Look at it like this. We have a ring $R=K[x_1,\ldots,x_n]$ and a ring $S=K[y]$. We want to treat the $x_i$s as polynomials in $y$, so we're looking for a mapping $f:R\to S$ that sends each $x_i$ to a polynomial in $y$ and satisfies a system of polynomial equations $P$ in the $x_i$. In other words, we want $f$ to be a ring homomorphism that sends $I(P)$ to 0. $f$, the map from $R$ to $S$, is the solution we seek.

Now we want to impose an additional condition: a subset $x_1,\ldots,x_i$ must map to irreducible elements $y_1,\ldots,y_i$ in $S$. Since $y_1,\ldots,y_i$ are irreducible, they are prime (in $S$), so we quotient with respect to their ideal $I$ and get a quotient ring $S/I$ that is an *integral domain*. We can also quotient $R$ by the ideal generated by $x_1,\ldots,x_i$ (call it $J$), and get $R/J$. $f$ can be similarly restricted, and now we have a homomorphism $\hat{f}: R/J \to S/I$. We can construct a Gröbner basis for $R/J$ by appending the $x_i$ that must be irreducible to the original system $P$, and reducing $P \cup \{x_1,\ldots,x_i\}$ to a Gröbner basis. This new Gröbner basis gives relationships satisfied by the *equivalence classes* in $S/I$. "Equal to zero" in this quotient system means "equal to zero or a multiple of an irreducible element" in the original system. However, if the quotient system is inconsistent, then the original system is also inconsistent, at least subject to the restriction that $x_1,\ldots,x_i$ must map to irreducibles.

Can we find additional relationships? Surprisingly, yes! We run this calculation with each irreducible individually. Pick one $x_1,\ldots,x_i$, call it $x_j$, compute a quotient Gröbner basis for $P \cup \{x_j\}$, take each polynomial in the quotient system's Gröbner basis and test to see if it's in the original system. If so, then it's really equal to zero. Otherwise, it's a multiple of $x_j$ and we can add that polynomial to the original system, equating it a term of the form $m x_j$, with $m$ a new indeterminate.

The augmented system will have extraneous zeros, at least if we require the irreducible polynomials to be non-zero. We can handle this by computing a primary decomposition and throwing away any primary components that include an irreducible element among their zeros. This is the ideal-theoretic equivalent of factoring a polynomial that must be equal to zero and throwing away factors that we know are non-zero. We can keep repeating these two processes (quotient ring basis and primary decomposition) until our ideal stabilizes.

**Example**

Consider the equation $af^2+bf+c=0$, with $f$ restricted to be irreducible.

Step 1: Form the system $\{af^2+bf+c, f\}$ and reduce to the Gröbner basis $\{f,c\}$. Of course $f$ is here; our interest is $c$. Since it isn't in the original ideal, it must be a multiple of $f$, so we add $c-mf$ our ideal to obtain

$$(af^2+bf+c, c-mf)$$

Step 2: A primary decomposition of this ideal gives two primary ideals, one of which is $(f,c)$. Since $f$ can't be zero, we throw it away and continue with the other primary ideal:

$$(af+b+m, ac+bm+m^2, c-mf)$$

Step 3: Back to the quotient calculation. Now our system is

$$\{af+b+m, ac+bm+m^2, c-mf, f\}$$

and we compute the Gröbner basis $\{f, c, b+m\}$. This implies that $b+m$ must also be a multiple of $f$, so we add $b+m-nf$ to our ideal, obtaining

$$(af+b+m, ac+bm+m^2, c-mf, b+m-nf)$$

Step 4: Another primary decomposition gives another extraneous ideal $(f,c,b+m)$. Throwing this away, we have

$$(a+n, fn-b-m, fm-c, bm+m^2-cn)$$

Step 5: A final quotient calculation, with the system

$$\{a+n, fn-b-m, fm-c, bm+m^2-cn, f\}$$

gives $\{f,c,b+m,a+n\}$, of which the only new element, $a+n$, reduces to zero.

So we've stabilized on

$$(a+n, fn-b-m, fm-c, bm+m^2-cn)$$

This ideal encodes all of the information I was able to extract in Polynomial constraints triggered by irreducibility