# Monic irreducible polynomials over infinite field

If $F$ is a countable field, then proving that $F$ has algebraic closure is quite simple: there can be at most countable number of monic irreducible polynomials over $F$, let they form the set $\mathcal{L}$. For each polynomial in $\mathcal{L}$, we get an algebraic extension. Writing $\mathcal{L}$ as countable ascending union of finite subsets, one can obtain algebraic closure of $F$.

Of course the argument will not work if $\mathcal{L}$ is uncountable, some variation in argument is required. This already has been appeared in some other questions on mathstack.

But this raised one question to me:

Question: Does there exists a field $F$ with the following three properties:

(1) $F$ is uncountable

(2) $F$ is not algebraically closed.

(3) The number of monic irreducible polynomials of degree $>1$ is countable.

Note that $\mathbb{R}$ satisfies (1) and (2) but not (3).

• For an irreducible polynomial $p(x)$, we can consider $p_{\lambda}(x)=\lambda^{-deg(p)}p(\lambda x)$. If $p(x)$ is irreducible, then I believe $p_{\lambda}(x)$ is also, and creates a $|F|$ number of polynomials. – Pax Kivimae Jun 22 '16 at 2:29
• I also expect it to be monic. is it monic? – p Groups Jun 22 '16 at 2:34
• If you start with a monic $p(x)$, then $p_\lambda(x)$ will also be monic. – Ted Shifrin Jun 22 '16 at 2:37
• perhaps you are right. – p Groups Jun 22 '16 at 2:41
• Your argument for the countable case is incomplete. Writing $L$ as the increasing union of its finite subsets does not help you; you need to pick an order in which to adjoin the roots of irreducible polynomials. Once you do this there's no need for $L$ to be countable; just use the axiom of choice to pick a well-ordering of $L$. – Qiaochu Yuan Jun 22 '16 at 4:03

## 1 Answer

There is no such field.

Take any uncountable field $F$ that is not algebraically closed.

Let $K$ be an algebraic closure of $F$.

Then $[K:F] > 1$ and hence $K \smallsetminus F$ is uncountable.

If there are countably many monic irreducible polynomials over $F$ of degree more than $1$, then there are countably many elements in $K \smallsetminus F$, since each of them has a minimal polynomial (which has finite degree), which gives a contradiction.