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Let $B = \mathbb R[x,y]$ where $x^2 + y^2 = 1$ which is called the coordinate ring of the unit circle.

I am trying to prove that $y$ is irreducible in $B.$

I have the following information about $B$:

1- $B$ is an integral domain.

2- $B = \mathbb R[x] \oplus \mathbb R[x]y$ and multiplication by $y$ is a map of $B$ as an $\mathbb R[x]$-module. And $y$ is integral over $\mathbb R[x]$ and so there is a ring inclusion $B \rightarrow M_2(\mathbb R[x]),$ called a representation of $B$ described below:

$$\phi:B \rightarrow M_{2}(\mathbb R[x]), \phi (a_{1} + a_{2}y) = \begin{pmatrix} a_{1} & a_{2}(1-x^2) \\ a_{2} & a_{1} \end{pmatrix} $$

3- Also, I know that $B^* = \mathbb R^*$

I am trying to use Eisenstein criterion for showing that $y$ is irreducible.

Here is my try:

We need to find a prime $p$ such that $p$ divides $0$ because we have only 1 coefficient which is the leading coefficient and every other coefficients are $0$ and any prime can divide zero trivially. Also any prime does not divide 1. But now I do not know why $p^2$ does not divide $a_0$ which is zero in our case ...I need $p^2$ to divide $a_0$ to fullfill the conditions for Eisenstein Criterion. Could anyone explain this to me please?

Also, is there any use for the fact that $y$ is integral in this proof?

user26857
  • 52,094
  • What about my question about Eisenstein criterion? –  Feb 25 '21 at 17:42
  • $y$ is not a polynomial! Then how to use Eisenstein? – user26857 Feb 25 '21 at 23:07
  • @user26857 $y$ is a polynomial so we can use Eisenstein ..... correct? –  Feb 25 '21 at 23:08
  • You seem to be very confused. The ring $B$ is a quotient of a polynomial ring, that is, $B=\mathbb R[X,Y]/(X^2+Y^2-1)$, and $y$ is the residue class of the indeterminate $Y$ modulo the ideal $(X^2+Y^2-1)$. – user26857 Feb 25 '21 at 23:11
  • I am sorry I was just asking. so we can use Eisenstein for showing that $X^2 + Y^2 -1$ is irreducible but not if we have little variables ..... correct @user26857 –  Feb 25 '21 at 23:13
  • No idea what are "little variables", but in this case Eisenstein can't be used. – user26857 Feb 25 '21 at 23:15
  • @user26857 I meant $x,y$ small variables(residue classes) –  Feb 25 '21 at 23:17

1 Answers1

1

Define $N:B\to\mathbb R[x]$ by $N(a_1+b_1y)=\det\phi(a_1+b_1y)$. Notice that $N$ behaves like a norm!

Let $z_i=a_i+b_iy$, $i=1,2$, such that $y=z_1z_2$. Then $N(y)=N(z_1)N(z_2)\Leftrightarrow x^2-1=N(z_1)N(z_2)$. Now we have the following cases:
(i) $\deg N(z_1)=0\Leftrightarrow z_1\in\mathbb R^*$,
(ii) $\deg N(z_1)=2$ $\Leftrightarrow$ $\deg N(z_2)=0$ $\Leftrightarrow$ $z_2\in\mathbb R^*$,
(iii) $\deg N(z_1)=\deg N(z_2)=1$. If $N(z_1)=x-1$, then $a_1^2(x)+b_1^2(x)(x^2-1)=x-1\Rightarrow x-1\mid a_1(x)\Rightarrow\exists a'_1\in\mathbb R[x]$ such that $a_1=(x-1)a'_1$ and plugging this in the foregoing equation we get $a'_1(x)^2(x-1)+b_1^2(x)(x+1)=1$. Looking now at the dominant coefficients of $a'_1$ and $b_1$ we find that one of these is zero (false!) or the sum of their square is zero (false!).

user26857
  • 52,094
  • I borrowed this technique from https://math.stackexchange.com/a/244487/121097 – user26857 Feb 25 '21 at 23:06
  • I guess in the fourth line from below there is a typo and it should be $x-1 | a_1(x)^2$ –  Feb 26 '21 at 00:31
  • what do you mean by the dominant coefficients?I do not understand why one of them is zero false and why sum of their square is zero false .... could you explain that please? –  Feb 26 '21 at 00:37
  • @mathmusic 1) $x-1$ is irreducible (it is a polynomial!) and as you probably know this means that it is also prime, and a prime $p\mid a^2\implies p\mid a$. 2) Dominant=Leading; in the given relation if $\deg a'_1\ne\deg b_1$, then the left hand side (LHS) has as the leading coefficient the square of one of the leading coefficients of $a'_1$, respectively $b_1$. If $\deg a'_1=\deg b_1$, then the leading coefficient of LHS is the sum of the squares of the leading coefficients of $a'_1$ and $b_1$. But the polynomial in LHS has degree $\ge 1$, and then its leading coefficient has to be $0$. – user26857 Feb 26 '21 at 09:00