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(N.B.: I have modified this answer to take into account the comments below about the case of characteristic $2$, when in fact, the discriminant is the square of an irreducible polynomial.)

The discriminant is defined by the property that, when $$ p(x) = x^d - s_1 x^{d-1} + \cdots + (-1)^ds_d = (x-t_1)(x-t_2)\cdots(x-t_d) $$ (i.e., when one substitutes the $i$th elementary symmetric function of the $t_k$ for $s_i$), the discriminant becomes $$ D(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j)^2. $$ Suppose there were a factorization $D(s_1,\ldots,s_d) = D_1(s_1,\ldots,s_d)D_1(s_1,\ldots,s_d)$ where each $D_i$ had positive degree. Then by unique factorization, when one substitutes as above, one must be able to write, for $a = 1,2$, $$ D_a(s_1,\ldots,s_d) = c_a \prod_{(i,j)\in S_a} (t_i-t_j) $$ where $c_1$ and $c_2$ are (nonzero) constants and $S_1$ and $S_2$ are disjoint nonempty subsets of the set of pairs $(i,j)$ in $\{1,\ldots,d\}$ whose union is the entire set of distinct pairs in this set. Thus, for example, $D_1(s_1,\ldots,s_d)$ will vanish when $t_i=t_j$ (for $i\not=j)$ only if $(i,j)$ or $(j,i)$ belongs to $S_1$. Since one can't detect which of the $t_i$ are equal using only $s_1,\ldots, s_d$, it follows that $S_1$ must contain either $(i,j)$ or $(j,i)$ for each distinct pair. The same argument applied to $D_2$ shows that $S_2$ also must contain either $(i,j)$ or $(j,i)$ for each distinct pair. Thus, one must have, for $a = 1,2$, $$ D_a(s_1,\ldots,s_d) = c'_a \prod_{i<j} (t_i-t_j) $$ for some constants $c'_1$ and $c'_2$.

However, when the characteristic of the field is not $2$, one cannot have a polynomial $E(s_1,\ldots,s_d)$ such that, after substitution, one obtains $$ E(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j) $$ since the left hand side cannot detect permutations in the $t_i$, whereas the right hand side will change sign when one makes an odd permutation of the $t_i$. Thus, $D$ is irreducible when the characteristic of the field is not $2$.

As Jarek Kuben pointed out in the comments below, when the characteristic of the field is $2$, the expression $$ F = \prod_{i<j} (t_i-t_j) = \prod_{i<j} (t_i+t_j) $$ is symmetric in the $t_i$, so $F$ can be written as a polynomial in the $s_i$. One then has $D = F^2$, so $D$ is a square. The above argument shows, however, that $F$ must be irreducible, since any factorization $D = D_1D_2$ has to have $D_1 = D_2 = F$ (up to constant multiples).

(N.B.: I have modified this answer to take into account the comments below about the case of characteristic $2$, when in fact, the discriminant is the square of an irreducible polynomial.)

The discriminant is defined by the property that, when $$ p(x) = x^d - s_1 x^{d-1} + \cdots + (-1)^ds_d = (x-t_1)(x-t_2)\cdots(x-t_d) $$ (i.e., when one substitutes the $i$th elementary symmetric function of the $t_k$ for $s_i$), the discriminant becomes $$ D(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j)^2. $$ Suppose there were a factorization $D(s_1,\ldots,s_d) = D_1(s_1,\ldots,s_d)D_2(s_1,\ldots,s_d)$ where each $D_i$ had positive degree. Then by unique factorization, when one substitutes as above, one must be able to write, for $a = 1,2$, $$ D_a(s_1,\ldots,s_d) = c_a \prod_{(i,j)\in S_a} (t_i-t_j) $$ where $c_1$ and $c_2$ are (nonzero) constants and $S_1$ and $S_2$ are disjoint nonempty subsets of the set of pairs $(i,j)$ in $\{1,\ldots,d\}$ whose union is the entire set of distinct pairs in this set. Thus, for example, $D_1(s_1,\ldots,s_d)$ will vanish when $t_i=t_j$ (for $i\not=j)$ only if $(i,j)$ or $(j,i)$ belongs to $S_1$. Since one can't detect which of the $t_i$ are equal using only $s_1,\ldots, s_d$, it follows that $S_1$ must contain either $(i,j)$ or $(j,i)$ for each distinct pair. The same argument applied to $D_2$ shows that $S_2$ also must contain either $(i,j)$ or $(j,i)$ for each distinct pair. Thus, one must have, for $a = 1,2$, $$ D_a(s_1,\ldots,s_d) = c'_a \prod_{i<j} (t_i-t_j) $$ for some constants $c'_1$ and $c'_2$.

