# When $K[x]$ is a field?

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?

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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 5 '12 at 9:21
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 5 '12 at 9:23
@zulon why not to count trivial ring as a field? It meets all requirements from the field definition. – Sergey Filkin Feb 5 '12 at 9:31
Actually by convention the trivial ring is not a field. There is some literature available at Wikipedia: en.wikipedia.org/wiki/Field_with_one_element – Najib Idrissi Feb 5 '12 at 9:34
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 5 '12 at 9:42

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