# Splitting Fields over arbitrary fields

I am currently learning field theory by myself (following Advanced Modern Algebra by Joseph Rotman at Chapter 3).

We suppose $$\mathbb{F}$$ is some arbitrary field and $$P\in\mathbb{F}[x]$$, the polynomial ring corresponding to $$\mathbb{F}$$. Suppose $$P$$ is irreducible. The Kronecker Theorem for field extensions claims that we can always construct a field $$E$$ where $$P$$ splits over $$E$$. We start by adjoining a root $$p_1$$ of $$P$$ to $$\mathbb{F}$$ to form some field extension $$\mathbb{F}_{p_1}$$ and I have no problems understanding why $$\mathbb{F}[x]/I_{P}$$ and $$\mathbb{F}_{p_1}$$ are isomorphic, where $$I_P$$ is the ideal generated by $$P$$ forming a quotient ring for $$\mathbb{F}[x]$$. This process makes sense if we have a polynomial in $$\mathbb{Q}[x]$$ since we can adjoin some root in $$\mathbb{C}$$ to $$\mathbb{F}$$ since the Fundamental Theorem of Algebra asserts such a root exists. We can pick some roots for quintics using iterative methods and adjoin some particular root to the field. However for some arbitrary collection of objects, together with operations which satisfy the field axioms, how are we defining such a root $$p_1$$? We know $$x+I_p$$ is a root of $$P$$ if we let $$P$$ be a polynomial with coefficients in $$\mathbb{F}[x]/I_P$$. But does that guarantee that $$p_1$$ is well defined, or is $$p_1$$ merely defined as some arbitrary element not in $$\mathbb{F}$$ satisfying $$P$$? I apologise in advance if this question seems elementary or silly or I am thinking in a too abstract or complicated way as I have not started my university education yet (but soon). Please kindly also correct any factual or conceptual errors in any assertions here. I would like to resolve this issue to move on to Galois Theory so I appreciate any help. Thank you.

you can artificially construct a field $$F_2$$ that is "literally" a superset of $$K_2$$ and define operations in such a way that $$K_2$$ is a subfield of it, by replacing $$K_1$$ by $$K_2$$ inside $$F_1$$. Assume that $$K_2$$ is "nominally" disjoint from $$F_1$$. Call $$\theta:K_2\to K_1$$ the isomorphism. Take the set $$F_2$$ to be $$(F_1\backslash K_1) \cup K_2$$ (we need this union to be disjoint) and define $$a+b$$ as $$a+_{F_1}b$$ (sum in $$F_1$$) if both elements are in $$F_1$$ (are not in $$K_2$$), as $$a+_{F_1}\theta(b)$$ is $$a$$ is in $$F_1$$ and $$b\in K_2$$ and similarly for the remaining 2 cases (I think you understood). Same thing for the product. If $$F_1$$ is not nominally disjoint from $$K_2$$ there are set theory techniques to produce a set (using for instance $$F_1 \times \{*\}$$) to make it such, modifying $$K_1$$ and $$\theta$$ accordingly. * is a symbol that does not appear as a (last) coordinate of any element of $$K_2$$ in any possible expression of it as an element of a cartesian product of sets. We can assume that such symbol exist, as there are limitations to what a set can be. Thinking of concrete cases one by one, this should not be a problem

• I don't understand your answer. The splitting field problem is to construct an ideal such that $F=K[x_1,\ldots,x_n]/I$ is a field, the $x_j$ are roots of $P$, and $P$ splits completely in there. Then for any other field $E$ containing $K$ where $P$ splits completely and generated over $K$ by its roots, it remains to show $E \cong F$. There is nothing else to do. The next step is the algebraic closure problem where we want the splitting field of some uncountable collection of polynomials (in which case the isomorphism $E \to F$ needs the axiom of choice) – reuns Aug 23 '19 at 8:07

You have a good understanding (especially since you haven't started university) but you seem a bit muddled, which is understandable when you first deal with roots in more abstract fields.

Proposition

For any irreducible polynomial $$f \in \mathbb{K}[x]$$, the field $$\mathbb{F}=\mathbb{K}[x]/\mathbb{K}[x]f$$ is given by $$\mathbb{F} = \mathbb{K}1_{\mathbb{F}} \oplus \mathbb{K}t \oplus \cdots \oplus \mathbb{K}t^{n-1},$$ where $$1_{\mathbb{F}} = 1+\mathbb{K}[x]f$$ and $$t=x+\mathbb{K}[x]f$$, and $$f(t)=0$$.

Definition

Given field extension $$\mathbb{K} \subseteq \mathbb{F}$$ and $$a_1,\dots,a_n$$, define $$\mathbb{K}(a_1,\dots,a_n) = \{ \frac{c}{d} \ | \ c,d \in \mathbb{K}[a_1,\dots,a_n], d(a_1,\dots,a_n) \}$$ which is the smallest subfield of $$\mathbb{F}$$ containing $$\mathbb{K}$$ and $$a_1,\dots,a_n$$.

