Determinant of symmetric matrix is an irreducible polynomial

This can also be done in a similar way as described in the post linked in the question for general square matrices. Letting $f\in\mathbb F[x_{ij}]_{1\le i\le j\le n}$ be the determinant, suppose that it decomposes as $f=gh$. We need to show that $g$ or $h$ is constant. This follows from the following properties of the determinant (when expressed as a linear combination of monomials in the $x_{ij}$),

1. For each $i=1,\ldots,n$, $f$ is first order in $x_{ii}$.
2. For each $1\le i < j\le n$, $f$ contains terms containing $x_{ij}$ but does not contain any terms containing $x_{ii}x_{ij}$ or $x_{jj}x_{ij}$.

I will make use of the following simple statements about factorisation in polynomial rings. (i) If a linear term in $k[x]$ factorizes into a product of two polynomials, then one is linear in $x$ and the other is independent of $x$. (ii) If $f=gh$ for $g\in k[x]\setminus k$ and $h\in k[y]\setminus k$ then $fg$ contains monomial terms containing $xy$.

Now, for the argument that if $f=gh$ then either $g$ or $h$ is constant. By property (1) above, for each $i$, either $g$ is first order in $x_{ii}$ and $h$ is independent of $x_{ii}$, or vice-versa.

Choose any $1\le i < j\le n$ such that $g$ is linear in $x_{ii}$. Then:

• $h$ is independent of $x_{ii}$.
• $h$ is also independent of $x_{ij}$ otherwise $gh$ will contain terms containing $x_{ii}x_{ij}$, contradicting (2).
• $g$ depends on $x_{ij}$ otherwise $gh$ would be independent of $x_{ij}$, contradicting (2).
• $g$ is linear in $x_{jj}$ otherwise $h$ would be linear in $x_{jj}$ and $gh$ would contain terms containing $x_{jj}x_{ij}$, contradicting (2).

Supposing that $g$ is linear in $x_{11}$ (wlog) the above argument can be applied to each $i < j$ to show that $h$ is independent of all the indeterminates $x_{ii},x_{jj},x_{ij}$ and, hence, is constant.