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Proving that $(ac - (b^2 + 1))$ is irreducible in $k[a,b,c].$

Here is my trial:

I usually work by degrees and taking the lower degree of my polynomial, but the thing here is that I have two variables of the same degree $a,c$ and even they are of degree $1.$ So I am a little bit confused in my case here.

Assume that $f(a,b,c) = ac - (b^2 + 1) = ac - b^2 -1 = g(a,b,c)h(a,b,c),$ we will look at the ring $k[a,b,c]$ as $k[a,b][c]$ i.e., a polynomial in $c$ whose coefficients come from $k[a,b].$ In this ring $f(a,b,c)$ is irreducible because $a,b^2 + 1$ are units and $f(a,b,c)$ is a linear polynomial. So either $g$ or $h$is a unit in $k[a,b][c].$

Without loss of generality, suppose $g(a,b,c)$ is a unit in $k[a,b][c]$. But then the only units there are elements of $k[a,b]$, so, $g(a,b,c) = e(a,b)$ for some polynomial $e(a,b) \in k[a,b].$

So, $f(a,b,c) = e(a,b)h(a,b,c).$ But then this means that $e(a,b)$ is a common divisor of $a,b^2 +1,$ but they are polynomials of different variables.

But then I do not know how to complete. Any help will be greatly appreciated!

I think it is not clear in my mind what should I do with this $e(a,b).$ Should I just say ok, from here $g$ is a unit in $k[x,y,z]$ and I am done? I know that I would have by this that $h(a,b,c)$ is a linear polynomial in $c, $ but how that say that $f$ is irreducible in $k[a,b,c]$?

EDIT:

A quick question, should I write $k[a,b][c]$ or $k(a,b)[c]$? which is more accurate?

Eric Wofsey
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1 Answers1

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Note that we can write $h(a,b,c)=h_n(a,b)c^n + \ldots + h_0(a,b)$ for some $n$ where each $h_i(a,b)$ are polynomials in $a,b$. From $ac-(b^2+1)=h(a,b,c)e(a,b)$ we have that $$h_1(a,b)e(a,b)=a, \quad h_0(a,b)e(a,b)=b^2+1 \ \text{ and } h_i(a,b)=0, \forall i\ge 2$$

Now use similar argument as you did with $g$, we see that $h_1(a,b)e(a,b)=a$ implies $h_1$ or $e$ is constant. If $e$ is constant then we are done. If $h_1(a,b)=k$ is constant then $e(a,b)=\dfrac{a}{k}$, hence, $h_0(a,b)\dfrac{a}{k}=b^2+1$ (contradiction).

mathmathmath
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  • To answer your quick question, when you say $a,b^2+1$ are units, you have to view them as elements in $k(a,b)$. – mathmathmath Apr 12 '21 at 00:43
  • you mean the field of rational functions in $2$ variables ? not the polynomial ring in 2 variables? –  Apr 12 '21 at 00:45
  • So until which line my argument make sense? –  Apr 12 '21 at 00:46
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    That's right. In polynomial ring $k[a,b]$, $a,b^2+1$ are not units. – mathmathmath Apr 12 '21 at 00:46
  • Ok. I see yourpoint. –  Apr 12 '21 at 00:47
  • Just change to "either $g$ or $h$ is a unit in $k(a,b)[c]$". – mathmathmath Apr 12 '21 at 00:47
  • No I do not mean my argument about the quick question ...... I mean my proof of the whole question until which line it is valid? –  Apr 12 '21 at 00:49
  • I meant the change from your proof. – mathmathmath Apr 12 '21 at 00:50
  • Ultimately, Do we want to show that $g$ is a unit in $k[a,b,c]$ not just a unit in $k[a,b]$? –  Apr 12 '21 at 00:50
  • Oh so you are saying that my all argument in the proof is fine but I should change what you mentioned? –  Apr 12 '21 at 00:52
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    Units in $k[a,b,c]$ are precisely constant. You want units in $k(a,b)[c]$, which are elements in $k(a,b)$ (since it is a field). So $g\in k(a,b)$, but you have assumed $g$ is a polynomial so $g\in k[a,b]$. – mathmathmath Apr 12 '21 at 00:52
  • Perfert! did you mean to say $h_i[a,b]$ not $h_i(a,b)$? –  Apr 12 '21 at 00:55
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    My $h_i$ are polynomials so $h_i(a,b)$ are just notations for polynomials. They have nothing to do with the field and so on. – mathmathmath Apr 12 '21 at 00:56
  • I do not see the contradiction in your last statement.$h_0 (a,b) = k(b^2 +1 )/a$ does not give any contradiction. –  Apr 12 '21 at 03:36
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    $h_0(a,b)$ is a POLYNOMIAL in a and b. – mathmathmath Apr 12 '21 at 04:16
  • Got you! sorry about that Should we say $e = 1 or h_1 = 1$ instead of saying that they are constants? –  Apr 12 '21 at 04:20
  • Also, why we need to "Now use similar argument as you did with g" ?does not we have just the polynomial a in that case, so everything is easy? –  Apr 12 '21 at 04:22
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    @mathmusic More conceptually the answer shows that a linear polynomial is reducible iff it is not primitive, i.e. its coef's have a nonunit common divisor (which is easily ruled out in OP since the lead coef $a$ is prime (so irreducible) and it doesn't divide the other coef, i.e. $,a\nmid b^2+1$ (else $0\mid 1$ by eval at $,a=0=b)\ \ $ – Bill Dubuque Apr 12 '21 at 20:23
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    @mathmusic Nore generally we can show it is prime (so irreducible) by this theorem. – Bill Dubuque Apr 12 '21 at 20:31
  • Thank you :) @BillDubuque –  Apr 20 '21 at 23:06