Ring of Polynomials over Reals is not Field

Theorem

Let $\R \sqbrk X$ be the ring of polynomials in an indeterminate $X$ over $\R$.


Then $\R \sqbrk X$ is not a field.


Proof

Consider the polynomial $x + 1$ in $\R \sqbrk X$.


There exists no polynomial $\map f x$ such that:

$\paren {x + 1} \map f x = 1$

This is because the left hand side has degree $1$, and the right hand side has degree $0$.

$\blacksquare$


Sources

  • 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $4$: Fields: $\S 15$. Examples of Fields: Example $19$
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