How to show that $x$ is irreducible in $\frac{\mathbb R[x,y]}{(x^2+y^2-1)}$?

Question

How to show that $x$ is irreducible in $\frac{\mathbb R[x,y]}{(x^2+y^2-1)}$?

I can show that x is not prime But unable to show x is irreducible.

Please can any one help me to solve this problem ANy Help will be appreciated

Answer 1

To show that $x$--or more appropriately, the equivalence class $[x]$ of $x$ is irreducible, we set $$[x]=uv$$ and show that one of $u$ or $v$ is a unit.

First, we need to think about what $u$ and $v$ can look like. The elements of $\Bbb R[x,y]$ are real polynomials in $x$ and $y.$ In the quotient, we can still think of them that way, but since $x^2+y^2-1=0,$ then we can always choose a representative such that no power of $y$ is greater than $1,$ by making the substitution $y^2\mapsto 1-x^2.$

Thus, there exists real univariate polynomials $p_1,q_1,p_2,q_2$ such that $$u=\bigl[p_1(x)+q_1(x)y\bigr]\tag{1}$$ and $$v=\bigl[p_2(x)+q_2(x)y\bigr].\tag{2}$$ This means that we have

\begin{eqnarray}[x] &=& \bigl[p_1(x)+q_1(x)y\bigr]\bigl[p_2(x)+q_2(x)y\bigr]\\ &=& \Bigl[\bigl(p_1(x)+q_1(x)y\bigr)\bigl(p_2(x)+q_2(x)y\bigr)\Bigr]\\ &=& \bigl[p_1(x)p_2(x)+p_1(x)q_2(x)y+p_2(x)q_1(x)y+q_1(x)q_2(x)y^2\bigr]\\ &=& \bigl[p_1(x)p_2(x)\bigr]+\bigl[p_1(x)q_2(x)+p_2(x)q_1(x)\bigr][y]+\bigl[q_1(x)q_2(x)\bigr]\bigl[y^2\bigr]\\ &=& \bigl[p_1(x)p_2(x)\bigr]+\bigl[p_1(x)q_2(x)+p_2(x)q_1(x)\bigr][y]+\bigl[q_1(x)q_2(x)\bigr]\bigl[1-x^2\bigr]\\ &=& \bigl[p_1(x)p_2(x)+(1-x^2)q_1(x)q_2(x)\bigr]+\bigl[p_1(x)q_2(x)+p_2(x)q_1(x)\bigr][y]\\ &=& \Bigl[p_1(x)p_2(x)+\bigl(1-x^2\bigr)q_1(x)q_2(x)+\bigl(p_1(x)q_2(x)+p_2(x)q_1(x)\bigr)y\Bigr]\end{eqnarray}

Put another way, $$p_1(x)p_2(x)+\bigl(1-x^2\bigr)q_1(x)q_2(x)+\bigl(p_1(x)q_2(x)+p_2(x)q_1(x)\bigr)y-x$$ is an element of the ideal generated by $x^2+y^2-1.$ Clearly, this can only occur if $\bigl(p_1(x)q_2(x)+p_2(x)q_1(x)\bigr)y=0,$ so we need $$p_1(x)q_2(x)+p_2(x)q_1(x)=0.\tag{3}$$

Thus, we're left with $p_1(x)p_2(x)+\bigl(1-x^2\bigr)q_1(x)q_2(x)-x$ as an element of the ideal generated by $x^2+y^2-1,$ which can only happen if $$p_1(x)p_2(x)+\bigl(1-x^2\bigr)q_1(x)q_2(x)-x=0.$$ (Do you see why?) Thus, $$x=p_1(x)p_2(x)+\bigl(1-x^2\bigr)q_1(x)q_2(x),\tag{4}$$ which can only happen if one of the polynomials $q_i$ is the zero polynomial. Without loss of generality, suppose $q_2=0.$ As a result, we have: $$v=\bigl[p_2(x)\bigr],\tag{2'}$$ $$p_2(x)q_1(x)=0,\tag{3'}$$ $$x=p_1(x)p_2(x).\tag{4'}$$ Now, by $(3'),$ one of $p_2,q_1$ must be the zero polynomial, but by $(4'),$ $p_2$ can't be the zero polynomial, so $q_1=0.$ Thus, we have $$u=\bigl[p_1(x)\bigr].\tag{1'}$$ Furthermore, by $(4'),$ one of $p_1,p_2$ must be a non-zero constant polynomial. Without loss of generality, suppose $p_2(x)=c$ for all $x,$ and some nonzero real $c.$ Thus, $v=[c],$ so $v$ is a unit, whence $x$ is irreducible, as desired.