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Let $F$ be a field, $E_1$ and $E_2$ are two distinct extension fields of $F$. Is it the case that we can always somehow find a field $G$ that contains both $E_1$ and $E_2$? In other words, could extensions of fields have different 'direction's such that they are incompatible?

Edit: I began to think about this problem while reading a proof. $F$ is a field. $a$ and $b$ are algebraic over $F$. $p(x)$ and $q(x)$ are two polynomials in $F[x]$ of minimum degree that respectively make $a$ and $b$ a zero. The proof claims that there is an extension $K$ of $F$ such that all distinct zeros of $p(x)$ and $q(x)$ lie in $K$. For a single polynomial, I know this kind of field exists because of the existence of splitting field, why it is true for two polynomials?

Watson
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qiang heng
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    A) This is a good question. The answer is that you can always find a field that contains subfields isomorphic to $E_1$ and $E_2$, but not necessarily as subsets. The term is the compositum of the extension fields. See, for example here. B) Wouldn't the splitting field of the product $p(x)q(x)$ give what you need? – Jyrki Lahtonen Sep 02 '18 at 08:33
  • Thank you! That's exactly what I was always confused about! – qiang heng Sep 02 '18 at 08:36
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    The following kind of problems occur. You could have $F=\Bbb{R}$, $E_1=\Bbb{C}$, $E_2=\Bbb{R}[x]/\langle x^2+1\rangle$. When we describe $E_1$ and $E_2$ as extensions of $\Bbb{R}$ we identify certain subsets within both with $\Bbb{R}$ in the well known way. But, those identifications don't immediately identify all the elements of $E_1$ with those of $E_2$. Without such extended identification it stands to reason that in some sense $\Bbb{R}$ is the intersection $E_1\cap E_2$. Yet, it is easy to show that any extension $\Omega$ of $\Bbb{R}$ can contain at most a single field isomorphic to either – Jyrki Lahtonen Sep 02 '18 at 08:38
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    (cont'd) Explaining that we really need that "up to isomorphism" part in many statements when dealing with field theory. – Jyrki Lahtonen Sep 02 '18 at 08:39
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    I agree that this is a good question, because it hints towards a subtlety that is often overlooked. Besides Jyrki's point that one often says or should say "up to isomorphism", another way of dealing with it, at least for algebraic extensions, is that one often fixes one algebraic closure $\bar K$ of a given field $K$, and then says that all algebraic extensions $L|K$ are to be taken within that algebraic closure. (And then with abstract arguments convinces oneself that if one had picked a different algebraic closure $\bar K '$, one would have got very isomorphic results.) – Torsten Schoeneberg Sep 02 '18 at 09:00
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    An example of how difficult these things can become if one is not careful is the question https://mathoverflow.net/q/82083/27465. – Torsten Schoeneberg Sep 02 '18 at 12:32
  • Is there a field that "naturally" contains both the the $2$-adic numbers and the $3$-adic numbers? They are, after all, both extensions of the rational numbers. – GEdgar Sep 04 '18 at 19:01

3 Answers3

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Consider field extensions $E_1/F$ and $E_2/F$. Then the tensor product $A=E_1\otimes_F E_2$ is a commutative ring, not necessarily a field though. Non-trivial commutative rings have maximal ideals, by a Zorn's lemma argument. Let $I$ be a maximal ideal of $A$. Then $K=A/I$ is a field. The map $x\mapsto \overline{x\otimes 1}\in A/I$ is a ring homomorphism $E_1\to K$. As $E_1$ is a field, this is an injective homomorphism, so we can think of $E_1$ being "contained" in $K$. Likewise $E_2$ is "contained" in $K$.

Beware though, the ideal $I$ may not be unique.

Angina Seng
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    As an alternative, factor out a prime ideal of $A$, rather than a maximal one, and take the field of fractions of the integral domain you get. – Angina Seng Sep 02 '18 at 12:25
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You could embed $F$ into its algebraic closure $\overline{F}$ (exists by Zorn's lemma, equivalent to the Axiom of Choice). Then both $E_1$ and $E_2$ are essentially (up to isomorphism) subfield of $\overline{F}$ and we can take the minimal subfield of $\overline{F}$ that contains $E_1 \cup E_2$. This is assuming algebraic extensions, the most interesting case, I think.

Henno Brandsma
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    However, if we are that careful, one should note that $\bar F$ is not unique either, only up to isomorphism fixing $F$. – Torsten Schoeneberg Sep 02 '18 at 11:34
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    And the outcome of the process may depend on the choice of subfields isomorphic to $E_1$ and $E_2$. Basically because that choice is not unique unless $E_i/F$ are Galois. Consider $F=\Bbb{Q}$, $E_1=F(\root3\of2)$, $E_2=F[x]/\langle x^3-2\rangle$. If you choose $E_1$ as the copy of $E_2$ sitting inside $\Bbb{C}$ you get $F(E_1\cup E_2)=E_1$. OTOH if you choose $F(e^{2\pi i/3}\root3\of2)$ as the copy of $E_2$, you get the degree six splitting field of $x^3-2$ inside $\Bbb{C}$. – Jyrki Lahtonen Sep 02 '18 at 12:44
  • @JyrkiLahtonen thanks for showing me it's more subtle than I thought. – Henno Brandsma Sep 02 '18 at 12:46
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For your specific example: Take $F$, then find the splitting field of $p(x)$. Let the splitting field be $G$. Now factorize $q(x)=\prod_{i=1}^{\ell} q_i(x)$ over $G$ into product of irreducible polynomials (Note that $q(x)$ may be irreducible over $F$ but over $G$ it might factorize)then extend $G$ to a splitting field of $q_1$ let it be $G_1$. Now factorize $\prod_{i=2}^{\ell} q_i(x)$ into product of irreducible polynomials over $G_1$ and extend to a splitting field and so on. Since the degree of $q(x)$ is finite, this process will stop. Then you have a field which contains all the zeros of $p(x),q(x)$.

Balaji sb
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