so?

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2019-05-09 11:31:13
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Consider $\mathbb{C}[x]$ the ring of polynomials with coefficients from $\mathbb{C}$. This is an instance of polynomial ring which is not an area, due to the fact that $x$ has no multiplicativ inverse.

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2019-05-10 10:13:32
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The devices of $D[x]$ are specifically the devices of $D$, when $D$ is a domain name.

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2019-05-10 09:59:13
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Because necessarily, the only polynomial that can have an adverse level is $0$, which is specified to have a level of $-\infty$. Non - absolutely no constants have level $0$. You after that have the level formula : $\deg (fg) = \deg (f) + \deg (g)$ for any kind of polynomials $f,g$. By examination, any kind of polynomial of level $n \geq 1$ would certainly require as an inverted a polynomial of level $-n$, which does not exist (i.e. what Agusti Roig claimed!) The set you desire does exist, nonetheless : it is called the area of sensible features , and also is specifically the set of proportions of polynomials. It is created similarly that the area of sensible numbers is from the ring of integers.

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2019-05-10 07:51:50
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Hint $\rm\quad\rm x \, f(x) = 1 \,$ in $\,\rm R[x]\ \Rightarrow \ 0 = 1 \,$ in $\,\rm R, \,$ by reviewing at $\rm\ x = 0$

Remark $\$ This has a really instructional global analysis : if $\rm\, x\,$ is a device in $\rm\, R[x]\,$ after that so also is every $\rm\, R$ - algebra component $\rm\, r,\,$ as adheres to by reviewing $\ \rm x \ f(x) = 1 \$ at $\rm\ x = r\,.\,$ Therefore to offer a counterexample it is adequate to show any kind of nonunit in any kind of $\rm R$ - algebra.  An all-natural selection is the nonunit $\,\rm 0\in R,\,$ which generates the above evidence.

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2019-05-10 06:52:30
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