However, when the characteristic of the field is not $2$, one cannot have a polynomial $E(s_1,\ldots,s_d)$ such that, after substitution, one obtains $$ E(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j) $$ since the left hand side cannot detect permutations in the $t_i$, whereas the right hand side will change sign when one makes an odd permutation of the $t_i$. Thus, $D$ is irreducible when the characteristic of the field is not $2$.

As Jarek Kuben pointed out in the comments below, when the characteristic of the field is $2$, the expression $$ F = \prod_{i<j} (t_i-t_j) = \prod_{i<j} (t_i+t_j) $$ is symmetric in the $t_i$, so $F$ can be written as a polynomial in the $s_i$. One then has $D = F^2$, so $D$ is a square. The above argument shows, however, that $F$ must be irreducible, since any factorization $D = D_1D_2$ has to have $D_1 = D_2 = F$ (up to constant multiples).

(N.B.: I have modified this answer to take into account the comments below about the case of characteristic $2$, when in fact, the discriminant is the square of an irreducible polynomial.)

The discriminant is defined by the property that, when $$ p(x) = x^d - s_1 x^{d-1} + \cdots + (-1)^ds_d = (x-t_1)(x-t_2)\cdots(x-t_d) $$ (i.e., when one substitutes the $i$th elementary symmetric function of the $t_k$ for $s_i$), the discriminant becomes $$ D(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j)^2. $$ Suppose there were a factorization $D(s_1,\ldots,s_d) = D_1(s_1,\ldots,s_d)D_1(s_1,\ldots,s_d)$ where each $D_i$ had positive degree. Then by unique factorization, when one substitutes as above, one must be able to write, for $a = 1,2$, $$ D_a(s_1,\ldots,s_d) = c_a \prod_{(i,j)\in S_a} (t_i-t_j) $$ where $c_1$ and $c_2$ are (nonzero) constants and $S_1$ and $S_2$ are disjoint nonempty subsets of the set of pairs $(i,j)$ in $\{1,\ldots,d\}$ whose union is the entire set of distinct pairs in this set. Thus, for example, $D_1(s_1,\ldots,s_d)$ will vanish when $t_i=t_j$ (for $i\not=j)$ only if $(i,j)$ or $(j,i)$ belongs to $S_1$. Since one can't detect which of the $t_i$ are equal using only $s_1,\ldots, s_d$, it follows that $S_1$ must contain either $(i,j)$ or $(j,i)$ for each distinct pair. The same argument applied to $D_2$ shows that $S_2$ also must contain either $(i,j)$ or $(j,i)$ for each distinct pair. Thus, one must have, for $a = 1,2$, $$ D_a(s_1,\ldots,s_d) = c'_a \prod_{i<j} (t_i-t_j) $$ for some constants $c'_1$ and $c'_2$.

However, when the characteristic of the field is not $2$, one cannot have a polynomial $E(s_1,\ldots,s_d)$ such that, after substitution, one obtains $$ E(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j) $$ since the left hand side cannot detect permutations in the $t_i$, whereas the right hand side will change sign when one makes an odd permutation of the $t_i$. Thus, $D$ is irreducible when the characteristic of the field is not $2$.

