A polynomial $p$ over a field $k$ is called irreducible if $p=fg$ for polynomials $f,g$ implies $f$ or $g$ are constant. One can consider the determinant of an $n\times n$ matrix to be a polynomial in $n^2$ variables. Does anyone know of a slick way to prove this polynomial is irreducible?

It feels like this should follow quite easily from basic properties of the determinant or an induction argument, but I cannot think of a nice proof. One consequence of this fact is that $GL_n$ is the complement of a hypersurface in $M_{n}$. Thanks.

  • 3
    $\begingroup$ 1) Can you represent a determinant in terms of smaller determinants? 2) Actually, $\mathrm{GL}_n$ is the complement to a hypersurface, although it can be made one in higher dimension. $\endgroup$ – Alexei Averchenko May 19 '12 at 15:52
  • $\begingroup$ Thanks for the hypersurface remark, it has been corrected. You can of course represent the determinant in terms of smaller determinants, but given something like $det_n = x_{11}det_{n-1,11} + ... + x_{1n}det_{n-1,1n}$ (with obvious notation used) it's not clear how to use this to show irreducibility as the $det_{n-1,1i}$ contain many of the same variables. $\endgroup$ – user31559 May 19 '12 at 15:58
  • $\begingroup$ Is slick a mathematical term I'm not aware of? $\endgroup$ – corsiKa May 19 '12 at 20:04
  • $\begingroup$ I meant slick as in 'elegant' or 'nice'. Perhaps it's not a common phrase universally. For example, one can interpret the determinant as a change in volume, so perhaps one could prove the irreducibility using some geometric argument. $\endgroup$ – user31559 May 20 '12 at 0:25

Deonote $p$ the determinant polyonomial. Observing that $p$ is of degree one in $x_{ij}$ for every $(i,j)$.

Now we can prove $p$ is irreducible. Suppose $p=fg$. Consider $x_{11}$, suppose $x_{11}$ appears in $f$, then $f$ is of degree one in $x_{11}$ and $g$ is of degree zero in $x_{11}$. Now consider $x_{1j}$, then $x_{1j}$ must appear in $f$, otherwise $g$ is of degree one in $x_{1j}$ and $f$ is degree zero in $x_{1j}$, then $fg=(ax_{11}+b)(cx_{1j}+d)=acx_{11}x_{1j}+bcx_{1j}+adx_{11}+bd\in k[\ldots][x_{11},x_{1j}]$, contradiction. So all $x_{1j}$ in $f$ for $j=1,\ldots,n$. Similar $x_{j1}$ are all in $f$. And since $x_{j1}$ is in $f$, it follows $x_{jk}$ are in $f$. Finally, all $x_{ij}$ are in $f$. And $g$ is a constant. We are done!

Edit: Contradiction: view $p$ be a polynomial of $x_{11},x_{1j}$, then $p=x_{11}h_1+x_{1j}h_2+h_3\in k[\ldots][x_{11},x_{1j}]$, where $h_1,h_2,h_3 \in k[\{x_{ij}\}\mid x_{ij}\neq x_{11},x_{1j}]$, i.e., they are "constant" about $x_{11},x_{1j}$, but $fg=acx_{11}x_{1j}+bcx_{1j}+adx_{11}+bd$, while $0\neq ac \in k[\{x_{ij}\}\mid x_{ij}\neq x_{11},x_{1j}]$ and $bc,ad,bd$ are "constant" about $x_{11},x_{1j}$(all the results come from the assumption $f$ is a polynomila of degree one in $x_{11}$ and of degree zero in $x_{1j}$ and $g$ is of degree one in $x_{1j}$ and of degree zero in $x_{11}$), so $p$ cannot equal to $fg$.

  • $\begingroup$ Dear wxu, could you explain why there is a contradiction ? $\endgroup$ – user18119 May 19 '12 at 19:39
  • $\begingroup$ Thanks! I'll accept your answer in a few days if no one else has a more elegant proof (your proof is quite nice). $\endgroup$ – user31559 May 20 '12 at 0:26
  • $\begingroup$ @wxu: I see. You mean $p$ has total degree one when viewed as polynomial in $x_{11}, x_{1j}$. Nice proof ! $\endgroup$ – user18119 May 20 '12 at 6:35
  • $\begingroup$ this is a nice proof which also help me this the problem that I posted today, thanks a lot.@wxu $\endgroup$ – user53800 Dec 27 '12 at 21:46

Here is a proof using a little bit of algebraic geometry (we assume k is algebrically closed). It can be seen that $V(det)$ is irreducible in $k^{n^2}$. It is for example the image of the morphism $M_n(k)\times M_n(k) \rightarrow V(det)$, $(P,Q)\mapsto PI_{n-1}Q$ and the source is irreducible.

Hence as $k[X_{ij}]$ is a UFD we have $I(V(det)) = \sqrt{(p)}$ where p is an irreducible polynomial. So $det = p^k$.

To show that $k=1$ we use the fact that the differential of the determinant at a matrice A is $D(det)_A(.) = Tr(Adj(A).)$ where Adj(A) is the adjugate matrix. For $A$ such that $rk(A)=n-1$ we have $rk(Adj(A))=1$ so $D(det)_A(.)\neq 0$ which forces $k=1$.


This answer is basically a proof from M.Bocher "Introduction to higher algebra" (Dover 2004) on pages 176-7.

  • $\begingroup$ "Above" does not really mean anything here, as the answers are sorted in various ways! $\endgroup$ – Mariano Suárez-Álvarez Jan 29 '18 at 22:28
  • $\begingroup$ it's better now :-) $\endgroup$ – Dima Pasechnik Jan 29 '18 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.