1
$\begingroup$

$F$ is a field, so by definition, $F$ is a commutative ring with unity in which every non-zero element is a unit. Then, $F[x]$ is a set of polynomials in which the coefficients come from $F$, so all of the non-zero coefficients have units in $F$.

This means we can have polynomials like $f=a_0+a_1x+a_2x^2+\cdots +a_nx^n$ where $a_k\in F,$ $0\le k\le n$.

If $F[x]$ were a field, every non-zero element would be a unit.

I'm not really sure where to go from here.

$\endgroup$
  • $\begingroup$ I don't think so. $\endgroup$ – TheMobiusLoops Oct 27 '13 at 2:33
  • $\begingroup$ It would be $n+1$. $\endgroup$ – TheMobiusLoops Oct 27 '13 at 2:35
  • $\begingroup$ So it can't have inverses because if it did, the degree would be 0. Right? $\endgroup$ – TheMobiusLoops Oct 27 '13 at 2:37
  • $\begingroup$ Geometrically, taking a ring $R$ and adding an indeterminate $R[x]$ is like "adding a dimension" to $R$. All fields have dimension $0$, and the above shows that for any ring (non-zero) $R$, $R[x]$ has dimension at least $1$. I could make this rigorous, but I don't want to--you already have a good rigorous answer, this is just intuition carries over to many settings. For example, you could ask if $\mathbb{Z}$ could ever be isomorphic to $F[x,y]$ for a field $F$, and the same intuition works. $\endgroup$ – Alex Youcis Oct 27 '13 at 3:16
1
$\begingroup$

You can "turn it into a field" by taking the quotient ring $F[x]/M$ for a maximal ideal $M$ of $F[x]$. This is sometimes called sending $M$ to zero.

This is actually another structure which is indeed a field, but it's not related isomorphically to $F$. So a better way to describe this is inducing a field using $F[x]$ and $M$.

$\endgroup$
1
$\begingroup$

$x$ is not zero in $F[x]$ and has no multiplicative inverse. So $F[x]$ cannot be a field.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.