Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question already has an answer here:

There's two notions of equivalent polynomials floating around, one saying that $f = g$ iff they're equivalent as maps, and the other saying $f = g$ iff they're equal on each coefficient when written in standard form.

I'm interested in polynomials over a finite field, irreducible polynomials and factoring so what type of equivalence should I use? For instance if we take map equivalence, then there are only a finite number of polynomials. And that makes a huge difference!

Please explain when it's okay to use what.

share|cite|improve this question

marked as duplicate by 91500, amWhy abstract-algebra May 14 '14 at 16:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The second definition is always the correct definition. The first definition is equivalent to the second definition if you allow yourself the freedom to pass to, say, finite extensions of your base field when plugging in numbers to a polynomial. – Qiaochu Yuan May 13 '14 at 3:42
up vote 6 down vote accepted

In my experience, in algebra, equality of polynomials is always equality of their coefficients.

When authors care only about functions induced by polynomials, they usually explicitly state that, or use term like "polynomial function".

One possible place when confusion could have arised (but it doesn't) is when you talk about regular functions defined on algebraic varieties. In classical algebraic geometry, these were just functions defined by polynomials, so in theory it could have happened that two different polynomials result in the same regular function on the affine space. The solution is that classical algebraic geometry always works with algebraically closed field as a base field, which must be infinite.

To be able to talk about varieties defined over non algebraically closed fields, one usually uses language of schemes, but regular functions defined on schemes aren't really "functions" anymore, so there's no problem either. Also, morphisms between schemes are also defined so that it's quite obvious that different polynomials (or actually different ring homomorphisms) give different morphisms, even if they're equal on point-set level.

share|cite|improve this answer

In abstract algebra polynomials over a ring $R$ are defined as formal sums $$ \sum_{k=0}^N a_k X^k $$ where $X$ is a formal variable and all $a_k\in R$. To make this precise, we can also model polynomials as sequences $(a_0, a_1, \dots)$ where all but finitely many $a_i$ are zero. Addition and multiplications is then given by $$ (a_0, a_1, \dots) + (b_0, b_1, \dots) := (a_0+a_1, b_0+b_1, \dots) $$ and $$(a_0, a_1, \dots) \cdot (b_0, b_1, \dots) := (c_0, c_1, \dots),\quad\text{where}\ c_k = \sum_{i=0}^k a_i b_{k-i}.$$ This corresponds to the usual coefficient-wise addition of polynomials and the Cauchy product formula $$ \left(\sum_{k=0}^\infty a_k X^k \right) \cdot \left(\sum_{k=0}^\infty b_k X^k \right) = \sum_{k=0}^\infty c_k X^k,\quad\mathrm{where}\ c_k=\sum_{i=0}^k a_i b_{k-i}. $$ Letting $X=(0,1,0,\dots)$ we obtain the usual representation of polynomials. So two polynomials are equal if the underlying tuples are equal which is the case if and only if all coefficients are equal.

Note that we can easily drop the "all but finitely many $a_i=0$" requirement and still get a ring structure with the above operations. This is the ring $R[[X]]$ of formal power series, while the ring of polynomials is usually denoted $R[X]$.

share|cite|improve this answer
What do you mean by let $X = (0,1, 0, \dots)$. Confusing! – Enjoys Math May 12 '14 at 16:13
Well I explained how every polynomial is modeled as a sequence of its coefficients. So since $X = 0\cdot X^0 + 1 \cdot X^1 + 0 \cdot X^2 + 0 \cdot X^3 + \cdots$ we have the tuple $(0,1,0,\dots)$ representing $X$. – Christoph May 12 '14 at 16:15

Algebraists employ formal (vs. functional) polynomials because this yields the greatest generality. Once one proves an identity in a polynomial ring $\rm\ R[x,y,z]\ $ then it will remain true for all specializations of $\rm\:x,y,z\:$ in any ring where the coefficients can be interpreted commutatively, i.e. any ring containing a central image of $\rm\:R,\:$ i.e. any $\rm\:R$-algebra. Thus we can prove once-and-for-all important polynomial identities such as the Binomial Theorem, Cramer's rule, Vieta's formula, etc. and later specialize the indeterminates as need be for applications in specific rings. This allows us to interpret such polynomial identities in the most universal ring-theoretic manner - in the greatest ring-theoretic generality.

For example, when we are solving recurrences over a finite field $\rm\,\mathbb F = \mathbb F_p\,$ it is helpful to employ "operator algebra", working with characteristic polynomials over $\rm\,\mathbb F,\,$ i.e. elements of the ring $\rm\,\mathbb F_p[S]\,$ for $\rm\,S\,$ is the shift operator $\rm\ S\ f(n)\, =\, f(n+1).\,$ They are not polynomial functions on $\rm\,\mathbb F_p,\,$ e.g. generally $\rm\ S^p \ne S\ $ since generally $\rm\ f(n+p) \ne f(n+1).\,$ But any polynomial identity of $\rm\,\mathbb F[x]\,$ specializes to this operator algebra by way of the evaluation map $\rm\,x\mapsto \,S.\,$ E.g. we might specialize universal polynomial factorization identities in order to factor the characteristic polynomial, e.g. difference of squares $\rm\ x^2\! - y^2 = (x\!-\!y)\ (x\!+\!y)\,$ $\,\Rightarrow\,$ $\rm\,S^2\!-\! c^2 = (S\!-\!c)\ (S\!+\! c)\ $ via $\rm\,x,y\mapsto S,c\,,\,$ or perhaps we may specialize a cyclotomic polynomial factorization, etc. This would not be possible if we instead employed the the much less general ring of polynomial functions over $\rm\,\mathbb F\,,\,$ since its specializations of $\rm\,x\,$ must satisfy $\rm\, x^p = x.\,$

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.