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If $K$ is a ring consisting only of zero, then $K[x]$ is a field (edit: from the comments below I learned that it's not). Are there another rings with this property? I think no. If $R$ contains $1 \neq 0$, then $1 \cdot x \in R[x]$ has no inverse because a degree of a product of two polynomials is a sum of degrees of the factors.

How do you think?

Sergey Filkin
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    Suppose $L$ is a field extension of $K$, and $a \in L$ is algebraic over $K$, then $K[a] = K(a)$ and thus $K[a]$ is a field.

    However, when $x$ is indeterminate or transcendental over $K$, then you are right .

     – sxd Feb 05 '12 at 09:21
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    If $K$ is the trivial ring (ie. it contains only 0) then it seems to me that $K[x]$ is also the trivial ring, which is not a field. – Najib Idrissi Feb 05 '12 at 09:23
  • @zulon why not to count trivial ring as a field? It meets all requirements from the field definition. – Sergey Filkin Feb 05 '12 at 09:31
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    Actually by convention the trivial ring is not a field. There is some literature available at Wikipedia: https://en.wikipedia.org/wiki/Field_with_one_element – Najib Idrissi Feb 05 '12 at 09:34
  • @DimitriSurinx I'm not familiar with field extensions, but when I will be, I will recall your comment. – Sergey Filkin Feb 05 '12 at 09:37
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    If $K$ is the zero ring, then the polynomial $K[X]$ is also the zero ring. The zero ring is a perfectly respectable and useful ring : it is the final object in the category of rings and corresponds to the empty scheme in algebraic geometry. But it is most emphatically not a field. Of course this is a convention, but every definition is a convention and the fact that the zero ring is not a field is universally adopted. – Georges Elencwajg Feb 05 '12 at 09:42
  • @zulon thank you for the link. Very interesting. – Sergey Filkin Feb 05 '12 at 09:42
  • See this question – Mikko Korhonen Feb 05 '12 at 13:17

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Yes, you are right.

$$R[X]=\{\sum_{i=1}^na_iX^i| a_i \in R ~~\text{and}~~ n \in \mathbb N\}$$

Let's call the elements in $R[X]$ a polynomial in $X$. A polynomial in $X$ cannot have negative/ fractional exponents in $X$, by definition.

And, as pointed out in the comments, even if $R=\{0\}$, the ring $R[X]$ is not a field. That amounts to saying, there is no field with a single element. Read the comments above for more enlightening dose of information.