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So the question goes: Let $$ A=\mathbb{R}[x]/\langle x^2-x+1\rangle . $$ Is A isomorphic to $ \mathbb{C} $ ?

The earlier parts of the question asked for me to a) find the reciprocal in A of $x+1+I$ and b) find $p(x)+I\in A$ such that $(p(x)+I)^2=-1+I$. I found the answers to both of these parts, but I cannot understand whether or not A would be isomorphic to the complex numbers. I want to say no because in part b) I found $p(x)=\sqrt {x^2-x}$ and this mimics how the complex numbers act, but it is not one-to-one, so I think it would be a homomorphism, but not an isomorphism. Is this the right way to look at this or am I misinterpreting something?

MJD
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Thomas L
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4 Answers4

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Map $\,x\in\Bbb R[x]\,$ to a root $\,\alpha$ of $\,f(x)= x^2\!-x+1\,$ in $\Bbb C.\,$ By $\,\alpha\not\in \Bbb R\,$ it is onto and has kernel $= (f).$

Bill Dubuque
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It would be indeed, since what we see is that the roots of $x^2-x+1=0$ are $\frac{1}{2}+\frac{\sqrt3}{2}i$ and $\frac{1}{2}-\frac{\sqrt3}{2}i$ let us take the "+" root.

Since the roots are not in $\mathbb{R}$ the polynomial is irreducible in $\mathbb{R}[x]$, now by taking $A=\mathbb{R}[x]/\langle x^2-x+1\rangle$ we look at $A$ as the set $$\left\{f\in\mathbb{R}[x]\middle\lvert x^2-x+1=0\right\} \\=\left\{f\in\mathbb{R}[x]\middle\lvert x=\frac{1}{2}+\frac{\sqrt3}{2}i\right\}\\=\left\{a+b\left(\frac{1}{2}+\frac{\sqrt3}{2}i\right)\middle \lvert a,b\in \mathbb{R}\right\} $$ which is isomorphic to $\mathbb{C}$.

We see that $\left\{a+b(\frac{1}{2}+\frac{\sqrt3}{2}i)\mid a,b\in \mathbb{R}\right\}$ is isomorphic to $\mathbb{C}$ by letting $b=\frac{2d}{\sqrt3}$ for some $d\in \mathbb{R}$, then letting $a=c-\frac{b}{2}$, for some $c\in \mathbb{R}$. So we have $A=\left\{a+b(\frac{1}{2}+\frac{\sqrt3}{2}i)\mid a,b\in \mathbb{R}\right\}=\{c+di\mid c,d\in \mathbb{R}\}$ which is isomorphic (or really even equivalent) to $\mathbb{C}$.

MJD
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Ellya
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  • I'm sorry, but I'm not understanding still. So we take the roots of the equation and use the positive root. We see they aren't in $\mathbb{R}$. I got that, but the following part is going over my head. We find the function f but is the expression $a+b(\frac{1}{2} + \frac{\sqrt3}{2}i)$ isomorphic to $\mathbb{C}$ because it is of the form a + bi? – Thomas L Mar 30 '14 at 20:10
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    indeed, but in the first bit, we are looking at functions $f\in \mathbb{R}[x]$, applying the fact that $x=\frac{1}{2}+\frac{\sqrt3}{2}i$ and seeing what happens. – Ellya Mar 30 '14 at 20:11
  • So was part b) of my answer also incorrect? Because it wants the square root of -1 would that mean it isn't ${\sqrt(x^2-x)}$? – Thomas L Mar 30 '14 at 20:15
  • I'm not sure, what is the $I$ here? – Ellya Mar 30 '14 at 20:19
  • $I$ is some identity element in the ring. – Thomas L Mar 30 '14 at 20:22
  • Thinking logically, the additive identity in the quotient ring is $x^2-x+1$ so we seek $p(x)$ s.t. $(p(x)+x^2-x+1)^2=x^2-x=-1$ so take $p(x)+x^2-x+1=\sqrt{-1}$ i.e. $p(x)=i$, but don't take this as gospel, this is off the top of my head. Also note that any multiple of $I$ is an identity element, so we could have $p(x)=ai$ for any $a\in\mathbb{R}$. – Ellya Mar 30 '14 at 20:30
  • Taking this into account we have $p(x)+I\in A$, now if $p(x)=ai$, and that holds for any $a\in \mathbb{R}$, then A mimics the structure of $\mathbb{C}$, and I think that the relationship is one to one. – Ellya Mar 30 '14 at 20:32
  • Very nicely written – Numbersandsoon Mar 31 '14 at 19:24
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Suppose $F$ is a proper algebraic extension of $\mathbb{R}$. Take $\alpha\in F$, $\alpha\notin\mathbb{R}$. Then $\alpha$ satisfies a quadratic polynomial over $\mathbb{R}$, say $f(x)=x^2+ax+b$, with $a^2-4b<0$, because these are the only non linear irreducible polynomials. If $\beta$ is the other root of $f$ (note that $\beta\in F$), we have $$ (\alpha-\beta)^2=a^2-4b $$ so $$ \left(\frac{\alpha-\beta}{\sqrt{4b-a^2}}\right)^2=-1 $$ Therefore there is an element $\gamma\in F$ such that $\gamma^2=-1$ and $\mathbb{R}[\gamma]$ is isomorphic to $\mathbb{C}$. Finally, $F$ is an algebraic extension of $\mathbb{R}[\gamma]$ which is algebraically closed, so $F=\mathbb{R}[\gamma]$.

egreg
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You know from whatever method you prefer (quadratic formula, calculus, etc.) that $x^2 - x + 1$ has no real roots. Because this is a degree 2 polynomial, that implies that this polynomial is irreducible over $\mathbb{R}$. Hence $\mathbb{R}[x]/(x^2 - x + 1)$ is a field extension of degree 2 over $\mathbb{R}$. But all quadratic extensions (degree 2) of a fixed field are isomorphic (prove it!). Hence this "new" field you found is isomorphic to your old familiar degree 2 extension over $\mathbb{R}$, namely $\mathbb{C} \cong \mathbb{R}/(x^2+1)$.

Supersingularity
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