Are the split-complex numbers a proper algebraic extension of the real numbers?

Question

Are the split-complex numbers, that is, $\mathbb R(j)$, where $j^2 = 1$, a proper extension of $\mathbb R$? I've been learning about how $F[x]/\langle p\rangle$, for $p$ in $F[x]$ and $F$ a field, is isomorphic to $F(a)$, if $p(a) = 0$. This got me thinking about $\mathbb R[x]/\langle x^2-1\rangle$, which is isomorphic to $\mathbb R(j)$, because both roots of $x^2-1$ are already in $\mathbb R$. So is there some notion in which $\mathbb R(j)$ is equivalent to $\mathbb R$, or more generally, that $F(a)$ is equivalent to $F$ if all roots of $p$ are already in $F$?

Answer 1

I think that the biggest misunderstanding here is that $X^2-1$ has two roots. Such property, i.e. polynomial $p(X)$ has at most $\deg(p(X))$ roots, is true over fields or more generally over integral domains (which can be treated as subrings of fields). But $\mathbb{R}(j)$ is neither, which can be easily seen, because $(j-1)\cdot(j+1)=0$ while none of them is zero.

We know that $\pm 1$ are roots of that polynomial, distinct if $K$ is not of characteristic $2$. But $K[X]/\langle p(X)\rangle$ may have more than two roots for that polynomial. In fact given $p(X)=X^2-1$ we already have two additional roots: $[X]$ and $[-X]$. So $X^2-1$ has at least four roots over $K[X]/\langle X^2-1\rangle $, distinct if $char(K)\neq 2$. Meaning $K[X]/\langle X^2-1\rangle $ is never a field.

In other words $x^2-1$ over $\mathbb{R}(j)$ has at least four roots: $-1$, $1$, $-j$ and $j$.

More generally, if $K$ is a field and $p(X)\in K[X]$ is a polynomial, then $K[X]/\langle p(X)\rangle$ is an algebra over $K$ of dimension $\deg(p(X))$. In particular if $p(X)$ is of degree at least two, then $K[X]/\langle p(X)\rangle$ is a proper extension of $K$. And it is a field if and only if $p(X)$ is an irreducible polynomial.

Answer 2

Quick notation discussion

Given a ring containing a field $F$ and an element $\alpha$,

$F[\alpha]$ is the smallest ring containing both $\alpha$ and $F$ (which sounds a lot like the polynomial ring $F[x]$, but $x$ is an indeterminate not an element of the field containing $F$)

If $\alpha$ and $F$ are both contained in a field then $F(\alpha)$ is the smallest field containing both $\alpha$ and $F$. If $\alpha$ does not reside in a common field with $F$ then this may not exist.

When $\alpha$ is an element of an algebraic extension of fields over $F$ we have $F[\alpha]=F(\alpha)$ but in general they may be different.

The questions in the post

So is there some notion in which $\mathbb R(j)$ is equivalent to $\mathbb R$, or more generally, that $F(a)$ is equivalent to $F$ if all roots of $p$ are already in $F$?

I'm going to replace $\mathbb R(j)$ with $\mathbb R[j]$ because (using the notation above) there is no field containing $j$ and $\mathbb R$. Any field containing $j$ would have four roots for $x^2-1$, which is not possible (only two are possible.)

Now indeed for any $\alpha\in \mathbb R$, one would have $\mathbb R[\alpha]=\mathbb R(\alpha)=\mathbb R$. But because you are introducing $x$ and passing to the quotient like this: $\mathbb R[x]/(x^2-1)$, $x$ becomes something necessarily new outside of $\mathbb R$. The only way it would become equivalent to something inside the image of $\mathbb R$ is if the polynomial down there were linear (like $ax+b$, $a,b\in\mathbb R$.)

Is $\mathbb R[x]/(x^2-1)$ a proper extension of $\mathbb R$?

Now, what is $j$? Well you have defined it as the image of $x$ in $\mathbb R[x]/(x^2-1)$. We already consider $\mathbb R$ as its own image in this polynomial ring, and one can even say that $1,j$ is a basis for $\mathbb R[j]$ so that it is two dimensional. Although $\mathbb R$ already contained all solutions to $x^2-1$ in $\mathbb R$ it did not contain the ones you introduced that were also solutions in $\mathbb R[j]$.

By introducing $x$ and passing to the quotient ring you have introduced a new solution to the equation $x^2-1=0$ in a ring containing $F$, in addition to any solutions that used to be there.

Given the discussion above, most people would consider it a proper extension, since it is $2$ dimensional as an $\mathbb R$ vector space and $\mathbb R$ is only one dimensional, and where it used to be a field it's now grown to something larger that is not a field.

Are the split-complex numbers a proper algebraic extension of the real numbers?

Normally "algebraic extension" refers to field extensions, and $R[j]$ is not a field. For more information, see this post