An example of a field $F$ such that $F^n$ uses element-wise operations, but $F$ is not a subfield of $F^n$? I'm going through The Linear Algebra a Beginning Graduate Student Ought to Know, and I came across this idea that I can't seem to understand. Suppose $F$ is a field, then he asks "is it possible to define multiplication in such a manner that $F^n$ will become a field naturally containing $F$ as a subfield?" The answer seems easy, I could just do elementwise multiplication (like we do addition), then for $(a_1,...,a_n),(b_1,...,b_n)\in F^n$, we have a that $(a_1,...,a_n)(b_1,...,b_2)=(a_1b_1,...,a_nb_n)$. Clearly multiplication is commutative and associative, every element has a multiplicative inverse except the identity, and it is distributive because $F$ is a field. However, he says "in general, the answer is negative."
Can you help me convince myself as to why this is true?
 A: If one of the entries of $x \in F^n$ is zero, then that element has no multiplicative inverse.
For an example of a field such that $F^n$ has no field structure, consider $\mathbb{C}^2$. If there is a field structure on $\mathbb{C}^2$ such that $\mathbb{C}$ is a subfield of $\mathbb{C}^2$, then it would be a degree 2 field extension. However, since $\mathbb{C}$ is algebraically closed, it has no degree 2 field extension.
A: Using elementwise multiplication, $F^n$ is never a field for $n>1$, so that won't do.
If you have a polynomial $f(x)\in F[x]$ which is irreducible and of degree $n$, the quotient $F[x]/(f(x))$ will be an $n$ dimensional vector space over $F$, so its underlying additive group is $F^n$.  The ring structure it gets as a quotient is in fact a field extension of $F$.
The problem is does a degree $n$ irreducible polynomial exist in $F[x]$?
It all depends on the field. For $\mathbb Q$, you can get an irreducible polynomial for any $n$.  For $\mathbb R$ you can only get $n\in \{1,2\}$. For $\mathbb C$ you can only get $n\in\{1\}$.