Remark

Now, suppose $$f \in \mathbb{K}[x]$$ with $$\mathrm{deg}f \geq 1$$. We can factorise $$f$$ into irreducibles over $$\mathbb{K}[x]$$ $$f = p_1 \cdots p_r$$. Since $$p_1 \in \mathbb{K}[x]$$ is irreducible, $$p_1$$ has a root $$t_1$$ in a larger field $$\mathbb{K}_1$$. We can factorise $$f$$ into irreducibles over $$\mathbb{K}_{1}[x]$$ $$f = (x - t_1) q_2 \cdots q_s$$. Since $$q_2 \in \mathbb{K}_{1}[x]$$ is irreducible, $$q_2$$ has a root $$t_2$$ in a larger field $$\mathbb{K}_2$$. We can factorise $$f$$ into irreducibles over $$\mathbb{K}_{2}[x]$$ $$f = (x -t_1)(x-t_2) h_2 \cdots h_l$$.

Continuing like this, we get a larger field $$\mathbb{F}$$ which splits $$f$$ into linear factors $$f = c (x-t_1) \cdots (x-t_n)$$ with $$c \in \mathbb{K}$$ and $$t_1,\dots,t_n \in \mathbb{F}$$. That fact that this is possible leads us to the following definition.

Definition

Let $$\mathbb{K}$$ be a field and let $$f \in \mathbb{K}[x] \setminus \{0\}$$.

We call a field $$\mathbb{F}$$, containing $$\mathbb{K}$$, a splitting field of $$f$$ over $$\mathbb{K}$$ if $$\exists c \in \mathbb{K}$$ $$\exists t_1,\dots,t_n \in \mathbb{F}$$ such that $$f = c (x-t_1) \dots (x-t_n)$$ and $$\mathbb{F} = \mathbb{K}(t_1,\dots,t_n)$$.

Remark

Notice that whilst, given $$t_1,\dots,t_n$$, $$\mathbb{F} = \mathbb{K}(t_1,\dots,t_n)$$ is completely and uniquely defined, the $$t_1,\dots,t_n$$ are definitely not uniquely defined by $$f \in \mathbb{K}[x]$$.

For example, take $$x^2 + 1 \in \mathbb{Q}[x]$$. We can say that its roots are $$(0,1)$$, $$(0,-1)$$ in $$\mathbb{C}$$ (as $$\mathbb{R}^2$$) or we could say that they are $$\bar{t}$$, $$-\bar{t}$$ in $$\mathbb{R}[t]/\mathbb{R}[t](t^2+1)$$.

What we do know is that the roots are unique up to renaming and that therefore $$f \in \mathbb{K}[x]$$ has only one splitting field up to field isomorphism (i.e. renaming).

• Thank you for your answer. I understand what you have written, but $\mathbb{F}$ is a larger field of which with the same field operations the collection $\{k+\mathbb{K} f\}$ for which $k\in\mathbb{K}$ is isomorphic to $\mathbb{K}$ by the natural map but does that imply that there exists a larger field having $\mathbb{K}$ as a subfield containing at least one root of $f$? – Herrweswegen Feb 15 '19 at 16:00
• Yes, my first proposition above says that $\mathbb{K} \cong \mathbb{K} 1_{\mathbb{F}} \subseteq \mathbb{F}$ and $\mathbb{F}$ contains at least one root of $f$. Some nit picking: there is a (subtle) difference between $k \in \mathbb{K}$ and $k \in \mathbb{K}[x]$, with one being an element of a field and the other being a constant polynomial, and canonical is probably a better word than natural as natural is a sensitive word for category theorists. – Joseph Martin Feb 15 '19 at 16:16
• Thank you for pointing out the terminology since some authors interchange them. I think what I do not understand is this. Given two isomorphic fields $\mathbb{K}_1,\mathbb{K}_2$ and we know there exists some larger field $\mathbb{F}_1$ containing $\mathbb{K}_1$ as a subfield and a solution to a polynomial $f\in\mathbb{K}_1[x]$ where $f=k_0 + k_1 x +\hdots$ does that means there exists a larger field $\mathbb{F}_2$ containing $\mathbb{K}_2$ as a subfield and a solution to $f^{'}\in\mathbb{K}_2[x]$ where $f=\pi(k_0) + \pi(k_1) x +\hdots$ – Herrweswegen Feb 15 '19 at 16:22
• where $\pi: \mathbb{K}_1\rightarrow\mathbb{K}_2$ is the isomorphism between these two fields? Sorry I gave to split up the comment (too long) – Herrweswegen Feb 15 '19 at 16:22
• Yes, this is obvious when you realise that a field isomorphism is simply a renaming of the elements; it is a bijection of sets so that the field operations and axioms are all preserved, the only thing that changes is the name for the elements. Whether you work with $f$ in $\mathbb{K}_{1}[x]$ or $\pi^*(f)$ in $\mathbb{K}_2$ makes no difference, they may as well be the same thing. (I have used the notation $\pi^*$ for applying $\pi$ to the coefficients of $f$). More nit picking: one should always write polynomials as having finitely many terms, otherwise they look like formal power series. – Joseph Martin Feb 15 '19 at 16:32