# Eisenstein Criterion

Why is it that the Eisenstein's Criterion would work when substituting $$x$$ with $$x + 1$$? Why is it OK to do this for polynomials in $$\mathbb{Q}[x]$$?

Thank you

Well, here is a proof.

Now if $f(x+1) = q(x)r(x)$, then $f(x) = f((x-1)+1) = q(x-1)r(x-1)$.

And if $f(x) = q(x)r(x)$, then $f(x+1) = q(x+1)r(x+1)$.

Notice also that $g(x)$, $g(x-1)$ and $g(x+1)$ always have the same degree. Thus we may conclude by above that $f(x)$ is irreducible if and only if $f(x+1)$ is irreducible.

With the same proof you can generalize this to prove that for all $c \in \mathbb{Z}$, $f(x)$ is irreducible if and only if $f(x+c)$ is irreducible.

Take $\varphi:\mathbb Q[x]\to\mathbb Q[x]$ which is identity on $\mathbb Q$ and $\varphi(x)=x+1$. Check that $\varphi$ is an automorphism of $\mathbb Q[x]$. Since every algebraic property is preserved by isomorphisms we can deduce that a polynomial $f\in\mathbb Q[x]$ is irreducible if and only if $\varphi(f)$ is irreducible. Moreover, note that $\varphi(f(x))=f(x+1)$.

I think the trick you are referring to is the fact that a polynomial $f(x)\in \mathbb{Q}[x]$ is irreducible if and only if the polynomial $f(x+1)$ is irreducible in $\mathbb{Q}[x]$. This can make applying the Eisenstein criterion possible when it is not initially, which is why it is useful to do this.

When working with polynomial functions you can substitute $$x$$ with $$x + 1$$ as it denotes a simple composition of two functions. To do this is in $$\mathbb{Q}[x]$$ you would need to have a corresponding definition of composition $$f \circ g$$ on $$\mathbb{Q}[x]$$.

(Below we use $$X$$ for the indeterminate to avoid confusion with $$x$$ in the ring $$R$$).

However, generally for a commutative unitary ring $$R$$ we can't just induce a composition operation from the ring of polynomial functions using the natural ring epimorphism $$\kappa : R[X] \rightarrow P(R)$$ (where $$P(R)$$ is the commutative unitary ring of polynomial functions on $$R$$) since in general $$\kappa$$ is not an isomorphism (eg taking $$R$$ as non-zero and finite, $$R[X]$$ is infinite but $$P(R)$$ is finite). But we can do this when $$R$$ is an infinite integral domain (eg in the present case of $$\mathbb{Q}[X]$$) for then $$\kappa$$ is an isomorphism because in an infinite integral domain polynomial function coefficients are unique apart from leading zeros.

$$\kappa$$ is the epimorphism which maps $$a_n X^n + \ldots + a_1 X + a_0$$ in $$R[X]$$ to the function $$x \mapsto a_n x^n + \ldots + a_1 x + a_0$$ in $$P(R)$$, so for the purposes of $$\mathbb{Q}[X]$$, where $$\kappa$$ becomes an isomorphism, we would just use $$\kappa$$ to mirror composition of polynomial functions.

So for example in $$\mathbb{Q}[X]$$ we would have $$(X^2 + 3X + 1) \circ (X^2 - 5) =$$ $$(X^2 - 5)^2 + 3(X^2 - 5) + 1 = X^4 - 7X^2 + 11$$, even though these are sequences in $$\mathbb{Q}$$ and not actual functions - we just mirrored the composition operation for polynomial functions in $$P(\mathbb{Q})$$. Generally we could use notations such as $$f(X^2 + 1)$$, $$f(g(X))$$ etc to denote compositions in $$R[X]$$, just as with functions.

In the general case where $$\kappa$$ is not an isomorphism we could still define composition on $$R[X]$$ using a notation $$\mathrm{EXPR}[v]$$ for a general expression in $$R[X]$$ involving only additions, subtractions, multiplications, non-negative integral powers, and bracketing, where $$v \in R[X]$$ is a single variable or 'placeholder', and every other term in the expression is a constant in $$R[X]$$ (and so identified with an element of $$R$$). Then $$\mathrm{EXPR}[v]$$ defines a parallel expression within $$R$$, with $$v$$ now ranging over $$R$$. Every $$f \in R[X]$$ can be written in a standard way as $$\mathrm{EXPR}[X] = a_n X^n + \ldots + a_1 X + a_0$$ for constants $$a_i$$, with any two such expressions differing only by leading zeros. Many other EXPR's giving $$f = \mathrm{EXPR}[X]$$ would be possible also.

We could then define $$f \circ g$$ in $$R[X]$$ as $$\mathrm{EXPR}[g]$$, where $$\mathrm{EXPR}[X]$$ is any standard expression for $$f$$, and then readily show $$f \circ g$$ equals $$\mathrm{EXPR}[g]$$ for any other expression of $$f$$.

Then we can show the following properties :

1. $$(f \circ g) \circ h = f \circ (g \circ h)$$, ie associative
2. $$(f + g) \circ h = f \circ h + g \circ h$$
3. $$(fg) \circ h = (f \circ h)(g \circ h)$$
4. $$\kappa(f \circ g) = \kappa(f) \circ \kappa(g)$$
5. $$\kappa(f)(x) = \mathrm{EXPR}[x]\; \forall x \in R$$, where $$\mathrm{EXPR}[X]$$ is any expression for $$f \in R[X]$$
6. $$\partial (f \circ g) \leq \partial f \cdot \partial g$$, with equality if $$R$$ is an integral domain

(properties 1 - 3 are the properties required in a composition ring).

Taking the definition of 'reducible' to be 'product of two lesser degree polynomials', we then obtain the following :

1. For any commutative unitary ring $$R$$, $$f \in R[X]$$, and $$a \in R$$ : $$f$$ irreducible $$\Leftrightarrow f(X + a)$$ irreducible
2. For any integral domain $$R$$, $$f \in R[X]$$, and $$a \in R \setminus \{0\}$$ : $$f(aX + b)$$ irreducible $$\Rightarrow f$$ irreducible.

In particular 1 and 2 would be applicable to $$\mathbb{Q}$$ or to any field, including finite fields where $$\kappa$$ is not an isomorphism.

Method 2 allows a simple 'change of variable' to $$Y$$ if we rename the indeterminate to $$Y$$ and informally substitute $$X = aY + b$$. Then we can investigate whether $$f(aY + b)$$ is irreducible in $$R[Y]$$, which if so would prove the original $$f$$ was irreducible in $$R[X]$$. Note the relation $$X = aY + b$$ does not make sense formally, it just allows the composition polynomial $$f(aY + b)$$ to be written out from the definition of $$f(X)$$.