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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

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4 Answers 4

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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.

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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 $\varphi(f(x))=f(x+1)$.

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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.

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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 $f(X)$.

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