As Jarek Kuben pointed out in the comments below, when the characteristic of the field is $2$, the expression $$ F = \prod_{i<j} (t_i-t_j) = \prod_{i<j} (t_i+t_j) $$ is symmetric in the $t_i$, so $F$ can be written as a polynomial in the $s_i$. One then has $D = F^2$, so $D$ is a square. The above argument shows, however, that $F$ must be irreducible, since any factorization $D = D_1D_2$ has to have $D_1 = D_2 = F$.

(N.B.: I have modified this answer to take into account the comments below about the case of characteristic $2$, when in fact, the discriminant is the square of an irreducible polynomial.)

The discriminant is defined by the property that, when $$ p(x) = x^d - s_1 x^{d-1} + \cdots + (-1)^ds_d = (x-t_1)(x-t_2)\cdots(x-t_d) $$ (i.e., when one substitutes the $i$th elementary symmetric function of the $t_k$ for $s_i$), the discriminant becomes $$ D(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j)^2. $$ Suppose there were a factorization $D(s_1,\ldots,s_d) = D_1(s_1,\ldots,s_d)D_1(s_1,\ldots,s_d)$ where each $D_i$ had positive degree. Then by unique factorization, when one substitutes as above, one must be able to write, for $a = 1,2$, $$ D_a(s_1,\ldots,s_d) = c_a \prod_{(i,j)\in S_a} (t_i-t_j) $$ where $c_1$ and $c_2$ are (nonzero) constants and $S_1$ and $S_2$ are disjoint nonempty subsets of the set of pairs $(i,j)$ in $\{1,\ldots,d\}$ whose union is the entire set of distinct pairs in this set. Thus, for example, $D_1(s_1,\ldots,s_d)$ will vanish when $t_i=t_j$ (for $i\not=j)$ only if $(i,j)$ or $(j,i)$ belongs to $S_1$. Since one can't detect which of the $t_i$ are equal using only $s_1,\ldots, s_d$, it follows that $S_1$ must contain either $(i,j)$ or $(j,i)$ for each distinct pair. The same argument applied to $D_2$ shows that $S_2$ also must contain either $(i,j)$ or $(j,i)$ for each distinct pair. Thus, one must have, for $a = 1,2$, $$ D_a(s_1,\ldots,s_d) = c'_a \prod_{i<j} (t_i-t_j) $$ for some constants $c'_1$ and $c'_2$.

However, when the characteristic of the field is not $2$, one cannot have a polynomial $E(s_1,\ldots,s_d)$ such that, after substitution, one obtains $$ E(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j) $$ since the left hand side cannot detect permutations in the $t_i$, whereas the right hand side will change sign when one makes an odd permutation of the $t_i$. Thus, $D$ is irreducible when the characteristic of the field is not $2$.

As Jarek Kuben pointed out in the comments below, when the characteristic of the field is $2$, the expression $$ F = \prod_{i<j} (t_i-t_j) = \prod_{i<j} (t_i+t_j) $$ is symmetric in the $t_i$, so $F$ can be written as a polynomial in the $s_i$. One then has $D = F^2$, so $D$ is a square. The above argument shows, however, that $F$ must be irreducible, since any factorization $D = D_1D_2$ has to have $D_1 = D_2 = F$ (up to constant multiples).

