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To solve the Dummit-Foote's exercise I've stuck here with this problem : (P-312)

Qn11. Show that $p(x)=x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$.

This is a polynomial of degree 2. But I can not think of zeros of this polynomial.

My another approach is : $p(x)=x^2-1\in \mathbb{Q}[x,y]/\langle y\rangle$ and also $p(x)=y^2-1\in \mathbb{Q}[x,y]/\langle x\rangle$ and $\langle x\rangle$, $\langle y\rangle$ are proper ideal of $\mathbb{Q}[x,y]$. Also, $p(x)=x^2-1$ can be factored with $(x+1)(x-1)$ in $\mathbb{Q}[x,y]/\langle y\rangle$. Then how I go ahead?

I think my attempt is stupid. Can any one suggest any references to understand/solve this?

Thanks in advance.

user3482749
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MAS
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3 Answers3

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Hint: consider $x^2+y^2-1\in\mathbb{Q}[y][x]$ and use a well-known irreducibility criterion holding over every PID.

egreg
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  • And what is that powerful irreducibility criterion that works over any PID? I'm asking just for my own learning. And does it work over any $R[x_1,\cdots,x_n]$ where $R$ is a PID? – stressed out Feb 14 '19 at 15:43
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    @stressedout Eisenstein's criterion. Only on a PID. – egreg Feb 14 '19 at 16:30
  • Thanks. The Wikipedia link doesn't discuss its generalization to PIDs, but it says something about generalizing it to $D[x]$ where $D$ is an integral domain. I hope I'm not bothering you by asking too many questions, but could you elaborate further? And the statement is for single variable polynomials. How does it apply to multivariable polynomials? – stressed out Feb 14 '19 at 16:43
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Just use the definition of irreducible element. Suppose $$ x^2+y^2-1 = p(x,y)\cdot q(x,y), $$ with $p,q\in \mathbb{Q}[x,y]$, so that $2 = \deg_x p + \deg_x q$ and $2 = \deg_y p + \deg_y q$. The only case to consider is when both $p$ and $q$ are linear in $x$ and $y$. We can write them explicitely and reach a contradiction when imposing that their product must be equal to $x^2+y^2-1$.

An approach which is not brute force is to look at $x^2+y^2-1$ as a polynomial in $\mathbb{Q}[y][x]$ and use one irreducibility criterion.

Alessio Del Vigna
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Hint: if a polynomial $p(x,y)$ over $K$ (any field, or in fact any domain) is reducible, then either for all $k\in K$, the polynomials $p(x,k)$ are reducible or for all $k\in K$, the polynomials $p(k,y)$ are reducible. This is clearly not true for $K=\mathbf Q$ and your $p$.

tomasz
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