Slick proof the determinant is an irreducible polynomial

Question

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.

Answer 1

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!

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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.

Answer 2

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

Answer 3

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 $k=1$.

Answer 4

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.