How about the following 'conceptual proof'? The discriminant is defined by the property that, when $$ p(x) = x^d - s_1 x^{d-1} + \cdots + (-1)^ds_d = (x-t_1)(x-t_2)\cdots(x-t_d) $$ (i.e., when one substitutes the $i$th elementary symmetric function of the $t_k$ for $s_i$), the discriminant becomes $$ D(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j)^2. $$ Suppose there were a factorization $D(s_1,\ldots,s_d) = D_1(s_1,\ldots,s_d)D_1(s_1,\ldots,s_d)$ where each $D_i$ had positive degree. Then by unique factorization, when one substitutes as above, one must be able to write $$ D_i(s_1,\ldots,s_d) = c_i \prod_{(i,j)\in S_i} (t_i-t_j) $$ where $c_1$ and $c_2$ are (nonzero) constants and $S_1$ and $S_2$ are disjoint nonempty subsets of the set of pairs $(i,j)$ in $\{1,\ldots,d\}$ whose union is the entire set of distinct pairs in this set. Thus, for example, $D_1(s_1,\ldots,s_d)$ will vanish when $t_i=t_j$ (for $i\not=j)$ only if $(i,j)$ or $(j,i)$ belongs to $S_1$. Since one can't detect which of the $t_i$ are equal using only $s_1,\ldots, s_d$, it follows that $S_1$ must contain either $(i,j)$ or $(j,i)$ for each distinct pair. The same argument applied to $D_2$ shows that $S_2$ also must contain either $(i,j)$ or $(j,i)$ for each distinct pair. Thus, one must have $$ D_i(s_1,\ldots,s_d) = c_i \prod_{i<j} (t_i-t_j) $$ for some constants $c_1$ and $c_2$. However, one cannot have a polynomial $E(s_1,\ldots,s_d)$ such that, after substitution, one obtains $$ E(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j) $$ since the left hand side cannot detect permutations in the $t_i$, whereas the right hand side will change sign when one makes an odd permutation of the $t_i$.

Thus, $D$ is irreducible.

(N.B.: I have modified this answer to take into account the comments below about the case of characteristic $2$, when in fact, the discriminant is the square of an irreducible polynomial.)

The discriminant is defined by the property that, when $$ p(x) = x^d - s_1 x^{d-1} + \cdots + (-1)^ds_d = (x-t_1)(x-t_2)\cdots(x-t_d) $$ (i.e., when one substitutes the $i$th elementary symmetric function of the $t_k$ for $s_i$), the discriminant becomes $$ D(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j)^2. $$ Suppose there were a factorization $D(s_1,\ldots,s_d) = D_1(s_1,\ldots,s_d)D_1(s_1,\ldots,s_d)$ where each $D_i$ had positive degree. Then by unique factorization, when one substitutes as above, one must be able to write, for $a = 1,2$, $$ D_a(s_1,\ldots,s_d) = c_a \prod_{(i,j)\in S_a} (t_i-t_j) $$ where $c_1$ and $c_2$ are (nonzero) constants and $S_1$ and $S_2$ are disjoint nonempty subsets of the set of pairs $(i,j)$ in $\{1,\ldots,d\}$ whose union is the entire set of distinct pairs in this set. Thus, for example, $D_1(s_1,\ldots,s_d)$ will vanish when $t_i=t_j$ (for $i\not=j)$ only if $(i,j)$ or $(j,i)$ belongs to $S_1$. Since one can't detect which of the $t_i$ are equal using only $s_1,\ldots, s_d$, it follows that $S_1$ must contain either $(i,j)$ or $(j,i)$ for each distinct pair. The same argument applied to $D_2$ shows that $S_2$ also must contain either $(i,j)$ or $(j,i)$ for each distinct pair. Thus, one must have, for $a = 1,2$, $$ D_a(s_1,\ldots,s_d) = c'_a \prod_{i<j} (t_i-t_j) $$ for some constants $c'_1$ and $c'_2$.

However, when the characteristic of the field is not $2$, one cannot have a polynomial $E(s_1,\ldots,s_d)$ such that, after substitution, one obtains $$ E(s_1,\ldots,s_d) = \prod_{i<j} (t_i-t_j) $$ since the left hand side cannot detect permutations in the $t_i$, whereas the right hand side will change sign when one makes an odd permutation of the $t_i$. Thus, $D$ is irreducible when the characteristic of the field is not $2$.

As Jarek Kuben pointed out in the comments below, when the characteristic of the field is $2$, the expression $$ F = \prod_{i<j} (t_i-t_j) = \prod_{i<j} (t_i+t_j) $$ is symmetric in the $t_i$, so $F$ can be written as a polynomial in the $s_i$. One then has $D = F^2$, so $D$ is a square. The above argument shows, however, that $F$ must be irreducible, since any factorization $D = D_1D_2$ has to have $D_1 = D_2 = F$.