If you have an injective linear map between two free modules of equal dimension, is the determinant of the matrix representing the map necessarily nonzero? If not is there an obvious counterexample? (Everything is over a multivariate polynomial ring over a field.)


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    $\begingroup$ For commutative domains such as multivariate polynomial rings over fields: free modules are flat, so you can tensor with the field of fractions of the ring while retaining an injective map. Then you have an injective linear map over a (large) field, so of course its determinant is nonzero. $\endgroup$ – Jack Schmidt Jun 22 '12 at 2:50
  • $\begingroup$ @Arturo: your first statement is false. Consider the inclusion $2 \mathbb{Z} \to \mathbb{Z}$. $\endgroup$ – Qiaochu Yuan Jun 22 '12 at 3:01
  • $\begingroup$ @Qiaochu: Hmmm... That's what I get for not checking Lam's book first... Thanks. $\endgroup$ – Arturo Magidin Jun 22 '12 at 3:02
  • $\begingroup$ @Jack: in infinite dimensions what would you mean by the determinant? $\endgroup$ – Qiaochu Yuan Jun 22 '12 at 3:40

Over an integral domain $D$ this is straightforward as Jack Schmidt says in the comments. Suppose $T : D^n \to D^n$ is a morphism such that $\det(T) = 0$. Then $T \otimes \text{Frac}(D) : \text{Frac}(D)^n \to \text{Frac}(D)^n$ has the same property, where $\text{Frac}(D)$ is the fraction field. Since we are now over a field, $T \otimes \text{Frac}(D)$ is not injective, and so it annihilates some nonzero vector in $\text{Frac}(D)^n$. Scaling this vector suitably puts it in $D^n$, and so $T$ is not injective.

Over a general commutative $R$ I have a proof for $n = 2$ (consider the adjugate) but it doesn't seem to generalize. I'll get back to you.

  • $\begingroup$ Thanks! This is sufficient for the case I'm considering. $\endgroup$ – MGN Jun 22 '12 at 4:22
  • $\begingroup$ I suppose this is still true without of the assumption that D is an integral domain. $\endgroup$ – YZhou Jun 22 '12 at 5:54
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    $\begingroup$ @Andrew: who knows? I can't find a proof or a counterexample. I'm actually about to post this as a separate question. $\endgroup$ – Qiaochu Yuan Jun 22 '12 at 5:57
  • $\begingroup$ Posted: math.stackexchange.com/questions/161523/… $\endgroup$ – Qiaochu Yuan Jun 22 '12 at 6:09
  • $\begingroup$ @QiaochuYuan Yuan:Would you please spend a minute to see whether my argument makes sense? I add an answer here. $\endgroup$ – YZhou Jun 22 '12 at 6:15

Let $T: A^n \to A^n$ be a morphism and let's use the same notation for its matrix, $T=(t_{ij})$. Let us assume that $det(T)=0$ and I will prove that $T$ fails to be injective.

There must be an integer $k$ such that there exists a minor of rank $k$ with nonzero determinant and any minor of larger rank has determinant zero. Without loss of generality, we can assume that the minor sits in the first $k$ rows and first $k$ columns. If $k=0$, then $T=0$, which is trivial. Since our task is to find a non-zero vector $x=(x_1,\cdots,x_n)$ killed by $T$ and inparticular our vector can be chosen to have the form $(x_1,\cdots,x_{k+1},0,\cdots,0)$, it suffices to assume that n=k+1.

Having adopted these assumptions, since $TT^*=det(T)I=0$, we just have to find a vector $x\neq 0$ such that $T^*x=0$, where $T^*$ is the adjugate matrix of $T$. Now $T^*$ is obviously nonzero since the first $k\times k$ minor of $T$ has nonzero determinant. Hence, such an $x$ must exist(assuming that the ring in question has a unit) and we are home!

  • $\begingroup$ I do not see why you can assume that $n = k+1$. Can you please move discussion of the question over arbitrary rings to math.stackexchange.com/questions/161523/… ? $\endgroup$ – Qiaochu Yuan Jun 22 '12 at 6:15
  • $\begingroup$ @Qiaochu Yuan: Suppose we have proved our statement in the case $n=k+1$. In a general case, denote the fisrt $(k+1)\times (k+1)$ entry by $A$ and assume that $AX=0$ for some nonzero $X=(x_1,\cdots,x_{k+1})$, then $TY=0$ where $0\neq Y:=(x_1,\cdots,x_{k+1},0,\cdots,0)$. Actually I borrowed the argument here:mathoverflow.net/questions/136/atiyah-macdonald-exercise-2-11 $\endgroup$ – YZhou Jun 22 '12 at 6:33
  • $\begingroup$ As far as I can tell, this does not follow (you have neglected the entries below the $(k+1) \times (k+1)$ subsquare in $T$). $\endgroup$ – Qiaochu Yuan Jun 22 '12 at 6:51
  • $\begingroup$ @QiaochuYuan:I am sorry that I made a terribly stupid mistake. Now I don't know the answer. By far we have known that the statement is true if it has a nonzero $(n-1)\times (n-1)$ minor. But I have no idea when all $(n-1)\times (n-1)$ minors are zero. This might be false. I guess. In the case of integral domains, one accutally first solves the equation in the field and then clear denominator. If the ring is not an integral domain, I guess this might kill the vector. $\endgroup$ – YZhou Jun 22 '12 at 15:38
  • $\begingroup$ @QiaochuYuan:I cannot find a counterexample either. Can we first write down a matrix with entries in $k[x_1,\cdots,x_n]$ and then kill some polynomial (e.g. the determinant of the matrix) to pass to the coordinate ring of a variety? This at least gives us some geometry intuition and might be a possible way to find a counterexample. Hope you can find the answer soon! $\endgroup$ – YZhou Jun 22 '12 at 15:42

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