Prove (or disprove) that a polynomial ring modulo an ideal is a field

Question

I have the following problem I'm working on: Prove or disprove that $\mathbb{R}[x,y]/\langle{x^2+y^2}\rangle$ is a field.

I know that this amounts to asking whether or not the polynomial $x^2 + y^2$ is irreducible in $\mathbb{R}[x,y]$ - if it is irreducible, then the ideal generated by the polynomial $x^2 + y^2$ is maximal in $\mathbb{R}[x,y]$, yielding that the quotient ring $\mathbb{R}[x,y]/\langle{x^2+y^2}\rangle$ is a field, whereas if it is reducible, the quotient ring has zero divisors and is not even an integral domain ($\Rightarrow$ it cannot be a field).

To this point, suppose it is in fact reducible. Then it must be the product of two linear polynomials in the variables $x$ and $y$

$\Rightarrow$ $x^2 + y^2 = (ax + by + c)(dx + ey + f)$ , where $a,b,c,d,e,f \in \mathbb{R}$

$\Rightarrow$ $x^2 + y^2 = adx^2 + dexy + afx + bdxy + bey^2 + bfy + cdx + cey + cf$

$\Rightarrow$ $ad = 1, be = 1, ae+bd = 0, af+cd = 0, bf + ce = 0, cf=0$.

From the resulting equations, one can try and get a contradiction to $x^2 + y^2$ being a product of two linear polynomials in the variables $x$ and $y$, or try and get a solution for the real coefficients that will make $x^2 + y^2$ factor.

I tried to achieve the former. If $cf = 0$, then $c = 0$ or $f = 0$ (using that $\mathbb{R}$ is an integral domain). If $c = 0$ and $f \neq 0$, then $bf = 0$ $\Rightarrow$ $b = 0$, yielding a contradiction to $be = 1$. If $c \neq 0$ and $f = 0$, $ce = 0$ $\Rightarrow$ $e = 0$, yielding a contradiction to $be = 1$. However, in the case that both $c = 0$ and $f = 0$, I'm not able to reach such a contradiction. Is my logic so far sound? How can I refine my argument from here?

Thanks!

Your problem is not tantamount to showing that $x^2+y^2$ is irreducible. In fact, no principal ideal can be maximal in the polynomial ring $k[x_1,\cdots, x_n]$ over the field $k$ for any $n\ge 2$. — , Oct 16 '19 at 19:01
$x^2+y^2$ is indeed irreducible in $\mathbf R[x,y]$, and this shows the ideal $(x^2+y^2)$ is prime since we're in a U.F.D. Hence the quotient is an integral domain. — Bernard, Oct 16 '19 at 19:29
Actually, $\mathbb{R}[x,y]/\langle{x^2+y^2}\rangle \cong \mathbb{C}[t]$ by $x \mapsto t, y \mapsto it$. — lhf, Oct 16 '19 at 22:40

score 5 · Accepted Answer · answered Oct 16 '19 at 19:04

5

You don't have to show (ir)reducibility because we work over a polynomial ring in strictly more than one variable.

Observe that $$(X^2 + Y^2) \subsetneq (X,Y)$$

and hence your ideal is not maximal. It follows that the quotient ring is not a field.

answered Oct 16 '19 at 19:04

J. De Ro

21,438

2

Hi EpsilonDelta. Could you explain why this proper inclusion holds? How can one see that $(X,Y)$ is not in fact the entire ring $\R[x,y]$ ? I'm trying to unpack the fact that our ideal is not maximal. – michiganbiker898 Oct 16 '19 at 22:57
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@mich See my answer for an simple way via evaluation (vs. more common degree arguments), and compare here. – Bill Dubuque Oct 17 '19 at 02:44
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For example, $X\notin (X^2+Y^2)$ (use a degree argument). – J. De Ro Oct 17 '19 at 05:14
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And $(X,Y)$ is not the entire ring because it does not contain constants. – J. De Ro Oct 17 '19 at 06:52

Bill Dubuque · Answer 2 · 2019-10-17T02:49:18.640

If suffices to show that $x$ is nonzero and noninvertible in the quotient ring $R$

If $x$ is invertible in $R\,$ then in $\Bbb R[x,y]\!:\ xf = 1 +(x^2\!+\!y^2)g\,$ so $\,0 = 1\,$ via eval at $\,x = 0 = y$

Similarly if $\,x=0\,$ in $R\,$ then in $\Bbb R[x,y]$ we have

$\ x = (x^2\!+\!y^2)f(x,y) \overset{\large y\ =\ x}\Rightarrow\ x = 2x^2f(x,x)$ $\overset{{\rm cancel}\ x} \Rightarrow 1 = 2xf(x,x)\,$ so $\,1 = 0\ $ via eval at $\,x=0$

Prove (or disprove) that a polynomial ring modulo an ideal is a field

2 Answers2