# Slick proof the determinant is an irreducible polynomial

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.

• 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. May 19, 2012 at 15:52
• 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. May 19, 2012 at 15:58
• Is slick a mathematical term I'm not aware of? May 19, 2012 at 20:04
• 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. May 20, 2012 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 the equality $$fg=(ax_{11}+b)(cx_{1j}+d)=acx_{11}x_{1j}+bcx_{1j}+adx_{11}+bd\in \mathbb{F}[x_{11}, x_{1j}, \dots]$$ leads to 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 $$p=x_{11}h_1+x_{1j}h_2+h_3\in \mathbb{F}[x_{11},x_{1j}, \dots],$$ where $$h_1,h_2,h_3 \in \mathbb{F}[\{x_{ij}\}\mid x_{ij}\neq x_{11},x_{1j}]$$, i.e., they are "constant" about $$x_{11},x_{1j}$$, but $$fg=acx_{11}x_{1j}+bcx_{1j}+adx_{11}+bd,$$ while $$0\neq ac \in \mathbb{F}[\{x_{ij}\}\mid x_{ij}\neq x_{11},x_{1j}]$$ and $$bc,ad,bd$$ are "constant" about $$x_{11},x_{1j}$$(all the results come from the assumption $$f$$ is a polynomial 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 $$fg$$ since the definition of the determinant.

• Dear wxu, could you explain why there is a contradiction ?
– user18119
May 19, 2012 at 19:39
• Thanks! I'll accept your answer in a few days if no one else has a more elegant proof (your proof is quite nice). May 20, 2012 at 0:26
• @wxu: I see. You mean $p$ has total degree one when viewed as polynomial in $x_{11}, x_{1j}$. Nice proof !
– user18119
May 20, 2012 at 6:35
• this is a nice proof which also help me this the problem that I posted today, thanks a lot.@wxu Dec 27, 2012 at 21:46
• Why can't $g$ contain $x_{11}$? Perhaps the product of the terms in $f$ and $g$ containing $x_{11}$ gets cancelled out by another such product.
– Tri
Oct 14, 2021 at 6:57

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

• "Above" does not really mean anything here, as the answers are sorted in various ways! Jan 29, 2018 at 22:28
• it's better now :-) Jan 29, 2018 at 23:17

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 $k=1$.

• How can you multiply $P$ and $I_{n-1}$, since these matrices have different sizes? Jan 27 at 17:45

Here is another algebraic argument which may be of interest, using induction on the size of the matrix:

Let $$A_{ij}$$ be the minor we get when we delete row $$i$$ and column $$j$$. If we expand along the first row of the matrix, we get $$\det A=(\det A_{11})x_{11}-(\det A_{12})x_{12}+\cdots\pm(\det A_{1n})x_{1n}.$$ None of the determinants $$\det A_{1i}$$ contains any variable $$x_{1i}$$, so this is a linear polynomial in the variables $$x_{1i}$$. Therefore it can only factor if the coefficients of the $$x_{1i}$$ have a common factor. But these coefficients are determinants of size $$n-1$$ so by induction they are all irreducible; since they are not scalar multiples of each other they have no (non-constant) common factor.