When Different Constructions Give the Same Field

Central Question

When are differently presented number systems genuinely the same as fields?

Why This Matters

Isomorphism separates mathematical structure from its presentation and shows why a defining equation is not itself a complete description of a field.

Mathematical Notes

Isomorphism means preserving arithmetic and the base field

An $F$-isomorphism fixes every element of $F$. An isomorphism between two extensions need not preserve the particular generators used to describe them.

Both real quotients below are copies of $\mathbb C$:

In the second quotient, $[x]/\sqrt2$ squares to $-1$. The change of generator reconciles the apparently different defining equations. Both maps are onto, and their kernels before quotienting are the displayed polynomial ideals.

Quadratic fields over $\mathbb Q$ behave differently

For nonsquares $d,e\in\mathbb Q^\times$,

To prove the nontrivial direction, the image of $\sqrt d$ must be $a+b\sqrt e$ and have square $d$. Comparing coefficients gives $2ab=0$. Since $d$ is not a rational square, $b\ne0$, hence $a=0$ and $d=b^2e$. Conversely, such a $b$ defines the isomorphism by $\sqrt d\mapsto b\sqrt e$.

Thus $\mathbb Q(\sqrt2)$ and $\mathbb Q(\sqrt8)$ are the same subfield of $\mathbb R$, but $\mathbb Q(\sqrt2)$, $\mathbb Q(\sqrt3)$ and $\mathbb Q(i)$ are pairwise nonisomorphic. Every field isomorphism between characteristic-zero fields fixes their prime subfield $\mathbb Q$, so this also rules out abstract field isomorphisms between these three.

Equal dimension does not determine field multiplication

All three are two-dimensional rational vector spaces. The missing information is how basis elements multiply. A vector-space isomorphism can send $\sqrt2$ to $\sqrt3$; a field isomorphism cannot, because it would turn the equation $2=(\sqrt2)^2$ into $2=3$.

Conjugate roots can generate different embedded subfields while still giving isomorphic extensions. With $\alpha=\sqrt[3]2$ and a primitive cube root of unity $\omega$, the fields $\mathbb Q(\alpha)$ and $\mathbb Q(\alpha\omega)$ are isomorphic over $\mathbb Q$ but are different subfields of $\mathbb C$.

Learning Prompt

Explore field isomorphisms through examples like $\mathbb R[x]/(x^2+1)$, $\mathbb R[x]/(x^2+2)$, and $\mathbb C$. Explain why apparently different constructions can produce isomorphic fields. Then compare this with $\mathbb Q(\sqrt2)$, $\mathbb Q(\sqrt3)$, and $\mathbb Q(i)$. Let me predict which fields are isomorphic before explaining.

Ideas

Examples

Questions

Connections

• Why Quotient by a Polynomial?
• Field Automorphisms
• Homeomorphism
• Week overview

One Thing That Surprised Me

Further Rabbit Holes

Sources

Ian Stewart, Galois Theory, fifth edition, CRC Press, §5.4; §13.1. Publisher and edition details.